Class 11: Transformation Formulae

$\displaystyle \textbf{Question 1: }\text{Express each of the following as the product of sines and cosines:}$
$\displaystyle \text{(i) }\sin12x+\sin4x\qquad\text{(ii) }\sin5x-\sin x\qquad\text{(iii) }\cos12x+\cos8x$
$\displaystyle \text{(iv) }\cos12x-\cos4x\qquad\text{(v) }\sin2x+\cos4x$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }\sin12x+\sin4x=2\sin\left(\frac{12x+4x}{2}\right)\cos\left(\frac{12x-4x}{2}\right)$
$\displaystyle =2\sin8x\cos4x$

$\displaystyle \text{(ii) }\sin5x-\sin x=2\cos\left(\frac{5x+x}{2}\right)\sin\left(\frac{5x-x}{2}\right)$
$\displaystyle =2\cos3x\sin2x$

$\displaystyle \text{(iii) }\cos12x+\cos8x=2\cos\left(\frac{12x+8x}{2}\right)\cos\left(\frac{12x-8x}{2}\right)$
$\displaystyle =2\cos10x\cos2x$

$\displaystyle \text{(iv) }\cos12x-\cos4x=-2\sin\left(\frac{12x+4x}{2}\right)\sin\left(\frac{12x-4x}{2}\right)$
$\displaystyle =-2\sin8x\sin4x$

$\displaystyle \text{(v) }\sin2x+\cos4x=\sin2x+\sin\left(\frac{\pi}{2}-4x\right)$
$\displaystyle =2\sin\left(\frac{2x+\frac{\pi}{2}-4x}{2}\right)\cos\left(\frac{2x-\frac{\pi}{2}+4x}{2}\right)$
$\displaystyle =2\sin\left(\frac{\pi}{4}-x\right)\cos\left(3x-\frac{\pi}{4}\right)$
$\displaystyle =2\sin\left(\frac{\pi}{4}-x\right)\cos\left(\frac{\pi}{4}-3x\right)$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Prove that:}$
$\displaystyle \text{(i) }\sin38^\circ+\sin22^\circ=\sin82^\circ\qquad\text{(ii) }\cos100^\circ+\cos20^\circ=\cos40^\circ$
$\displaystyle \text{(iii) }\sin50^\circ+\sin10^\circ=\cos20^\circ\qquad\text{(iv) }\sin23^\circ+\sin37^\circ=\cos7^\circ$
$\displaystyle \text{(v) }\sin105^\circ+\cos105^\circ=\cos45^\circ\qquad\text{(vi) }\sin40^\circ+\sin20^\circ=\cos10^\circ$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) LHS}=\sin38^\circ+\sin22^\circ$
$\displaystyle =2\sin\left(\frac{38^\circ+22^\circ}{2}\right)\cos\left(\frac{38^\circ-22^\circ}{2}\right)$
$\displaystyle =2\sin30^\circ\cos8^\circ$
$\displaystyle =2\times\frac{1}{2}\cos8^\circ$
$\displaystyle =\cos8^\circ$
$\displaystyle =\sin82^\circ=\text{RHS. Hence proved.}$

$\displaystyle \text{(ii) LHS}=\cos100^\circ+\cos20^\circ$
$\displaystyle =2\cos\left(\frac{100^\circ+20^\circ}{2}\right)\cos\left(\frac{100^\circ-20^\circ}{2}\right)$
$\displaystyle =2\cos60^\circ\cos40^\circ$
$\displaystyle =2\times\frac{1}{2}\cos40^\circ$
$\displaystyle =\cos40^\circ=\text{RHS. Hence proved.}$

$\displaystyle \text{(iii) LHS}=\sin50^\circ+\sin10^\circ$
$\displaystyle =2\sin\left(\frac{50^\circ+10^\circ}{2}\right)\cos\left(\frac{50^\circ-10^\circ}{2}\right)$
$\displaystyle =2\sin30^\circ\cos20^\circ$
$\displaystyle =2\times\frac{1}{2}\cos20^\circ$
$\displaystyle =\cos20^\circ=\text{RHS. Hence proved.}$

$\displaystyle \text{(iv) LHS}=\sin23^\circ+\sin37^\circ$
$\displaystyle =2\sin\left(\frac{23^\circ+37^\circ}{2}\right)\cos\left(\frac{23^\circ-37^\circ}{2}\right)$
$\displaystyle =2\sin30^\circ\cos(-7^\circ)$
$\displaystyle =2\times\frac{1}{2}\cos7^\circ$
$\displaystyle =\cos7^\circ=\text{RHS. Hence proved.}$

$\displaystyle \text{(v) LHS}=\sin105^\circ+\cos105^\circ$
$\displaystyle =\sin105^\circ+\cos(90^\circ+15^\circ)$
$\displaystyle =\sin105^\circ-\sin15^\circ$
$\displaystyle =2\cos\left(\frac{105^\circ+15^\circ}{2}\right)\sin\left(\frac{105^\circ-15^\circ}{2}\right)$
$\displaystyle =2\cos60^\circ\sin45^\circ$
$\displaystyle =2\times\frac{1}{2}\times\frac{1}{\sqrt{2}}$
$\displaystyle =\frac{1}{\sqrt{2}}$
$\displaystyle =\cos45^\circ=\text{RHS. Hence proved.}$

