Prove the following identities \displaystyle (1-25)

\displaystyle \textbf{Question 1: }\text{Prove that }\sqrt{\frac{1-\cos2x}{1+\cos2x}}=\tan x,\text{ where }\tan x\ge0.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sqrt{\frac{1-\cos2x}{1+\cos2x}}
\displaystyle =\sqrt{\frac{2\sin^2x}{2\cos^2x}}
\displaystyle =\sqrt{\frac{\sin^2x}{\cos^2x}}
\displaystyle =\left|\frac{\sin x}{\cos x}\right|
\displaystyle =|\tan x|
\displaystyle =\tan x\qquad[\because \tan x\ge0]
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Prove that }\frac{\sin2x}{1-\cos2x}=\cot x.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin2x}{1-\cos2x}
\displaystyle =\frac{2\sin x\cos x}{2\sin^2x}
\displaystyle =\frac{\cos x}{\sin x}
\displaystyle =\cot x
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Prove that }\frac{\sin2x}{1+\cos2x}=\tan x.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin2x}{1+\cos2x}
\displaystyle =\frac{2\sin x\cos x}{2\cos^2x}
\displaystyle =\frac{\sin x}{\cos x}
\displaystyle =\tan x
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Prove that }\sqrt{2+\sqrt{2+2\cos4x}}=2\cos x,\text{ where }0<x<\frac{\pi}{4}.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sqrt{2+\sqrt{2+2\cos4x}}
\displaystyle =\sqrt{2+\sqrt{2(1+\cos4x)}}
\displaystyle =\sqrt{2+\sqrt{2\times2\cos^22x}}
\displaystyle =\sqrt{2+2\cos2x}\qquad\left[\because 0<2x<\frac{\pi}{2}\right]
\displaystyle =\sqrt{2(1+\cos2x)}
\displaystyle =\sqrt{2\times2\cos^2x}
\displaystyle =2\cos x\qquad\left[\because 0<x<\frac{\pi}{4}\right]
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Prove that }\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x}=\tan x.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x}
\displaystyle =\frac{2\sin^2x+2\sin x\cos x}{2\cos^2x+2\sin x\cos x}
\displaystyle =\frac{2\sin x(\sin x+\cos x)}{2\cos x(\sin x+\cos x)}
\displaystyle =\frac{\sin x}{\cos x}
\displaystyle =\tan x
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Prove that }\frac{\sin x+\sin2x}{1+\cos x+\cos2x}=\tan x.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin x+\sin2x}{1+\cos x+\cos2x}
\displaystyle =\frac{\sin x+2\sin x\cos x}{\cos x+(1+\cos2x)}
\displaystyle =\frac{\sin x(1+2\cos x)}{\cos x+2\cos^2x}
\displaystyle =\frac{\sin x(1+2\cos x)}{\cos x(1+2\cos x)}
\displaystyle =\frac{\sin x}{\cos x}
\displaystyle =\tan x
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Prove that }\frac{\cos2x}{1+\sin2x}=\tan\left(\frac{\pi}{4}-x\right).
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\cos2x}{1+\sin2x}
\displaystyle =\frac{\cos^2x-\sin^2x}{\sin^2x+\cos^2x+2\sin x\cos x}
\displaystyle =\frac{(\cos x-\sin x)(\cos x+\sin x)}{(\cos x+\sin x)^2}
\displaystyle =\frac{\cos x-\sin x}{\cos x+\sin x}
\displaystyle =\frac{\frac{1}{\sqrt2}(\cos x-\sin x)}{\frac{1}{\sqrt2}(\cos x+\sin x)}
\displaystyle =\frac{\sin\left(\frac{\pi}{4}-x\right)}{\cos\left(\frac{\pi}{4}-x\right)}
\displaystyle =\tan\left(\frac{\pi}{4}-x\right)
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Prove that }\frac{\cos x}{1-\sin x}=\tan\left(\frac{\pi}{4}+\frac{x}{2}\right).
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\cos x}{1-\sin x}
\displaystyle =\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}}
\displaystyle =\frac{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^2}
\displaystyle =\frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}}
\displaystyle =\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}
\displaystyle =\frac{\tan\frac{\pi}{4}+\tan\frac{x}{2}}{1-\tan\frac{\pi}{4}\tan\frac{x}{2}}
\displaystyle =\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Prove that }\cos^2\frac{\pi}{8}+\cos^2\frac{3\pi}{8}+\cos^2\frac{5\pi}{8}+\cos^2\frac{7\pi}{8}=2.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\cos^2\frac{\pi}{8}+\cos^2\frac{3\pi}{8}+\cos^2\frac{5\pi}{8}+\cos^2\frac{7\pi}{8}
\displaystyle =\cos^2\frac{\pi}{8}+\cos^2\frac{3\pi}{8}+\cos^2\left(\pi-\frac{3\pi}{8}\right)+\cos^2\left(\pi-\frac{\pi}{8}\right)
\displaystyle =\cos^2\frac{\pi}{8}+\cos^2\frac{3\pi}{8}+\cos^2\frac{3\pi}{8}+\cos^2\frac{\pi}{8}
\displaystyle =2\left(\cos^2\frac{\pi}{8}+\cos^2\frac{3\pi}{8}\right)
\displaystyle =2\left(\cos^2\frac{\pi}{8}+\cos^2\left(\frac{\pi}{2}-\frac{\pi}{8}\right)\right)
\displaystyle =2\left(\cos^2\frac{\pi}{8}+\sin^2\frac{\pi}{8}\right)
\displaystyle =2
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Prove that }\sin^2\frac{\pi}{8}+\sin^2\frac{3\pi}{8}+\sin^2\frac{5\pi}{8}+\sin^2\frac{7\pi}{8}=2.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sin^2\frac{\pi}{8}+\sin^2\frac{3\pi}{8}+\sin^2\frac{5\pi}{8}+\sin^2\frac{7\pi}{8}
\displaystyle =\sin^2\frac{\pi}{8}+\sin^2\frac{3\pi}{8}+\sin^2\left(\pi-\frac{3\pi}{8}\right)+\sin^2\left(\pi-\frac{\pi}{8}\right)
\displaystyle =\sin^2\frac{\pi}{8}+\sin^2\frac{3\pi}{8}+\sin^2\frac{3\pi}{8}+\sin^2\frac{\pi}{8}
\displaystyle =2\left(\sin^2\frac{\pi}{8}+\sin^2\frac{3\pi}{8}\right)
\displaystyle =2\left(\sin^2\frac{\pi}{8}+\sin^2\left(\frac{\pi}{2}-\frac{\pi}{8}\right)\right)
\displaystyle =2\left(\sin^2\frac{\pi}{8}+\cos^2\frac{\pi}{8}\right)
\displaystyle =2
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Prove that }(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2=4\cos^2\left(\frac{\alpha-\beta}{2}\right).