$\displaystyle \text{(vi) LHS}=\sin40^\circ+\sin20^\circ$
$\displaystyle =2\sin\left(\frac{40^\circ+20^\circ}{2}\right)\cos\left(\frac{40^\circ-20^\circ}{2}\right)$
$\displaystyle =2\sin30^\circ\cos10^\circ$
$\displaystyle =2\times\frac{1}{2}\cos10^\circ$
$\displaystyle =\cos10^\circ=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Prove that:}$
$\displaystyle \text{(i) }\cos55^\circ+\cos65^\circ+\cos175^\circ=0$
$\displaystyle \text{(ii) }\sin50^\circ-\sin70^\circ+\sin10^\circ=0$
$\displaystyle \text{(iii) }\cos80^\circ+\cos40^\circ-\cos20^\circ=0$
$\displaystyle \text{(iv) }\cos20^\circ+\cos100^\circ+\cos140^\circ=0$
$\displaystyle \text{(v) }\sin\frac{5\pi}{18}-\cos\frac{4\pi}{9}=\sqrt{3}\sin\frac{\pi}{9}$
$\displaystyle \text{(vi) }\cos\frac{\pi}{12}-\sin\frac{\pi}{12}=\frac{1}{\sqrt{2}}$
$\displaystyle \text{(vii) }\sin80^\circ-\cos70^\circ=\cos50^\circ$
$\displaystyle \text{(viii) }\sin51^\circ+\cos81^\circ=\cos21^\circ$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) LHS}=\cos55^\circ+\cos65^\circ+\cos175^\circ$
$\displaystyle =\cos55^\circ+\cos65^\circ+\cos(180^\circ-5^\circ)$
$\displaystyle =\cos55^\circ+\cos65^\circ-\cos5^\circ$
$\displaystyle =2\cos\left(\frac{55^\circ+65^\circ}{2}\right)\cos\left(\frac{55^\circ-65^\circ}{2}\right)-\cos5^\circ$
$\displaystyle =2\cos60^\circ\cos(-5^\circ)-\cos5^\circ$
$\displaystyle =2\times\frac{1}{2}\cos5^\circ-\cos5^\circ$
$\displaystyle =0=\text{RHS. Hence proved.}$

$\displaystyle \text{(ii) LHS}=\sin50^\circ-\sin70^\circ+\sin10^\circ$
$\displaystyle =2\cos\left(\frac{50^\circ+70^\circ}{2}\right)\sin\left(\frac{50^\circ-70^\circ}{2}\right)+\sin10^\circ$
$\displaystyle =2\cos60^\circ\sin(-10^\circ)+\sin10^\circ$
$\displaystyle =-2\times\frac{1}{2}\sin10^\circ+\sin10^\circ$
$\displaystyle =0=\text{RHS. Hence proved.}$

$\displaystyle \text{(iii) LHS}=\cos80^\circ+\cos40^\circ-\cos20^\circ$
$\displaystyle =2\cos\left(\frac{80^\circ+40^\circ}{2}\right)\cos\left(\frac{80^\circ-40^\circ}{2}\right)-\cos20^\circ$
$\displaystyle =2\cos60^\circ\cos20^\circ-\cos20^\circ$
$\displaystyle =2\times\frac{1}{2}\cos20^\circ-\cos20^\circ$
$\displaystyle =0=\text{RHS. Hence proved.}$

$\displaystyle \text{(iv) LHS}=\cos20^\circ+\cos100^\circ+\cos140^\circ$
$\displaystyle =2\cos\left(\frac{20^\circ+100^\circ}{2}\right)\cos\left(\frac{20^\circ-100^\circ}{2}\right)+\cos(180^\circ-40^\circ)$
$\displaystyle =2\cos60^\circ\cos(-40^\circ)-\cos40^\circ$
$\displaystyle =2\times\frac{1}{2}\cos40^\circ-\cos40^\circ$
$\displaystyle =0=\text{RHS. Hence proved.}$

$\displaystyle \text{(v) LHS}=\sin\frac{5\pi}{18}-\cos\frac{4\pi}{9}$
$\displaystyle =\sin\frac{5\pi}{18}-\sin\left(\frac{\pi}{2}-\frac{4\pi}{9}\right)$
$\displaystyle =\sin\frac{5\pi}{18}-\sin\frac{\pi}{18}$
$\displaystyle =2\cos\left(\frac{\frac{5\pi}{18}+\frac{\pi}{18}}{2}\right)\sin\left(\frac{\frac{5\pi}{18}-\frac{\pi}{18}}{2}\right)$
$\displaystyle =2\cos\frac{\pi}{6}\sin\frac{\pi}{9}$
$\displaystyle =2\times\frac{\sqrt{3}}{2}\sin\frac{\pi}{9}$
$\displaystyle =\sqrt{3}\sin\frac{\pi}{9}=\text{RHS. Hence proved.}$

$\displaystyle \text{(vi) LHS}=\cos\frac{\pi}{12}-\sin\frac{\pi}{12}$
$\displaystyle =\cos\frac{\pi}{12}-\cos\left(\frac{\pi}{2}-\frac{\pi}{12}\right)$
$\displaystyle =\cos\frac{\pi}{12}-\cos\frac{5\pi}{12}$
$\displaystyle =-2\sin\left(\frac{\frac{\pi}{12}+\frac{5\pi}{12}}{2}\right)\sin\left(\frac{\frac{\pi}{12}-\frac{5\pi}{12}}{2}\right)$
$\displaystyle =-2\sin\frac{\pi}{4}\sin\left(-\frac{\pi}{6}\right)$
$\displaystyle =2\times\frac{1}{\sqrt{2}}\times\frac{1}{2}$
$\displaystyle =\frac{1}{\sqrt{2}}=\text{RHS. Hence proved.}$

$\displaystyle \text{(vii) LHS}=\sin80^\circ-\cos70^\circ$
$\displaystyle =\sin80^\circ-\sin20^\circ$
$\displaystyle =2\cos\left(\frac{80^\circ+20^\circ}{2}\right)\sin\left(\frac{80^\circ-20^\circ}{2}\right)$
$\displaystyle =2\cos50^\circ\sin30^\circ$
$\displaystyle =2\times\cos50^\circ\times\frac{1}{2}$
$\displaystyle =\cos50^\circ=\text{RHS. Hence proved.}$

$\displaystyle \text{(viii) LHS}=\sin51^\circ+\cos81^\circ$
$\displaystyle =\sin51^\circ+\sin9^\circ$
$\displaystyle =2\sin\left(\frac{51^\circ+9^\circ}{2}\right)\cos\left(\frac{51^\circ-9^\circ}{2}\right)$
$\displaystyle =2\sin30^\circ\cos21^\circ$
$\displaystyle =2\times\frac{1}{2}\cos21^\circ$
$\displaystyle =\cos21^\circ=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Prove that:}$
$\displaystyle \text{(i) }\cos\left(\frac{3\pi}{4}+x\right)-\cos\left(\frac{3\pi}{4}-x\right)=-\sqrt{2}\sin x$
$\displaystyle \text{(ii) }\cos\left(\frac{\pi}{4}+x\right)+\cos\left(\frac{\pi}{4}-x\right)=\sqrt{2}\cos x$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) LHS}=\cos\left(\frac{3\pi}{4}+x\right)-\cos\left(\frac{3\pi}{4}-x\right)$
$\displaystyle =-2\sin\left(\frac{\left(\frac{3\pi}{4}+x\right)+\left(\frac{3\pi}{4}-x\right)}{2}\right)$
$\displaystyle \qquad\times\sin\left(\frac{\left(\frac{3\pi}{4}+x\right)-\left(\frac{3\pi}{4}-x\right)}{2}\right)$
$\displaystyle =-2\sin\frac{3\pi}{4}\sin x$
$\displaystyle =-2\times\frac{1}{\sqrt{2}}\sin x$
$\displaystyle =-\sqrt{2}\sin x=\text{RHS. Hence proved.}$

$\displaystyle \text{(ii) LHS}=\cos\left(\frac{\pi}{4}+x\right)+\cos\left(\frac{\pi}{4}-x\right)$
$\displaystyle =2\cos\left(\frac{\left(\frac{\pi}{4}+x\right)+\left(\frac{\pi}{4}-x\right)}{2}\right)$
$\displaystyle \qquad\times\cos\left(\frac{\left(\frac{\pi}{4}+x\right)-\left(\frac{\pi}{4}-x\right)}{2}\right)$
$\displaystyle =2\cos\frac{\pi}{4}\cos x$
$\displaystyle =2\times\frac{1}{\sqrt{2}}\cos x$
$\displaystyle =\sqrt{2}\cos x=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Prove that:}$
$\displaystyle \text{(i) }\sin65^\circ+\cos65^\circ=\sqrt{2}\cos20^\circ$
$\displaystyle \text{(ii) }\sin47^\circ+\cos77^\circ=\cos17^\circ$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) LHS}=\sin65^\circ+\cos65^\circ$
$\displaystyle =\sin65^\circ+\cos(90^\circ-25^\circ)$
$\displaystyle =\sin65^\circ+\sin25^\circ$
$\displaystyle =2\sin\left(\frac{65^\circ+25^\circ}{2}\right)\cos\left(\frac{65^\circ-25^\circ}{2}\right)$
$\displaystyle =2\sin45^\circ\cos20^\circ$
$\displaystyle =2\times\frac{1}{\sqrt{2}}\cos20^\circ$
$\displaystyle =\sqrt{2}\cos20^\circ=\text{RHS. Hence proved.}$

$\displaystyle \text{(ii) LHS}=\sin47^\circ+\cos77^\circ$
$\displaystyle =\sin47^\circ+\cos(90^\circ-13^\circ)$
$\displaystyle =\sin47^\circ+\sin13^\circ$
$\displaystyle =2\sin\left(\frac{47^\circ+13^\circ}{2}\right)\cos\left(\frac{47^\circ-13^\circ}{2}\right)$
$\displaystyle =2\sin30^\circ\cos17^\circ$
$\displaystyle =2\times\frac{1}{2}\cos17^\circ$
$\displaystyle =\cos17^\circ=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Prove that:}$
$\displaystyle \text{(i) }\cos3A+\cos5A+\cos7A+\cos15A=4\cos4A\cos5A\cos6A$
$\displaystyle \text{(ii) }\cos A+\cos3A+\cos5A+\cos7A=4\cos A\cos2A\cos4A$
$\displaystyle \text{(iii) }\sin A+\sin2A+\sin4A+\sin5A$
$\displaystyle \qquad=4\cos\frac{A}{2}\cos\frac{3A}{2}\sin3A$
$\displaystyle \text{(iv) }\sin3A+\sin2A-\sin A=4\sin A\cos\frac{A}{2}\cos\frac{3A}{2}$
$\displaystyle \text{(v) }\cos20^\circ\cos100^\circ+\cos100^\circ\cos140^\circ$
$\displaystyle \qquad-\cos140^\circ\cos200^\circ=-\frac{3}{4}$
$\displaystyle \text{(vi) }\sin\frac{x}{2}\sin\frac{7x}{2}+\sin\frac{3x}{2}\sin\frac{11x}{2}=\sin2x\sin5x$
$\displaystyle \text{(vii) }\cos x\cos\frac{x}{2}-\cos3x\cos\frac{9x}{2}=\sin4x\sin\frac{7x}{2}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) LHS}=\cos3A+\cos5A+\cos7A+\cos15A$
$\displaystyle =(\cos5A+\cos3A)+(\cos15A+\cos7A)$
$\displaystyle =2\cos4A\cos A+2\cos11A\cos4A$
$\displaystyle =2\cos4A(\cos11A+\cos A)$
$\displaystyle =2\cos4A\left[2\cos\left(\frac{11A+A}{2}\right)\cos\left(\frac{11A-A}{2}\right)\right]$
$\displaystyle =2\cos4A(2\cos6A\cos5A)$
$\displaystyle =4\cos4A\cos5A\cos6A=\text{RHS. Hence proved.}$

$\displaystyle \text{(ii) LHS}=\cos A+\cos3A+\cos5A+\cos7A$
$\displaystyle =(\cos3A+\cos A)+(\cos7A+\cos5A)$
$\displaystyle =2\cos2A\cos A+2\cos6A\cos A$
$\displaystyle =2\cos A(\cos6A+\cos2A)$
$\displaystyle =2\cos A\left[2\cos\left(\frac{6A+2A}{2}\right)\cos\left(\frac{6A-2A}{2}\right)\right]$
$\displaystyle =2\cos A(2\cos4A\cos2A)$
$\displaystyle =4\cos A\cos2A\cos4A=\text{RHS. Hence proved.}$