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2
\displaystyle =\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta+\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta
\displaystyle =2+2\cos\alpha\cos\beta+2\sin\alpha\sin\beta
\displaystyle =2+2\cos(\alpha-\beta)
\displaystyle =2\left[1+\cos(\alpha-\beta)\right]
\displaystyle =2\times2\cos^2\left(\frac{\alpha-\beta}{2}\right)
\displaystyle =4\cos^2\left(\frac{\alpha-\beta}{2}\right)
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Prove that }\sin^2\left(\frac{\pi}{8}+\frac{x}{2}\right)-\sin^2\left(\frac{\pi}{8}-\frac{x}{2}\right)=\frac{1}{\sqrt{2}}\sin x.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sin^2\left(\frac{\pi}{8}+\frac{x}{2}\right)-\sin^2\left(\frac{\pi}{8}-\frac{x}{2}\right)
\displaystyle =\frac{1}{2}\left[1-\cos2\left(\frac{\pi}{8}+\frac{x}{2}\right)\right]-\frac{1}{2}\left[1-\cos2\left(\frac{\pi}{8}-\frac{x}{2}\right)\right]
\displaystyle =\frac{1}{2}\left[\cos\left(\frac{\pi}{4}-x\right)-\cos\left(\frac{\pi}{4}+x\right)\right]
\displaystyle =\frac{1}{2}\left[-2\sin\left(\frac{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}+x\right)}{2}\right)\sin\left(\frac{\left(\frac{\pi}{4}-x\right)-\left(\frac{\pi}{4}+x\right)}{2}\right)\right]
\displaystyle =\frac{1}{2}\left[-2\sin\frac{\pi}{4}\sin(-x)\right]
\displaystyle =\sin\frac{\pi}{4}\sin x
\displaystyle =\frac{1}{\sqrt{2}}\sin x
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Prove that }1+\cos^22x=2\left(\cos^4x+\sin^4x\right).
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=1+\cos^22x
\displaystyle =1+\left(\cos^2x-\sin^2x\right)^2
\displaystyle =1+\cos^4x+\sin^4x-2\cos^2x\sin^2x
\displaystyle =\left(\cos^2x+\sin^2x\right)^2+\cos^4x+\sin^4x-2\cos^2x\sin^2x
\displaystyle =\cos^4x+\sin^4x+2\cos^2x\sin^2x+\cos^4x+\sin^4x-2\cos^2x\sin^2x
\displaystyle =2\left(\cos^4x+\sin^4x\right)
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Prove that }\cos^32x+3\cos2x=4\left(\cos^6x-\sin^6x\right).
\displaystyle \text{Answer:}
\displaystyle \text{RHS}=4\left(\cos^6x-\sin^6x\right)
\displaystyle =4\left[\left(\cos^2x\right)^3-\left(\sin^2x\right)^3\right]
\displaystyle =4\left(\cos^2x-\sin^2x\right)\left(\cos^4x+\sin^4x+\sin^2x\cos^2x\right)
\displaystyle =4\cos2x\left(\cos^4x+\sin^4x+\sin^2x\cos^2x\right)
\displaystyle =4\cos2x\left[\left(\cos^2x-\sin^2x\right)^2+3\sin^2x\cos^2x\right]
\displaystyle =4\cos2x\left(\cos^22x+3\sin^2x\cos^2x\right)
\displaystyle =4\cos2x\left[\cos^22x+3\left(\frac{1-\cos2x}{2}\right)\left(\frac{1+\cos2x}{2}\right)\right]
\displaystyle =4\cos2x\left[\cos^22x+\frac{3}{4}\left(1-\cos^22x\right)\right]
\displaystyle =\cos2x\left[4\cos^22x+3\left(1-\cos^22x\right)\right]
\displaystyle =\cos2x\left(\cos^22x+3\right)
\displaystyle =\cos^32x+3\cos2x
\displaystyle =\text{LHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Prove that }(\sin3x+\sin x)\sin x+(\cos3x-\cos x)\cos x=0.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(\sin3x+\sin x)\sin x+(\cos3x-\cos x)\cos x
\displaystyle =2\sin2x\cos x\sin x-2\sin2x\sin x\cos x
\displaystyle =2\sin2x\sin x\cos x-2\sin2x\sin x\cos x
\displaystyle =0
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Prove that }\cos^2\left(\frac{\pi}{4}-x\right)-\sin^2\left(\frac{\pi}{4}-x\right)=\sin2x.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\cos^2\left(\frac{\pi}{4}-x\right)-\sin^2\left(\frac{\pi}{4}-x\right)
\displaystyle =\cos2\left(\frac{\pi}{4}-x\right)
\displaystyle =\cos\left(\frac{\pi}{2}-2x\right)
\displaystyle =\sin2x
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Prove that }\cos4x=1-8\cos^2x+8\cos^4x.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\cos4x
\displaystyle =\cos(2\cdot2x)
\displaystyle =2\cos^22x-1
\displaystyle =2\left(2\cos^2x-1\right)^2-1
\displaystyle =2\left(4\cos^4x-4\cos^2x+1\right)-1
\displaystyle =8\cos^4x-8\cos^2x+1
\displaystyle =1-8\cos^2x+8\cos^4x
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Prove that }\sin4x=4\sin x\cos^3x-4\cos x\sin^3x.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sin4x
\displaystyle =\sin(2\cdot2x)
\displaystyle =2\sin2x\cos2x
\displaystyle =2(2\sin x\cos x)(\cos^2x-\sin^2x)
\displaystyle =4\sin x\cos x(\cos^2x-\sin^2x)
\displaystyle =4\sin x\cos^3x-4\sin^3x\cos x
\displaystyle =4\sin x\cos^3x-4\cos x\sin^3x
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Prove that }3(\sin x-\cos x)^4+6(\sin x+\cos x)^2
\displaystyle \qquad+4(\sin^6x+\cos^6x)=13.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=3(\sin x-\cos x)^4+6(\sin x+\cos x)^2+4(\sin^6x+\cos^6x)
\displaystyle =3\left(\sin^4x-4\sin^3x\cos x+6\sin^2x\cos^2x-4\sin x\cos^3x+\cos^4x\right)
\displaystyle \qquad+6\left(\sin^2x+2\sin x\cos x+\cos^2x\right)+4(\sin^6x+\cos^6x)
\displaystyle =3\left[\sin^4x+\cos^4x-4\sin x\cos x(\sin^2x+\cos^2x)+6\sin^2x\cos^2x\right]
\displaystyle \qquad+6(1+2\sin x\cos x)+4(\sin^6x+\cos^6x)
\displaystyle =3\left(\sin^4x+\cos^4x-4\sin x\cos x+6\sin^2x\cos^2x\right)
\displaystyle \qquad+6(1+2\sin x\cos x)+4(\sin^6x+\cos^6x)
\displaystyle \text{Now, }\sin^6x+\cos^6x=(\sin^2x+\cos^2x)(\sin^4x-\sin^2x\cos^2x+\cos^4x)
\displaystyle =\sin^4x-\sin^2x\cos^2x+\cos^4x
\displaystyle \therefore \text{LHS}=3\sin^4x+3\cos^4x-12\sin x\cos x+18\sin^2x\cos^2x
\displaystyle \qquad+6+12\sin x\cos x+4\sin^4x-4\sin^2x\cos^2x+4\cos^4x
\displaystyle =7\sin^4x+7\cos^4x+14\sin^2x\cos^2x+6
\displaystyle =7\left(\sin^4x+\cos^4x+2\sin^2x\cos^2x\right)+6
\displaystyle =7(\sin^2x+\cos^2x)^2+6
\displaystyle =7+6
\displaystyle =13
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Prove that }2(\sin^6x+\cos^6x)-3(\sin^4x+\cos^4x)+1=0.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=2(\sin^6x+\cos^6x)-3(\sin^4x+\cos^4x)+1
\displaystyle =2\left[(\sin^2x)^3+(\cos^2x)^3\right]-3(\sin^4x+\cos^4x)+1
\displaystyle =2(\sin^2x+\cos^2x)(\sin^4x-\sin^2x\cos^2x+\cos^4x)
\displaystyle \qquad-3(\sin^4x+\cos^4x)+1
\displaystyle =2(\sin^4x-\sin^2x\cos^2x+\cos^4x)-3(\sin^4x+\cos^4x)+1
\displaystyle =-\sin^4x-\cos^4x-2\sin^2x\cos^2x+1
\displaystyle =-(\sin^2x+\cos^2x)^2+1
\displaystyle =-1+1
\displaystyle =0
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Prove that }\cos^6x-\sin^6x=\cos2x\left(1-\frac{1}{4}\sin^22x\right).