$\displaystyle \text{(iii) LHS}=\sin A+\sin2A+\sin4A+\sin5A$
$\displaystyle =(\sin2A+\sin A)+(\sin5A+\sin4A)$
$\displaystyle =2\sin\frac{3A}{2}\cos\frac{A}{2}+2\sin\frac{9A}{2}\cos\frac{A}{2}$
$\displaystyle =2\cos\frac{A}{2}\left(\sin\frac{3A}{2}+\sin\frac{9A}{2}\right)$
$\displaystyle =2\cos\frac{A}{2}\left(2\sin3A\cos\frac{3A}{2}\right)$
$\displaystyle =4\cos\frac{A}{2}\cos\frac{3A}{2}\sin3A$
$\displaystyle =\text{RHS. Hence proved.}$

$\displaystyle \text{(iv) LHS}=\sin3A+\sin2A-\sin A$
$\displaystyle =(\sin3A-\sin A)+\sin2A$
$\displaystyle =2\cos\left(\frac{3A+A}{2}\right)\sin\left(\frac{3A-A}{2}\right)+\sin2A$
$\displaystyle =2\sin A\cos2A+2\sin A\cos A$
$\displaystyle =2\sin A(\cos2A+\cos A)$
$\displaystyle =2\sin A\left[2\cos\left(\frac{2A+A}{2}\right)\cos\left(\frac{2A-A}{2}\right)\right]$
$\displaystyle =4\sin A\cos\frac{3A}{2}\cos\frac{A}{2}$
$\displaystyle =\text{RHS. Hence proved.}$

$\displaystyle \text{(v) LHS}=\cos20^\circ\cos100^\circ+\cos100^\circ\cos140^\circ$
$\displaystyle \qquad-\cos140^\circ\cos200^\circ$
$\displaystyle =\frac{1}{2}\big[\cos120^\circ+\cos80^\circ+\cos240^\circ+\cos40^\circ$
$\displaystyle \qquad-\cos340^\circ-\cos60^\circ\big]$
$\displaystyle =\frac{1}{2}\big[-\tfrac{1}{2}+\cos80^\circ-\tfrac{1}{2}+\cos40^\circ-\cos20^\circ-\tfrac{1}{2}\big]$
$\displaystyle =\frac{1}{2}\big[\cos80^\circ+\cos40^\circ-\cos20^\circ-\tfrac{3}{2}\big]$
$\displaystyle =\frac{1}{2}\big[2\cos60^\circ\cos20^\circ-\cos20^\circ-\tfrac{3}{2}\big]$
$\displaystyle =\frac{1}{2}\big[\cos20^\circ-\cos20^\circ-\tfrac{3}{2}\big]$
$\displaystyle =-\frac{3}{4}=\text{RHS. Hence proved.}$

$\displaystyle \text{(vi) LHS}=\sin\frac{x}{2}\sin\frac{7x}{2}+\sin\frac{3x}{2}\sin\frac{11x}{2}$
$\displaystyle =\frac{1}{2}\left[\cos3x-\cos4x+\cos4x-\cos7x\right]$
$\displaystyle =\frac{1}{2}(\cos3x-\cos7x)$
$\displaystyle =-\frac{1}{2}(\cos7x-\cos3x)$
$\displaystyle =-\frac{1}{2}\left[-2\sin\left(\frac{7x+3x}{2}\right)\sin\left(\frac{7x-3x}{2}\right)\right]$
$\displaystyle =\sin5x\sin2x=\text{RHS. Hence proved.}$

$\displaystyle \text{(vii) LHS}=\cos x\cos\frac{x}{2}-\cos3x\cos\frac{9x}{2}$
$\displaystyle =\frac{1}{2}\left[\cos\frac{3x}{2}+\cos\frac{x}{2}-\cos\frac{15x}{2}-\cos\frac{3x}{2}\right]$
$\displaystyle =\frac{1}{2}\left(\cos\frac{x}{2}-\cos\frac{15x}{2}\right)$
$\displaystyle =-\sin\left(\frac{\frac{x}{2}+\frac{15x}{2}}{2}\right)$
$\displaystyle \qquad\times\sin\left(\frac{\frac{x}{2}-\frac{15x}{2}}{2}\right)$
$\displaystyle =-\sin4x\sin\left(-\frac{7x}{2}\right)$
$\displaystyle =\sin4x\sin\frac{7x}{2}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Prove that:}$
$\displaystyle \text{(i) }\frac{\sin A+\sin3A}{\cos A-\cos3A}=\cot A\qquad$
$\displaystyle \text{(ii) }\frac{\sin9A-\sin7A}{\cos7A-\cos9A}=\cot8A$
$\displaystyle \text{(iii) }\frac{\sin A-\sin B}{\cos A+\cos B}=\tan\left(\frac{A-B}{2}\right)$
$\displaystyle \text{(iv) }\frac{\sin A+\sin B}{\sin A-\sin B}$
$\displaystyle \qquad=\tan\left(\frac{A+B}{2}\right)\cot\left(\frac{A-B}{2}\right)$
$\displaystyle \text{(v) }\frac{\cos A+\cos B}{\cos B-\cos A}$
$\displaystyle \qquad=\cot\left(\frac{A+B}{2}\right)\cot\left(\frac{A-B}{2}\right)$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) LHS}=\frac{\sin A+\sin3A}{\cos A-\cos3A}$
$\displaystyle =\frac{2\sin\left(\frac{A+3A}{2}\right)\cos\left(\frac{A-3A}{2}\right)}{-2\sin\left(\frac{A+3A}{2}\right)\sin\left(\frac{A-3A}{2}\right)}$
$\displaystyle =\frac{2\sin2A\cos(-A)}{-2\sin2A\sin(-A)}$
$\displaystyle =\frac{2\sin2A\cos A}{2\sin2A\sin A}$
$\displaystyle =\frac{\cos A}{\sin A}$
$\displaystyle =\cot A=\text{RHS. Hence proved.}$

$\displaystyle \text{(ii) LHS}=\frac{\sin9A-\sin7A}{\cos7A-\cos9A}$
$\displaystyle =\frac{2\cos\left(\frac{9A+7A}{2}\right)\sin\left(\frac{9A-7A}{2}\right)}{-2\sin\left(\frac{7A+9A}{2}\right)\sin\left(\frac{7A-9A}{2}\right)}$
$\displaystyle =\frac{2\cos8A\sin A}{-2\sin8A\sin(-A)}$
$\displaystyle =\frac{2\cos8A\sin A}{2\sin8A\sin A}$
$\displaystyle =\frac{\cos8A}{\sin8A}$
$\displaystyle =\cot8A=\text{RHS. Hence proved.}$