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\cos^6x-\sin^6x
\displaystyle =(\cos^2x)^3-(\sin^2x)^3
\displaystyle =(\cos^2x-\sin^2x)\left(\cos^4x+\sin^2x\cos^2x+\sin^4x\right)
\displaystyle =\cos2x\left(\cos^4x+\sin^2x\cos^2x+\sin^4x\right)
\displaystyle =\cos2x\left(\cos^4x+2\sin^2x\cos^2x+\sin^4x-\sin^2x\cos^2x\right)
\displaystyle =\cos2x\left[(\cos^2x+\sin^2x)^2-\sin^2x\cos^2x\right]
\displaystyle =\cos2x\left(1-\sin^2x\cos^2x\right)
\displaystyle =\cos2x\left[1-\frac{1}{4}(2\sin x\cos x)^2\right]
\displaystyle =\cos2x\left(1-\frac{1}{4}\sin^22x\right)
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Prove that }\tan\left(\frac{\pi}{4}+x\right)+\tan\left(\frac{\pi}{4}-x\right)=2\sec2x.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\tan\left(\frac{\pi}{4}+x\right)+\tan\left(\frac{\pi}{4}-x\right)
\displaystyle =\frac{\tan\frac{\pi}{4}+\tan x}{1-\tan\frac{\pi}{4}\tan x}+\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}
\displaystyle =\frac{1+\tan x}{1-\tan x}+\frac{1-\tan x}{1+\tan x}
\displaystyle =\frac{(1+\tan x)^2+(1-\tan x)^2}{1-\tan^2x}
\displaystyle =\frac{1+2\tan x+\tan^2x+1-2\tan x+\tan^2x}{1-\tan^2x}
\displaystyle =\frac{2(1+\tan^2x)}{1-\tan^2x}
\displaystyle =\frac{2\left(\frac{1}{\cos^2x}\right)}{\frac{\cos^2x-\sin^2x}{\cos^2x}}
\displaystyle =\frac{2}{\cos^2x-\sin^2x}
\displaystyle =\frac{2}{\cos2x}
\displaystyle =2\sec2x
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Prove that }\cot^2x-\tan^2x=4\cot2x\,\mathrm{cosec}\,2x.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\cot^2x-\tan^2x
\displaystyle =\frac{\cos^2x}{\sin^2x}-\frac{\sin^2x}{\cos^2x}
\displaystyle =\frac{\cos^4x-\sin^4x}{\sin^2x\cos^2x}
\displaystyle =\frac{(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)}{\sin^2x\cos^2x}
\displaystyle =\frac{\cos^2x-\sin^2x}{\sin^2x\cos^2x}
\displaystyle =\frac{4\cos2x}{(2\sin x\cos x)^2}
\displaystyle =\frac{4\cos2x}{\sin^22x}
\displaystyle =4\left(\frac{\cos2x}{\sin2x}\right)\left(\frac{1}{\sin2x}\right)
\displaystyle =4\cot2x\,\mathrm{cosec}\,2x
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Prove that }\cos4x-\cos4\alpha=8(\cos x-\cos\alpha)(\cos x+\cos\alpha)
\displaystyle \qquad\times(\cos x-\sin\alpha)(\cos x+\sin\alpha).
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\cos4x-\cos4\alpha
\displaystyle =(2\cos^22x-1)-(2\cos^22\alpha-1)
\displaystyle =2(\cos^22x-\cos^22\alpha)
\displaystyle =2(\cos2x+\cos2\alpha)(\cos2x-\cos2\alpha)
\displaystyle =2(2\cos^2x-1+1-2\sin^2\alpha)
\displaystyle \qquad\times(2\cos^2x-1-2\cos^2\alpha+1)
\displaystyle =8(\cos^2x-\sin^2\alpha)(\cos^2x-\cos^2\alpha)
\displaystyle =8(\cos x-\sin\alpha)(\cos x+\sin\alpha)
\displaystyle \qquad\times(\cos x-\cos\alpha)(\cos x+\cos\alpha)
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Prove that }\sin3x+\sin2x-\sin x=4\sin x\cos\frac{x}{2}\cos\frac{3x}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sin3x+\sin2x-\sin x
\displaystyle =(\sin3x-\sin x)+\sin2x
\displaystyle =2\cos2x\sin x+2\sin x\cos x
\displaystyle =2\sin x(\cos2x+\cos x)
\displaystyle =2\sin x\left(2\cos\frac{3x}{2}\cos\frac{x}{2}\right)
\displaystyle =4\sin x\cos\frac{x}{2}\cos\frac{3x}{2}
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Prove that }\tan82\frac{1}{2}^{\circ}=(\sqrt3+\sqrt2)(\sqrt2+1)
\displaystyle \qquad=\sqrt2+\sqrt3+\sqrt4+\sqrt6.