$\displaystyle \text{(iii) LHS}=\frac{\sin A-\sin B}{\cos A+\cos B}$
$\displaystyle =\frac{2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)}{2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}$
$\displaystyle =\frac{\sin\left(\frac{A-B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)}$
$\displaystyle =\tan\left(\frac{A-B}{2}\right)=\text{RHS. Hence proved.}$

$\displaystyle \text{(iv) LHS}=\frac{\sin A+\sin B}{\sin A-\sin B}$
$\displaystyle =\frac{2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}{2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)}$
$\displaystyle =\frac{\sin\left(\frac{A+B}{2}\right)}{\cos\left(\frac{A+B}{2}\right)}$
$\displaystyle \qquad\times\frac{\cos\left(\frac{A-B}{2}\right)}{\sin\left(\frac{A-B}{2}\right)}$
$\displaystyle =\tan\left(\frac{A+B}{2}\right)\cot\left(\frac{A-B}{2}\right)$
$\displaystyle =\text{RHS. Hence proved.}$

$\displaystyle \text{(v) LHS}=\frac{\cos A+\cos B}{\cos B-\cos A}$
$\displaystyle =\frac{2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}{-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{B-A}{2}\right)}$
$\displaystyle =\frac{2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}{2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)}$
$\displaystyle =\frac{\cos\left(\frac{A+B}{2}\right)}{\sin\left(\frac{A+B}{2}\right)}$
$\displaystyle \qquad\times\frac{\cos\left(\frac{A-B}{2}\right)}{\sin\left(\frac{A-B}{2}\right)}$
$\displaystyle =\cot\left(\frac{A+B}{2}\right)\cot\left(\frac{A-B}{2}\right)$
$\displaystyle =\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Prove that:}$
$\displaystyle \text{(i) }\frac{\sin A+\sin3A+\sin5A}{\cos A+\cos3A+\cos5A}=\tan3A$
$\displaystyle \text{(ii) }\frac{\cos3A+2\cos5A+\cos7A}{\cos A+2\cos3A+\cos5A}=\frac{\cos5A}{\cos3A}$
$\displaystyle \text{(iii) }\frac{\cos4A+\cos3A+\cos2A}{\sin4A+\sin3A+\sin2A}=\cot3A$
$\displaystyle \text{(iv) }\frac{\sin3A+\sin5A+\sin7A+\sin9A}{\cos3A+\cos5A+\cos7A+\cos9A}=\tan6A$
$\displaystyle \text{(v) }\frac{\sin5A-\sin7A+\sin8A-\sin4A}{\cos4A+\cos7A-\cos5A-\cos8A}=\cot6A$
$\displaystyle \text{(vi) }\frac{\sin5A\cos2A-\sin6A\cos A}{\sin A\sin2A-\cos2A\cos3A}=\tan A$
$\displaystyle \text{(vii) }\frac{\sin11A\sin A+\sin7A\sin3A}{\cos11A\sin A+\cos7A\sin3A}=\tan8A$
$\displaystyle \text{(viii) }\frac{\sin3A\cos4A-\sin A\cos2A}{\sin4A\sin A+\cos6A\cos A}=\tan2A$
$\displaystyle \text{(ix) }\frac{\sin A\sin2A+\sin3A\sin6A}{\sin A\cos2A+\sin3A\cos6A}=\tan5A$
$\displaystyle \text{(x) }\frac{\sin A+2\sin3A+\sin5A}{\sin3A+2\sin5A+\sin7A}=\frac{\sin3A}{\sin5A}$
$\displaystyle \text{(xi) }\frac{\sin(\theta+\phi)-2\sin\theta+\sin(\theta-\phi)}{\cos(\theta+\phi)-2\cos\theta+\cos(\theta-\phi)}=\tan\theta$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) LHS}=\frac{(\sin5A+\sin A)+\sin3A}{(\cos5A+\cos A)+\cos3A}$
$\displaystyle =\frac{2\sin3A\cos2A+\sin3A}{2\cos3A\cos2A+\cos3A}$
$\displaystyle =\frac{\sin3A(2\cos2A+1)}{\cos3A(2\cos2A+1)}=\tan3A=\text{RHS. Hence proved.}$

$\displaystyle \text{(ii) LHS}=\frac{(\cos7A+\cos3A)+2\cos5A}{(\cos5A+\cos A)+2\cos3A}$
$\displaystyle =\frac{2\cos5A\cos2A+2\cos5A}{2\cos3A\cos2A+2\cos3A}$
$\displaystyle =\frac{2\cos5A(\cos2A+1)}{2\cos3A(\cos2A+1)}=\frac{\cos5A}{\cos3A}=\text{RHS. Hence proved.}$

$\displaystyle \text{(iii) LHS}=\frac{(\cos4A+\cos2A)+\cos3A}{(\sin4A+\sin2A)+\sin3A}$
$\displaystyle =\frac{2\cos3A\cos A+\cos3A}{2\sin3A\cos A+\sin3A}$
$\displaystyle =\frac{\cos3A(2\cos A+1)}{\sin3A(2\cos A+1)}=\cot3A=\text{RHS. Hence proved.}$

$\displaystyle \text{(iv) LHS}=\frac{(\sin9A+\sin3A)+(\sin7A+\sin5A)}{(\cos9A+\cos3A)+(\cos7A+\cos5A)}$
$\displaystyle =\frac{2\sin6A(\cos3A+\cos A)}{2\cos6A(\cos3A+\cos A)}=\tan6A=\text{RHS. Hence proved.}$

$\displaystyle \text{(v) LHS}=\frac{-(\sin7A-\sin5A)+(\sin8A-\sin4A)}{-(\cos7A-\cos5A)-(\cos8A-\cos4A)}$
$\displaystyle =\frac{2\cos6A(-\sin A+\sin2A)}{2\sin6A(-\sin A+\sin2A)}=\cot6A=\text{RHS. Hence proved.}$