\displaystyle \text{Answer:}
\displaystyle \text{We know that }\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}
\displaystyle \therefore \tan7\frac{1}{2}^{\circ}=\frac{\sin15^\circ}{1+\cos15^\circ}
\displaystyle =\frac{\sin(45^\circ-30^\circ)}{1+\cos(45^\circ-30^\circ)}
\displaystyle =\frac{\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ}{1+\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ}
\displaystyle =\frac{\frac{1}{\sqrt2}\cdot\frac{\sqrt3}{2}-\frac{1}{\sqrt2}\cdot\frac{1}{2}}{1+\frac{1}{\sqrt2}\cdot\frac{\sqrt3}{2}+\frac{1}{\sqrt2}\cdot\frac{1}{2}}
\displaystyle =\frac{\frac{\sqrt3-1}{2\sqrt2}}{\frac{2\sqrt2+\sqrt3+1}{2\sqrt2}}
\displaystyle =\frac{\sqrt3-1}{2\sqrt2+\sqrt3+1}
\displaystyle \therefore \cot7\frac{1}{2}^{\circ}=\frac{2\sqrt2+\sqrt3+1}{\sqrt3-1}
\displaystyle \tan82\frac{1}{2}^{\circ}=\cot7\frac{1}{2}^{\circ}
\displaystyle =\frac{2\sqrt2+\sqrt3+1}{\sqrt3-1}
\displaystyle =\frac{(2\sqrt2+\sqrt3+1)(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}
\displaystyle =\frac{2\sqrt6+2\sqrt2+3+2\sqrt3+1}{2}
\displaystyle =\sqrt6+\sqrt2+\sqrt3+2
\displaystyle =\sqrt2+\sqrt3+\sqrt4+\sqrt6
\displaystyle \text{Also, }(\sqrt3+\sqrt2)(\sqrt2+1)
\displaystyle =\sqrt6+\sqrt3+2+\sqrt2
\displaystyle =\sqrt2+\sqrt3+\sqrt4+\sqrt6
\displaystyle \therefore \tan82\frac{1}{2}^{\circ}=(\sqrt3+\sqrt2)(\sqrt2+1)
\displaystyle \qquad=\sqrt2+\sqrt3+\sqrt4+\sqrt6.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Prove that }\cot\frac{\pi}{8}=\sqrt2+1.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\cot\frac{\pi}{8}=\cot22\frac{1}{2}^{\circ}
\displaystyle \text{We know that }\cot2x=\frac{\cot^2x-1}{2\cot x}
\displaystyle \therefore \cot45^\circ=\frac{\cot^222\frac{1}{2}^{\circ}-1}{2\cot22\frac{1}{2}^{\circ}}
\displaystyle \therefore 1=\frac{\cot^222\frac{1}{2}^{\circ}-1}{2\cot22\frac{1}{2}^{\circ}}
\displaystyle \therefore 2\cot22\frac{1}{2}^{\circ}=\cot^222\frac{1}{2}^{\circ}-1
\displaystyle \text{Let }\cot22\frac{1}{2}^{\circ}=a
\displaystyle \therefore a^2-2a-1=0
\displaystyle \therefore (a-1)^2=2
\displaystyle \therefore a-1=\pm\sqrt2
\displaystyle \therefore a=1\pm\sqrt2
\displaystyle \text{Since }22\frac{1}{2}^{\circ}\text{ lies in the first quadrant, }\cot22\frac{1}{2}^{\circ}>0
\displaystyle \therefore a=1+\sqrt2
\displaystyle \therefore \cot\frac{\pi}{8}=\cot22\frac{1}{2}^{\circ}=1+\sqrt2
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{(i) If }\cos x=-\frac{3}{5}\text{ and }x\text{ lies in the third quadrant, find}
\displaystyle \cos\frac{x}{2},\ \sin\frac{x}{2}\text{ and }\sin2x.
\displaystyle \text{(ii) If }\cos x=-\frac{3}{5}\text{ and }x\text{ lies in the second quadrant, find }\sin2x
\displaystyle \text{and }\sin\frac{x}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given }\cos x=-\frac{3}{5}\text{ and }x\text{ lies in the third quadrant.}
\displaystyle \cos x=2\cos^2\frac{x}{2}-1
\displaystyle -\frac{3}{5}=2\cos^2\frac{x}{2}-1
\displaystyle 2\cos^2\frac{x}{2}=1-\frac{3}{5}=\frac{2}{5}
\displaystyle \cos^2\frac{x}{2}=\frac{1}{5}
\displaystyle \cos\frac{x}{2}=\pm\frac{1}{\sqrt5}
\displaystyle \text{Since }x\text{ lies in the third quadrant, }\frac{x}{2}\text{ lies in the second quadrant.}
\displaystyle \therefore \cos\frac{x}{2}=-\frac{1}{\sqrt5}=-\frac{\sqrt5}{5}
\displaystyle \text{Also, }\cos x=1-2\sin^2\frac{x}{2}
\displaystyle -\frac{3}{5}=1-2\sin^2\frac{x}{2}
\displaystyle 2\sin^2\frac{x}{2}=1+\frac{3}{5}=\frac{8}{5}
\displaystyle \sin^2\frac{x}{2}=\frac{4}{5}
\displaystyle \sin\frac{x}{2}=\pm\frac{2}{\sqrt5}
\displaystyle \text{Since }\frac{x}{2}\text{ lies in the second quadrant, }\sin\frac{x}{2}>0.
\displaystyle \therefore \sin\frac{x}{2}=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}
\displaystyle \sin x=\pm\sqrt{1-\cos^2x}
\displaystyle =\pm\sqrt{1-\left(-\frac{3}{5}\right)^2}
\displaystyle =\pm\sqrt{\frac{16}{25}}=\pm\frac{4}{5}
\displaystyle \text{Since }x\text{ lies in the third quadrant, }\sin x<0.
\displaystyle \therefore \sin x=-\frac{4}{5}
\displaystyle \sin2x=2\sin x\cos x
\displaystyle =2\left(-\frac{4}{5}\right)\left(-\frac{3}{5}\right)
\displaystyle =\frac{24}{25}
\displaystyle \therefore \cos\frac{x}{2}=-\frac{\sqrt5}{5},\quad\sin\frac{x}{2}=\frac{2\sqrt5}{5},\quad\sin2x=\frac{24}{25}.
\displaystyle \\

\displaystyle \text{(ii) Given }\cos x=-\frac{3}{5}\text{ and }x\text{ lies in the second quadrant.}
\displaystyle \sin x=\pm\sqrt{1-\cos^2x}
\displaystyle =\pm\sqrt{1-\left(-\frac{3}{5}\right)^2}
\displaystyle =\pm\frac{4}{5}
\displaystyle \text{Since }x\text{ lies in the second quadrant, }\sin x>0.
\displaystyle \therefore \sin x=\frac{4}{5}
\displaystyle \sin2x=2\sin x\cos x
\displaystyle =2\left(\frac{4}{5}\right)\left(-\frac{3}{5}\right)
\displaystyle =-\frac{24}{25}
\displaystyle \text{Also, }\sin^2\frac{x}{2}=\frac{1-\cos x}{2}
\displaystyle =\frac{1-\left(-\frac{3}{5}\right)}{2}
\displaystyle =\frac{4}{5}
\displaystyle \sin\frac{x}{2}=\pm\frac{2}{\sqrt5}
\displaystyle \text{Since }x\text{ lies in the second quadrant, }\frac{x}{2}\text{ lies in the first quadrant.}
\displaystyle \therefore \sin\frac{x}{2}=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}
\displaystyle \therefore \sin2x=-\frac{24}{25}\quad\text{and}\quad\sin\frac{x}{2}=\frac{2\sqrt5}{5}.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{If }\sin x=\frac{\sqrt5}{3}\text{ and }x\text{ lies in the second quadrant, find}
\displaystyle \cos\frac{x}{2},\ \sin\frac{x}{2}\text{ and }\tan\frac{x}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\sin x=\frac{\sqrt5}{3}
\displaystyle \cos x=\pm\sqrt{1-\sin^2x}
\displaystyle =\pm\sqrt{1-\frac{5}{9}}
\displaystyle =\pm\frac{2}{3}
\displaystyle \text{Since }x\text{ lies in the second quadrant, }\cos x<0.
\displaystyle \therefore \cos x=-\frac{2}{3}
\displaystyle \text{We know that }\sin^2\frac{x}{2}=\frac{1-\cos x}{2}
\displaystyle \therefore \sin^2\frac{x}{2}=\frac{1-\left(-\frac{2}{3}\right)}{2}
\displaystyle =\frac{5}{6}
\displaystyle \therefore \sin\frac{x}{2}=\pm\sqrt{\frac{5}{6}}
\displaystyle \text{Since }x\text{ lies in the second quadrant, }\frac{x}{2}\text{ lies in the first quadrant.}
\displaystyle \therefore \sin\frac{x}{2}=\sqrt{\frac{5}{6}}=\frac{\sqrt{30}}{6}
\displaystyle \text{Also, }\cos^2\frac{x}{2}=\frac{1+\cos x}{2}
\displaystyle \therefore \cos^2\frac{x}{2}=\frac{1+\left(-\frac{2}{3}\right)}{2}
\displaystyle =\frac{1}{6}
\displaystyle \therefore \cos\frac{x}{2}=\pm\frac{1}{\sqrt6}
\displaystyle \text{Since }\frac{x}{2}\text{ lies in the first quadrant, }\cos\frac{x}{2}>0.