$\displaystyle \text{(vi) LHS}=\frac{\sin3A-\sin5A}{-(\cos3A+\cos5A)}$
$\displaystyle =\frac{-(\sin5A-\sin3A)}{-(\cos5A+\cos3A)}$
$\displaystyle =\frac{2\sin A\cos4A}{2\cos4A\cos A}=\tan A=\text{RHS. Hence proved.}$

$\displaystyle \text{(vii) LHS}=\frac{\cos10A-\cos12A+\cos4A-\cos10A}{\sin12A-\sin10A+\sin10A-\sin4A}$
$\displaystyle =\frac{-(\cos12A-\cos4A)}{\sin12A-\sin4A}$
$\displaystyle =\frac{2\sin8A\sin4A}{2\sin4A\cos8A}=\tan8A=\text{RHS. Hence proved.}$

$\displaystyle \text{(viii) LHS}=\frac{\sin7A-\sin3A}{\cos3A+\cos7A}$
$\displaystyle =\frac{2\sin2A\cos5A}{2\cos5A\cos2A}=\tan2A=\text{RHS. Hence proved.}$

$\displaystyle \text{(ix) LHS}=\frac{\cos A-\cos9A}{\sin9A-\sin A}$
$\displaystyle =\frac{2\sin5A\sin4A}{2\sin4A\cos5A}=\tan5A=\text{RHS. Hence proved.}$

$\displaystyle \text{(x) LHS}=\frac{(\sin A+\sin5A)+2\sin3A}{(\sin3A+\sin7A)+2\sin5A}$
$\displaystyle =\frac{2\sin3A\cos2A+2\sin3A}{2\sin5A\cos2A+2\sin5A}$
$\displaystyle =\frac{2\sin3A(\cos2A+1)}{2\sin5A(\cos2A+1)}$
$\displaystyle =\frac{\sin3A}{\sin5A}=\text{RHS. Hence proved.}$

$\displaystyle \text{(xi) LHS}=\frac{[\sin(\theta+\phi)+\sin(\theta-\phi)]-2\sin\theta}{[\cos(\theta+\phi)+\cos(\theta-\phi)]-2\cos\theta}$
$\displaystyle =\frac{2\sin\theta\cos\phi-2\sin\theta}{2\cos\theta\cos\phi-2\cos\theta}$
$\displaystyle =\frac{2\sin\theta(\cos\phi-1)}{2\cos\theta(\cos\phi-1)}=\tan\theta=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Prove that:}$
$\displaystyle \text{(i) }\sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)$
$\displaystyle \qquad=4\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\beta+\gamma}{2}\right)\sin\left(\frac{\gamma+\alpha}{2}\right)$
$\displaystyle \text{(ii) }\cos(A+B+C)+\cos(A-B+C)+\cos(A+B-C)$
$\displaystyle \qquad+\cos(-A+B+C)=4\cos A\cos B\cos C$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) LHS}=\sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)$
$\displaystyle =(\sin\alpha+\sin\beta)+\{\sin\gamma-\sin(\alpha+\beta+\gamma)\}$
$\displaystyle =2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)$
$\displaystyle \qquad+2\cos\left(\frac{\alpha+\beta+2\gamma}{2}\right)\sin\left(-\frac{\alpha+\beta}{2}\right)$
$\displaystyle =2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)$
$\displaystyle \qquad-2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha+\beta+2\gamma}{2}\right)$
$\displaystyle =2\sin\left(\frac{\alpha+\beta}{2}\right)\left\{\cos\left(\frac{\alpha-\beta}{2}\right)-\cos\left(\frac{\alpha+\beta+2\gamma}{2}\right)\right\}$
$\displaystyle =2\sin\left(\frac{\alpha+\beta}{2}\right)\left\{-2\sin\left(\frac{\alpha+\gamma}{2}\right)\sin\left(-\frac{\beta+\gamma}{2}\right)\right\}$
$\displaystyle =4\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha+\gamma}{2}\right)\sin\left(\frac{\beta+\gamma}{2}\right)$
$\displaystyle =4\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\beta+\gamma}{2}\right)\sin\left(\frac{\gamma+\alpha}{2}\right)$
$\displaystyle =\text{RHS. Hence proved.}$

$\displaystyle \text{(ii) LHS}=\cos(A+B+C)+\cos(A-B+C)+\cos(A+B-C)$
$\displaystyle \qquad+\cos(-A+B+C)$
$\displaystyle =[\cos(A+B+C)+\cos(A-B+C)]$
$\displaystyle \qquad+[\cos(A+B-C)+\cos(-A+B+C)]$
$\displaystyle =2\cos\left(\frac{A+B+C+A-B+C}{2}\right)$
$\displaystyle \qquad\times\cos\left(\frac{A+B+C-A+B-C}{2}\right)$
$\displaystyle \qquad+2\cos\left(\frac{A+B-C-A+B+C}{2}\right)$
$\displaystyle \qquad\times\cos\left(\frac{A+B-C+A-B-C}{2}\right)$
$\displaystyle =2\cos(A+C)\cos B+2\cos B\cos(A-C)$
$\displaystyle =2\cos B[\cos(A+C)+\cos(A-C)]$
$\displaystyle =2\cos B\left[2\cos\left(\frac{A+C+A-C}{2}\right)\cos\left(\frac{A+C-A+C}{2}\right)\right]$
$\displaystyle =4\cos B\cos A\cos C$
$\displaystyle =4\cos A\cos B\cos C=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{If }\cos A+\cos B=\frac{1}{2}\text{ and }\sin A+\sin B=\frac{1}{4},$
$\displaystyle \text{prove that }\tan\left(\frac{A+B}{2}\right)=\frac{1}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\cos A+\cos B=\frac{1}{2}\text{ and }\sin A+\sin B=\frac{1}{4}.$
$\displaystyle \text{Dividing }\sin A+\sin B=\frac{1}{4}\text{ by }\cos A+\cos B=\frac{1}{2},$
$\displaystyle \frac{\sin A+\sin B}{\cos A+\cos B}=\frac{\frac{1}{4}}{\frac{1}{2}}$
$\displaystyle \frac{2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}{2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}=\frac{1}{2}$
$\displaystyle \frac{\sin\left(\frac{A+B}{2}\right)}{\cos\left(\frac{A+B}{2}\right)}=\frac{1}{2}$
$\displaystyle \therefore \tan\left(\frac{A+B}{2}\right)=\frac{1}{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{If }\mathrm{cosec}A+\sec A=\mathrm{cosec}B+\sec B,\text{ then prove that}$
$\displaystyle \tan A\tan B=\cot\left(\frac{A+B}{2}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\mathrm{cosec}A+\sec A=\mathrm{cosec}B+\sec B.$
$\displaystyle \Rightarrow \sec A-\sec B=\mathrm{cosec}B-\mathrm{cosec}A$
$\displaystyle \Rightarrow \frac{1}{\cos A}-\frac{1}{\cos B}=\frac{1}{\sin B}-\frac{1}{\sin A}$
$\displaystyle \Rightarrow \frac{\cos B-\cos A}{\cos A\cos B}=\frac{\sin A-\sin B}{\sin A\sin B}$
$\displaystyle \Rightarrow \frac{\sin A\sin B}{\cos A\cos B}=\frac{\sin A-\sin B}{\cos B-\cos A}$
$\displaystyle \Rightarrow \tan A\tan B=\frac{\sin A-\sin B}{\cos B-\cos A}$
$\displaystyle =\frac{2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)}{2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)}$
$\displaystyle =\frac{\cos\left(\frac{A+B}{2}\right)}{\sin\left(\frac{A+B}{2}\right)}$
$\displaystyle =\cot\left(\frac{A+B}{2}\right).$
$\displaystyle \therefore \tan A\tan B=\cot\left(\frac{A+B}{2}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{If }\sin2A=\lambda\sin2B,\text{ prove that}$
$\displaystyle \frac{\tan(A+B)}{\tan(A-B)}=\frac{\lambda+1}{\lambda-1}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\sin2A=\lambda\sin2B.$
$\displaystyle \Rightarrow \lambda=\frac{\sin2A}{\sin2B}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{\lambda+1}{\lambda-1}=\frac{\sin2A+\sin2B}{\sin2A-\sin2B}$
$\displaystyle =\frac{2\sin(A+B)\cos(A-B)}{2\cos(A+B)\sin(A-B)}$
$\displaystyle =\frac{\sin(A+B)}{\cos(A+B)}\times\frac{\cos(A-B)}{\sin(A-B)}$
$\displaystyle =\frac{\tan(A+B)}{\tan(A-B)}$
$\displaystyle \therefore \frac{\tan(A+B)}{\tan(A-B)}=\frac{\lambda+1}{\lambda-1}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Prove that:}$
$\displaystyle \text{(i) }\frac{\cos(A+B+C)+\cos(-A+B+C)+\cos(A-B+C)+\cos(A+B-C)}{\sin(A+B+C)+\sin(-A+B+C)+\sin(A-B+C)+\sin(-A-B+C)}=\cot C$
$\displaystyle \text{(ii) }\sin(B-C)\cos(A-D)+\sin(C-A)\cos(B-D)$
$\displaystyle \qquad+\sin(A-B)\cos(C-D)=0$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) LHS}=\frac{\cos(A+B+C)+\cos(-A+B+C)+\cos(A-B+C)+\cos(A+B-C)}{\sin(A+B+C)+\sin(-A+B+C)+\sin(A-B+C)+\sin(-A-B+C)}$
$\displaystyle =\frac{[\cos(A+B+C)+\cos(-A+B+C)]+[\cos(A-B+C)+\cos(A+B-C)]}{[\sin(A+B+C)+\sin(-A+B+C)]+[\sin(A-B+C)+\sin(-A-B+C)]}$
$\displaystyle =\frac{2\cos(B+C)\cos A+2\cos C\cos(A-B)}{2\sin(B+C)\cos A+2\sin(C-B)\cos A}$
$\displaystyle =\frac{2\cos A[\cos(B+C)+\cos(C-B)]}{2\cos A[\sin(B+C)+\sin(C-B)]}$
$\displaystyle =\frac{\cos(B+C)+\cos(C-B)}{\sin(B+C)+\sin(C-B)}$
$\displaystyle =\frac{2\cos C\cos B}{2\sin C\cos B}$
$\displaystyle =\frac{\cos C}{\sin C}$
$\displaystyle =\cot C=\text{RHS. Hence proved.}$