\displaystyle \therefore \cos\frac{x}{2}=\frac{1}{\sqrt6}=\frac{\sqrt6}{6}
\displaystyle \tan\frac{x}{2}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}
\displaystyle =\frac{\sqrt{\frac{5}{6}}}{\frac{1}{\sqrt6}}
\displaystyle =\sqrt5
\displaystyle \therefore \cos\frac{x}{2}=\frac{\sqrt6}{6},\quad\sin\frac{x}{2}=\frac{\sqrt{30}}{6},\quad\tan\frac{x}{2}=\sqrt5.
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{(i) If }0\leq x\leq\pi,\ x\text{ lies in the second quadrant and }\sin x=\frac14,
\displaystyle \text{find }\cos\frac{x}{2},\ \sin\frac{x}{2}\text{ and }\tan\frac{x}{2}.
\displaystyle \text{(ii) If }\cos x=\frac45\text{ and }x\text{ is acute, find }\tan2x.
\displaystyle \text{(iii) If }\sin x=\frac45\text{ and }0<x<\frac{\pi}{2},\text{ find }\sin4x.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given }\sin x=\frac14
\displaystyle \cos x=\pm\sqrt{1-\sin^2x}
\displaystyle =\pm\sqrt{1-\frac{1}{16}}
\displaystyle =\pm\frac{\sqrt{15}}{4}
\displaystyle \text{Since }x\text{ lies in the second quadrant, }\cos x<0.
\displaystyle \therefore \cos x=-\frac{\sqrt{15}}{4}
\displaystyle \cos^2\frac{x}{2}=\frac{1+\cos x}{2}
\displaystyle =\frac{1-\frac{\sqrt{15}}{4}}{2}
\displaystyle =\frac{4-\sqrt{15}}{8}
\displaystyle \therefore \cos\frac{x}{2}=\pm\sqrt{\frac{4-\sqrt{15}}{8}}
\displaystyle \text{Since }x\text{ lies in the second quadrant, }\frac{x}{2}\text{ lies in the first quadrant.}
\displaystyle \therefore \cos\frac{x}{2}=\sqrt{\frac{4-\sqrt{15}}{8}}
\displaystyle \sin^2\frac{x}{2}=\frac{1-\cos x}{2}
\displaystyle =\frac{1+\frac{\sqrt{15}}{4}}{2}
\displaystyle =\frac{4+\sqrt{15}}{8}
\displaystyle \therefore \sin\frac{x}{2}=\pm\sqrt{\frac{4+\sqrt{15}}{8}}
\displaystyle \text{Since }\frac{x}{2}\text{ lies in the first quadrant, }\sin\frac{x}{2}>0.
\displaystyle \therefore \sin\frac{x}{2}=\sqrt{\frac{4+\sqrt{15}}{8}}
\displaystyle \tan\frac{x}{2}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}
\displaystyle =\sqrt{\frac{4+\sqrt{15}}{4-\sqrt{15}}}
\displaystyle =\sqrt{\frac{(4+\sqrt{15})^2}{16-15}}
\displaystyle =4+\sqrt{15}
\displaystyle \therefore \cos\frac{x}{2}=\sqrt{\frac{4-\sqrt{15}}{8}},\quad\sin\frac{x}{2}=\sqrt{\frac{4+\sqrt{15}}{8}}
\displaystyle \text{and }\tan\frac{x}{2}=4+\sqrt{15}.
\displaystyle \\

\displaystyle \text{(ii) Given }\cos x=\frac45\text{ and }x\text{ is acute.}
\displaystyle \sin x=\sqrt{1-\cos^2x}
\displaystyle =\sqrt{1-\left(\frac45\right)^2}
\displaystyle =\frac35
\displaystyle \therefore \tan x=\frac{\sin x}{\cos x}=\frac{\frac35}{\frac45}=\frac34
\displaystyle \tan2x=\frac{2\tan x}{1-\tan^2x}
\displaystyle =\frac{2\times\frac34}{1-\left(\frac34\right)^2}
\displaystyle =\frac{\frac32}{\frac{7}{16}}
\displaystyle =\frac{24}{7}
\displaystyle \therefore \tan2x=\frac{24}{7}.
\displaystyle \\

\displaystyle \text{(iii) Given }\sin x=\frac45\text{ and }0<x<\frac{\pi}{2}.
\displaystyle \cos x=\sqrt{1-\sin^2x}
\displaystyle =\sqrt{1-\frac{16}{25}}
\displaystyle =\frac35
\displaystyle \sin2x=2\sin x\cos x
\displaystyle =2\times\frac45\times\frac35
\displaystyle =\frac{24}{25}
\displaystyle \cos2x=\cos^2x-\sin^2x
\displaystyle =\left(\frac35\right)^2-\left(\frac45\right)^2
\displaystyle =\frac{9}{25}-\frac{16}{25}
\displaystyle =-\frac{7}{25}
\displaystyle \sin4x=2\sin2x\cos2x
\displaystyle =2\times\frac{24}{25}\times\left(-\frac{7}{25}\right)
\displaystyle =-\frac{336}{625}
\displaystyle \therefore \sin4x=-\frac{336}{625}.
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{If }\tan x=\frac{b}{a},\text{ where }a>b>0,\text{ find the value of}
\displaystyle \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}.
\displaystyle \text{Answer:}
\displaystyle \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}
\displaystyle =\frac{a+b}{\sqrt{(a-b)(a+b)}}+\frac{a-b}{\sqrt{(a+b)(a-b)}}
\displaystyle =\frac{a+b+a-b}{\sqrt{a^2-b^2}}
\displaystyle =\frac{2a}{\sqrt{a^2-b^2}}
\displaystyle =\frac{2}{\sqrt{1-\frac{b^2}{a^2}}}
\displaystyle =\frac{2}{\sqrt{1-\tan^2x}}
\displaystyle =\frac{2}{\sqrt{\frac{\cos^2x-\sin^2x}{\cos^2x}}}
\displaystyle =\frac{2\cos x}{\sqrt{\cos^2x-\sin^2x}}
\displaystyle =\frac{2\cos x}{\sqrt{\cos2x}}
\displaystyle \therefore \text{The required value is }\frac{2\cos x}{\sqrt{\cos2x}}.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{If }\tan A=\frac17\text{ and }\tan B=\frac13,\text{ show that }\cos2A=\sin4B.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\tan A=\frac17\text{ and }\tan B=\frac13
\displaystyle \cos2A=\frac{1-\tan^2A}{1+\tan^2A}
\displaystyle =\frac{1-\left(\frac17\right)^2}{1+\left(\frac17\right)^2}
\displaystyle =\frac{49-1}{49+1}
\displaystyle =\frac{48}{50}
\displaystyle =\frac{24}{25}
\displaystyle \text{Also, }\sin4B=2\sin2B\cos2B
\displaystyle =2\left(\frac{2\tan B}{1+\tan^2B}\right)\left(\frac{1-\tan^2B}{1+\tan^2B}\right)
\displaystyle =\frac{4\tan B(1-\tan^2B)}{(1+\tan^2B)^2}
\displaystyle =\frac{4\left(\frac13\right)\left[1-\left(\frac13\right)^2\right]}{\left[1+\left(\frac13\right)^2\right]^2}
\displaystyle =\frac{4\left(\frac13\right)\left(\frac89\right)}{\left(\frac{10}{9}\right)^2}
\displaystyle =\frac{\frac{32}{27}}{\frac{100}{81}}
\displaystyle =\frac{32}{27}\times\frac{81}{100}
\displaystyle =\frac{24}{25}
\displaystyle \therefore \cos2A=\sin4B.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{Prove that }\cos7^\circ\cos14^\circ\cos28^\circ\cos56^\circ
\displaystyle =\frac{\sin68^\circ}{16\cos83^\circ}.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\cos7^\circ\cos14^\circ\cos28^\circ\cos56^\circ
\displaystyle =\frac{2\sin7^\circ\cos7^\circ\cos14^\circ\cos28^\circ\cos56^\circ}{2\sin7^\circ}
\displaystyle =\frac{\sin14^\circ\cos14^\circ\cos28^\circ\cos56^\circ}{2\sin7^\circ}
\displaystyle =\frac{2\sin14^\circ\cos14^\circ\cos28^\circ\cos56^\circ}{4\sin7^\circ}
\displaystyle =\frac{\sin28^\circ\cos28^\circ\cos56^\circ}{4\sin7^\circ}
\displaystyle =\frac{2\sin28^\circ\cos28^\circ\cos56^\circ}{8\sin7^\circ}
\displaystyle =\frac{\sin56^\circ\cos56^\circ}{8\sin7^\circ}
\displaystyle =\frac{2\sin56^\circ\cos56^\circ}{16\sin7^\circ}
\displaystyle =\frac{\sin112^\circ}{16\sin7^\circ}
\displaystyle =\frac{\sin(180^\circ-68^\circ)}{16\sin(90^\circ-83^\circ)}
\displaystyle =\frac{\sin68^\circ}{16\cos83^\circ}
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{Prove that }\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{8\pi}{15}\cos\frac{16\pi}{15}
\displaystyle =\frac{1}{16}.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{8\pi}{15}\cos\frac{16\pi}{15}
\displaystyle =\cos24^\circ\cos48^\circ\cos96^\circ\cos192^\circ
\displaystyle =\frac{2\sin24^\circ\cos24^\circ\cos48^\circ\cos96^\circ\cos192^\circ}{2\sin24^\circ}
\displaystyle =\frac{2\sin48^\circ\cos48^\circ\cos96^\circ\cos192^\circ}{4\sin24^\circ}
\displaystyle =\frac{2\sin96^\circ\cos96^\circ\cos192^\circ}{8\sin24^\circ}
\displaystyle =\frac{2\sin192^\circ\cos192^\circ}{16\sin24^\circ}
\displaystyle =\frac{\sin384^\circ}{16\sin24^\circ}
\displaystyle =\frac{\sin(360^\circ+24^\circ)}{16\sin24^\circ}
\displaystyle =\frac{\sin24^\circ}{16\sin24^\circ}
\displaystyle =\frac{1}{16}
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{Prove that }\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{8\pi}{15}=-\frac{1}{16}.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{8\pi}{15}
\displaystyle =\cos12^\circ\cos24^\circ\cos48^\circ\cos96^\circ
\displaystyle =\frac{2\sin12^\circ\cos12^\circ\cos24^\circ\cos48^\circ\cos96^\circ}{2\sin12^\circ}
\displaystyle =\frac{2\sin24^\circ\cos24^\circ\cos48^\circ\cos96^\circ}{4\sin12^\circ}
\displaystyle =\frac{2\sin48^\circ\cos48^\circ\cos96^\circ}{8\sin12^\circ}
\displaystyle =\frac{2\sin96^\circ\cos96^\circ}{16\sin12^\circ}
\displaystyle =\frac{\sin192^\circ}{16\sin12^\circ}
\displaystyle =\frac{\sin(180^\circ+12^\circ)}{16\sin12^\circ}
\displaystyle =\frac{-\sin12^\circ}{16\sin12^\circ}
\displaystyle =-\frac{1}{16}
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{Prove that }\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}
\displaystyle \qquad\times\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}=\frac{1}{64}.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}
\displaystyle =\frac{2\sin\frac{\pi}{65}\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}}{2\sin\frac{\pi}{65}}
\displaystyle =\frac{2\sin\frac{2\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}}{4\sin\frac{\pi}{65}}
\displaystyle =\frac{2\sin\frac{4\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}}{8\sin\frac{\pi}{65}}
\displaystyle =\frac{2\sin\frac{8\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}}{16\sin\frac{\pi}{65}}
\displaystyle =\frac{2\sin\frac{16\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}}{32\sin\frac{\pi}{65}}
\displaystyle =\frac{2\sin\frac{32\pi}{65}\cos\frac{32\pi}{65}}{64\sin\frac{\pi}{65}}
\displaystyle =\frac{\sin\frac{64\pi}{65}}{64\sin\frac{\pi}{65}}
\displaystyle =\frac{\sin\left(\pi-\frac{\pi}{65}\right)}{64\sin\frac{\pi}{65}}
\displaystyle =\frac{\sin\frac{\pi}{65}}{64\sin\frac{\pi}{65}}
\displaystyle =\frac{1}{64}
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{If }2\tan\alpha=3\tan\beta,\text{ prove that }\tan(\alpha-\beta)
\displaystyle =\frac{\sin2\beta}{5-\cos2\beta}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }2\tan\alpha=3\tan\beta
\displaystyle \therefore \tan\alpha=\frac{3}{2}\tan\beta
\displaystyle \text{LHS}=\tan(\alpha-\beta)
\displaystyle =\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}
\displaystyle =\frac{\frac{3}{2}\tan\beta-\tan\beta}{1+\frac{3}{2}\tan^2\beta}
\displaystyle =\frac{\tan\beta}{2+3\tan^2\beta}
\displaystyle =\frac{\frac{\sin\beta}{\cos\beta}}{2+3\frac{\sin^2\beta}{\cos^2\beta}}
\displaystyle =\frac{\sin\beta\cos\beta}{2\cos^2\beta+3\sin^2\beta}
\displaystyle =\frac{\sin\beta\cos\beta}{2(\cos^2\beta+\sin^2\beta)+\sin^2\beta}
\displaystyle =\frac{\sin\beta\cos\beta}{2+\sin^2\beta}
\displaystyle =\frac{2\sin\beta\cos\beta}{4+2\sin^2\beta}
\displaystyle =\frac{\sin2\beta}{4+(1-\cos2\beta)}
\displaystyle =\frac{\sin2\beta}{5-\cos2\beta}
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{If }\sin\alpha+\sin\beta=a\text{ and }\cos\alpha+\cos\beta=b,\text{ prove that:}
\displaystyle \text{(i) }\sin(\alpha+\beta)=\frac{2ab}{a^2+b^2}
\displaystyle \text{(ii) }\cos(\alpha-\beta)=\frac{a^2+b^2-2}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given }\sin\alpha+\sin\beta=a
\displaystyle \therefore a=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}
\displaystyle \text{Also, }\cos\alpha+\cos\beta=b
\displaystyle \therefore b=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}
\displaystyle \therefore 2ab=8\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha+\beta}{2}\cos^2\frac{\alpha-\beta}{2}
\displaystyle \text{Also, }a^2+b^2
\displaystyle =4\cos^2\frac{\alpha-\beta}{2}\left(\sin^2\frac{\alpha+\beta}{2}+\cos^2\frac{\alpha+\beta}{2}\right)
\displaystyle =4\cos^2\frac{\alpha-\beta}{2}
\displaystyle \therefore \frac{2ab}{a^2+b^2}
\displaystyle =\frac{8\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha+\beta}{2}\cos^2\frac{\alpha-\beta}{2}}{4\cos^2\frac{\alpha-\beta}{2}}
\displaystyle =2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha+\beta}{2}
\displaystyle =\sin(\alpha+\beta)
\displaystyle \therefore \sin(\alpha+\beta)=\frac{2ab}{a^2+b^2}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \text{(ii) Squaring and adding the given equations,}
\displaystyle (\sin\alpha+\sin\beta)^2+(\cos\alpha+\cos\beta)^2=a^2+b^2
\displaystyle \sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta+\cos^2\alpha+\cos^2\beta
\displaystyle \qquad+2\cos\alpha\cos\beta=a^2+b^2
\displaystyle 2+2(\sin\alpha\sin\beta+\cos\alpha\cos\beta)=a^2+b^2
\displaystyle 2+2\cos(\alpha-\beta)=a^2+b^2
\displaystyle 2\cos(\alpha-\beta)=a^2+b^2-2
\displaystyle \therefore \cos(\alpha-\beta)=\frac{a^2+b^2-2}{2}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{If }2\tan\frac{\alpha}{2}=\tan\frac{\beta}{2},\text{ prove that}
\displaystyle \cos\alpha=\frac{3+5\cos\beta}{5+3\cos\beta}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }2\tan\frac{\alpha}{2}=\tan\frac{\beta}{2}
\displaystyle \text{RHS}=\frac{3+5\cos\beta}{5+3\cos\beta}
\displaystyle \text{Using }\cos\beta=\frac{1-\tan^2\frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}},
\displaystyle \text{RHS}=\frac{3+5\left(\frac{1-\tan^2\frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}\right)}{5+3\left(\frac{1-\tan^2\frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}\right)}
\displaystyle =\frac{3\left(1+\tan^2\frac{\beta}{2}\right)+5\left(1-\tan^2\frac{\beta}{2}\right)}{5\left(1+\tan^2\frac{\beta}{2}\right)+3\left(1-\tan^2\frac{\beta}{2}\right)}
\displaystyle =\frac{3+3\tan^2\frac{\beta}{2}+5-5\tan^2\frac{\beta}{2}}{5+5\tan^2\frac{\beta}{2}+3-3\tan^2\frac{\beta}{2}}
\displaystyle =\frac{8-2\tan^2\frac{\beta}{2}}{8+2\tan^2\frac{\beta}{2}}
\displaystyle \text{Since }\tan\frac{\beta}{2}=2\tan\frac{\alpha}{2},
\displaystyle \text{RHS}=\frac{8-2\left(2\tan\frac{\alpha}{2}\right)^2}{8+2\left(2\tan\frac{\alpha}{2}\right)^2}
\displaystyle =\frac{8-8\tan^2\frac{\alpha}{2}}{8+8\tan^2\frac{\alpha}{2}}
\displaystyle =\frac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}
\displaystyle =\cos\alpha=\text{LHS}
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{If }\cos x=\frac{\cos\alpha+\cos\beta}{1+\cos\alpha\cos\beta},\text{ prove that}
\displaystyle \tan\frac{x}{2}=\pm\tan\frac{\alpha}{2}\tan\frac{\beta}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\cos x=\frac{\cos\alpha+\cos\beta}{1+\cos\alpha\cos\beta}
\displaystyle \Rightarrow \frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}=\frac{\cos\alpha+\cos\beta}{1+\cos\alpha\cos\beta}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{\left(1-\tan^2\frac{x}{2}\right)+\left(1+\tan^2\frac{x}{2}\right)}{\left(1-\tan^2\frac{x}{2}\right)-\left(1+\tan^2\frac{x}{2}\right)}
\displaystyle =\frac{\cos\alpha+\cos\beta+1+\cos\alpha\cos\beta}{\cos\alpha+\cos\beta-1-\cos\alpha\cos\beta}
\displaystyle \Rightarrow \frac{2}{-2\tan^2\frac{x}{2}}=\frac{(1+\cos\alpha)(1+\cos\beta)}{-(1-\cos\alpha)(1-\cos\beta)}
\displaystyle \Rightarrow \frac{1}{\tan^2\frac{x}{2}}=\frac{(1+\cos\alpha)(1+\cos\beta)}{(1-\cos\alpha)(1-\cos\beta)}
\displaystyle \Rightarrow \tan^2\frac{x}{2}=\frac{(1-\cos\alpha)(1-\cos\beta)}{(1+\cos\alpha)(1+\cos\beta)}
\displaystyle \Rightarrow \tan^2\frac{x}{2}=\frac{\left(2\sin^2\frac{\alpha}{2}\right)\left(2\sin^2\frac{\beta}{2}\right)}{\left(2\cos^2\frac{\alpha}{2}\right)\left(2\cos^2\frac{\beta}{2}\right)}
\displaystyle \Rightarrow \tan^2\frac{x}{2}=\frac{4\sin^2\frac{\alpha}{2}\sin^2\frac{\beta}{2}}{4\cos^2\frac{\alpha}{2}\cos^2\frac{\beta}{2}}
\displaystyle \Rightarrow \tan^2\frac{x}{2}=\tan^2\frac{\alpha}{2}\tan^2\frac{\beta}{2}
\displaystyle \therefore \tan\frac{x}{2}=\pm\tan\frac{\alpha}{2}\tan\frac{\beta}{2}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{If }\sec(x+\alpha)+\sec(x-\alpha)=2\sec x\text{ and }\cos\alpha\ne1,
\displaystyle \text{prove that }\cos x=\pm\sqrt{2}\cos\frac{\alpha}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\sec(x+\alpha)+\sec(x-\alpha)=2\sec x
\displaystyle \Rightarrow \frac{1}{\cos(x+\alpha)}+\frac{1}{\cos(x-\alpha)}=\frac{2}{\cos x}
\displaystyle \Rightarrow \frac{\cos(x+\alpha)+\cos(x-\alpha)}{\cos(x+\alpha)\cos(x-\alpha)}=\frac{2}{\cos x}
\displaystyle \Rightarrow \frac{2\cos x\cos\alpha}{\cos^2x\cos^2\alpha-\sin^2x\sin^2\alpha}=\frac{2}{\cos x}
\displaystyle \Rightarrow \cos^2x\cos\alpha=\cos^2x\cos^2\alpha-\sin^2x\sin^2\alpha
\displaystyle \Rightarrow \cos^2x\cos\alpha=\cos^2x\cos^2\alpha-(1-\cos^2x)\sin^2\alpha
\displaystyle \Rightarrow \cos^2x\cos\alpha=\cos^2x(\cos^2\alpha+\sin^2\alpha)-\sin^2\alpha
\displaystyle \Rightarrow \cos^2x\cos\alpha=\cos^2x-\sin^2\alpha
\displaystyle \Rightarrow \cos^2x(1-\cos\alpha)=\sin^2\alpha
\displaystyle \Rightarrow \cos^2x(1-\cos\alpha)=(1-\cos\alpha)(1+\cos\alpha)
\displaystyle \text{Since }\cos\alpha\ne1,\text{ we have }1-\cos\alpha\ne0.
\displaystyle \Rightarrow \cos^2x=1+\cos\alpha
\displaystyle \Rightarrow \cos^2x=2\cos^2\frac{\alpha}{2}
\displaystyle \therefore \cos x=\pm\sqrt{2}\cos\frac{\alpha}{2}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{If }\cos\alpha+\cos\beta=\frac{1}{3}\text{ and }\sin\alpha+\sin\beta=\frac{1}{4},
\displaystyle \text{prove that }\cos\frac{\alpha-\beta}{2}=\pm\frac{5}{24}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\cos\alpha+\cos\beta=\frac{1}{3}\text{ and }\sin\alpha+\sin\beta=\frac{1}{4}
\displaystyle \text{Squaring and adding the two equations,}
\displaystyle (\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2=\frac{1}{9}+\frac{1}{16}
\displaystyle \Rightarrow \cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta
\displaystyle \qquad+\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta=\frac{25}{144}
\displaystyle \Rightarrow 2+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)=\frac{25}{144}
\displaystyle \Rightarrow 2+2\cos(\alpha-\beta)=\frac{25}{144}
\displaystyle \Rightarrow 2\cos(\alpha-\beta)=\frac{25}{144}-2
\displaystyle \Rightarrow 2\cos(\alpha-\beta)=-\frac{263}{144}
\displaystyle \Rightarrow \cos(\alpha-\beta)=-\frac{263}{288}
\displaystyle \text{Using }\cos^2\frac{\alpha-\beta}{2}=\frac{1+\cos(\alpha-\beta)}{2},
\displaystyle \cos^2\frac{\alpha-\beta}{2}=\frac{1-\frac{263}{288}}{2}
\displaystyle =\frac{\frac{25}{288}}{2}=\frac{25}{576}
\displaystyle \therefore \cos\frac{\alpha-\beta}{2}=\pm\frac{5}{24}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{If }\alpha\text{ and }\beta\text{ are acute angles such that }\sin\alpha=\frac{4}{5}
\displaystyle \text{and }\cos\beta=\frac{5}{13},\text{ prove that }\cos\frac{\alpha-\beta}{2}=\frac{8}{\sqrt{65}}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\sin\alpha=\frac{4}{5}
\displaystyle \text{Since }\alpha\text{ is acute, }\cos\alpha>0.
\displaystyle \therefore \cos\alpha=\sqrt{1-\sin^2\alpha}
\displaystyle =\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}
\displaystyle \text{Also, }\cos\beta=\frac{5}{13}
\displaystyle \text{Since }\beta\text{ is acute, }\sin\beta>0.
\displaystyle \therefore \sin\beta=\sqrt{1-\cos^2\beta}
\displaystyle =\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}
\displaystyle \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta
\displaystyle =\frac{3}{5}\times\frac{5}{13}+\frac{4}{5}\times\frac{12}{13}
\displaystyle =\frac{15+48}{65}=\frac{63}{65}
\displaystyle \text{Using }\cos^2\frac{\alpha-\beta}{2}=\frac{1+\cos(\alpha-\beta)}{2},
\displaystyle \cos^2\frac{\alpha-\beta}{2}=\frac{1+\frac{63}{65}}{2}
\displaystyle =\frac{\frac{128}{65}}{2}=\frac{64}{65}
\displaystyle \text{Since }\alpha\text{ and }\beta\text{ are acute, }-\frac{\pi}{2}<\frac{\alpha-\beta}{2}<\frac{\pi}{2}.
\displaystyle \therefore \cos\frac{\alpha-\beta}{2}>0
\displaystyle \therefore \cos\frac{\alpha-\beta}{2}=\sqrt{\frac{64}{65}}=\frac{8}{\sqrt{65}}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{If }a\cos2x+b\sin2x=c\text{ has }\alpha\text{ and }\beta\text{ as its roots,}
\displaystyle \text{prove that:}
\displaystyle \text{(i) }\tan\alpha+\tan\beta=\frac{2b}{a+c}
\displaystyle \text{(ii) }\tan\alpha\tan\beta=\frac{c-a}{c+a}
\displaystyle \text{(iii) }\tan(\alpha+\beta)=\frac{b}{a}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a\cos2x+b\sin2x=c
\displaystyle \text{Using }\cos2x=\frac{1-\tan^2x}{1+\tan^2x}\text{ and }\sin2x=\frac{2\tan x}{1+\tan^2x},
\displaystyle a\left(\frac{1-\tan^2x}{1+\tan^2x}\right)+b\left(\frac{2\tan x}{1+\tan^2x}\right)=c
\displaystyle \Rightarrow a(1-\tan^2x)+2b\tan x=c(1+\tan^2x)
\displaystyle \Rightarrow a-a\tan^2x+2b\tan x=c+c\tan^2x
\displaystyle \Rightarrow (a+c)\tan^2x-2b\tan x+(c-a)=0
\displaystyle \text{Since }\alpha\text{ and }\beta\text{ are roots, }\tan\alpha\text{ and }\tan\beta
\displaystyle \text{are the roots of this quadratic equation.}
\displaystyle \text{(i) By the sum of roots,}
\displaystyle \tan\alpha+\tan\beta=-\frac{-2b}{a+c}
\displaystyle \therefore \tan\alpha+\tan\beta=\frac{2b}{a+c}
\displaystyle \text{(ii) By the product of roots,}
\displaystyle \therefore \tan\alpha\tan\beta=\frac{c-a}{a+c}
\displaystyle \text{(iii) }\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}
\displaystyle =\frac{\frac{2b}{a+c}}{1-\frac{c-a}{a+c}}
\displaystyle =\frac{\frac{2b}{a+c}}{\frac{a+c-c+a}{a+c}}
\displaystyle =\frac{\frac{2b}{a+c}}{\frac{2a}{a+c}}
\displaystyle \therefore \tan(\alpha+\beta)=\frac{b}{a}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{If }\cos\alpha+\cos\beta=0=\sin\alpha+\sin\beta,\text{ prove that}
\displaystyle \cos2\alpha+\cos2\beta=-2\cos(\alpha+\beta).
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\cos\alpha+\cos\beta=0\text{ and }\sin\alpha+\sin\beta=0
\displaystyle \therefore (\cos\alpha+\cos\beta)^2=(\sin\alpha+\sin\beta)^2
\displaystyle \Rightarrow \cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta
\displaystyle =\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta
\displaystyle \Rightarrow (\cos^2\alpha-\sin^2\alpha)+(\cos^2\beta-\sin^2\beta)
\displaystyle =-2(\cos\alpha\cos\beta-\sin\alpha\sin\beta)
\displaystyle \Rightarrow \cos2\alpha+\cos2\beta=-2\cos(\alpha+\beta)
\displaystyle \text{Hence proved.}
\displaystyle \\

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