$\displaystyle \text{(ii) LHS}=\sin(B-C)\cos(A-D)+\sin(C-A)\cos(B-D)$
$\displaystyle \qquad+\sin(A-B)\cos(C-D)$
$\displaystyle =\frac{1}{2}\big[2\sin(B-C)\cos(A-D)+2\sin(C-A)\cos(B-D)$
$\displaystyle \qquad+2\sin(A-B)\cos(C-D)\big]$
$\displaystyle =\frac{1}{2}\big[\sin(A+B-C-D)+\sin(B+D-A-C)$
$\displaystyle \qquad+\sin(B+C-A-D)+\sin(C+D-A-B)$
$\displaystyle \qquad+\sin(A+C-B-D)+\sin(A+D-B-C)\big]$
$\displaystyle =\frac{1}{2}\big[\sin(A+B-C-D)-\sin(A+B-C-D)$
$\displaystyle \qquad+\sin(B+D-A-C)-\sin(B+D-A-C)$
$\displaystyle \qquad-\sin(A+D-B-C)+\sin(A+D-B-C)\big]$
$\displaystyle =\frac{1}{2}[0]$
$\displaystyle =0=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{If }\frac{\cos(A-B)}{\cos(A+B)}+\frac{\cos(C+D)}{\cos(C-D)}=0,\text{ prove that}$
$\displaystyle \tan A\tan B\tan C\tan D=-1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\frac{\cos(A-B)}{\cos(A+B)}+\frac{\cos(C+D)}{\cos(C-D)}=0.$
$\displaystyle \frac{\cos A\cos B+\sin A\sin B}{\cos A\cos B-\sin A\sin B}+\frac{\cos C\cos D-\sin C\sin D}{\cos C\cos D+\sin C\sin D}=0$
$\displaystyle \Rightarrow \frac{1+\tan A\tan B}{1-\tan A\tan B}+\frac{1-\tan C\tan D}{1+\tan C\tan D}=0$
$\displaystyle \Rightarrow (1+\tan A\tan B)(1+\tan C\tan D)$
$\displaystyle \qquad+(1-\tan C\tan D)(1-\tan A\tan B)=0$
$\displaystyle \Rightarrow 1+\tan A\tan B+\tan C\tan D+\tan A\tan B\tan C\tan D$
$\displaystyle \qquad+1-\tan A\tan B-\tan C\tan D+\tan A\tan B\tan C\tan D=0$
$\displaystyle \Rightarrow 2+2\tan A\tan B\tan C\tan D=0$
$\displaystyle \Rightarrow \tan A\tan B\tan C\tan D=-1.$
$\displaystyle \therefore \tan A\tan B\tan C\tan D=-1.\text{ Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{If }\cos(\alpha+\beta)\sin(\gamma+\delta)=\cos(\alpha-\beta)\sin(\gamma-\delta),$
$\displaystyle \text{then prove that }\cot\alpha\cot\beta\cot\gamma=\cot\delta.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\cos(\alpha+\beta)\sin(\gamma+\delta)=\cos(\alpha-\beta)\sin(\gamma-\delta).$
$\displaystyle \Rightarrow(\cos\alpha\cos\beta-\sin\alpha\sin\beta)(\sin\gamma\cos\delta+\cos\gamma\sin\delta)$
$\displaystyle \qquad=(\cos\alpha\cos\beta+\sin\alpha\sin\beta)(\sin\gamma\cos\delta-\cos\gamma\sin\delta)$
$\displaystyle \text{Dividing both sides by }\sin\alpha\sin\beta\sin\gamma\sin\delta,$
$\displaystyle (\cot\alpha\cot\beta-1)(\cot\delta+\cot\gamma)$
$\displaystyle \qquad=(\cot\alpha\cot\beta+1)(\cot\delta-\cot\gamma)$
$\displaystyle \Rightarrow \cot\alpha\cot\beta\cot\delta+\cot\alpha\cot\beta\cot\gamma-\cot\delta-\cot\gamma$
$\displaystyle \qquad=\cot\alpha\cot\beta\cot\delta-\cot\alpha\cot\beta\cot\gamma+\cot\delta-\cot\gamma$
$\displaystyle \Rightarrow 2\cot\alpha\cot\beta\cot\gamma=2\cot\delta$
$\displaystyle \therefore \cot\alpha\cot\beta\cot\gamma=\cot\delta.\text{ Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{If }y\sin\phi=x\sin(2\theta+\phi),\text{ prove that}$
$\displaystyle (x+y)\cot(\theta+\phi)=(y-x)\cot\theta.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }y\sin\phi=x\sin(2\theta+\phi).$
$\displaystyle \Rightarrow \frac{y}{x}=\frac{\sin(2\theta+\phi)}{\sin\phi}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{y-x}{y+x}=\frac{\sin(2\theta+\phi)-\sin\phi}{\sin(2\theta+\phi)+\sin\phi}$
$\displaystyle =\frac{2\cos\left(\frac{2\theta+2\phi}{2}\right)\sin\left(\frac{2\theta}{2}\right)}{2\sin\left(\frac{2\theta+2\phi}{2}\right)\cos\left(\frac{2\theta}{2}\right)}$
$\displaystyle =\frac{\sin\theta\cos(\theta+\phi)}{\cos\theta\sin(\theta+\phi)}$
$\displaystyle =\tan\theta\cot(\theta+\phi)$
$\displaystyle \Rightarrow \frac{y-x}{y+x}=\tan\theta\cot(\theta+\phi)$
$\displaystyle \Rightarrow \frac{y-x}{y+x}\cot\theta=\cot(\theta+\phi)$
$\displaystyle \therefore (y-x)\cot\theta=(y+x)\cot(\theta+\phi).$
$\displaystyle \therefore (x+y)\cot(\theta+\phi)=(y-x)\cot\theta.\text{ Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{If }\cos(A+B)\sin(C-D)=\cos(A-B)\sin(C+D),\text{ prove that}$
$\displaystyle \tan A\tan B\tan C+\tan D=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\cos(A+B)\sin(C-D)=\cos(A-B)\sin(C+D).$
$\displaystyle \Rightarrow(\cos A\cos B-\sin A\sin B)(\sin C\cos D-\cos C\sin D)$
$\displaystyle \qquad=(\cos A\cos B+\sin A\sin B)(\sin C\cos D+\cos C\sin D).$
$\displaystyle \text{Dividing both sides by }\cos A\cos B\cos C\cos D,$
$\displaystyle (1-\tan A\tan B)(\tan C-\tan D)=(1+\tan A\tan B)(\tan C+\tan D).$
$\displaystyle \Rightarrow \tan C-\tan D-\tan A\tan B\tan C+\tan A\tan B\tan D$
$\displaystyle \qquad=\tan C+\tan D+\tan A\tan B\tan C+\tan A\tan B\tan D.$
$\displaystyle \Rightarrow -2\tan D=2\tan A\tan B\tan C.$
$\displaystyle \Rightarrow \tan A\tan B\tan C+\tan D=0.$
$\displaystyle \therefore \tan A\tan B\tan C+\tan D=0.\text{ Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{If }x\cos\theta=y\cos\left(\theta+\frac{2\pi}{3}\right)$
$\displaystyle \qquad=z\cos\left(\theta+\frac{4\pi}{3}\right),\text{ prove that }xy+yz+zx=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }x\cos\theta=y\cos\left(\theta+\frac{2\pi}{3}\right)=z\cos\left(\theta+\frac{4\pi}{3}\right)=k.$

$\displaystyle \text{Case I: }k\ne0$
$\displaystyle \text{Then all the three cosine factors are non-zero.}$
$\displaystyle x=\frac{k}{\cos\theta},\qquad y=\frac{k}{\cos\left(\theta+\frac{2\pi}{3}\right)},\qquad z=\frac{k}{\cos\left(\theta+\frac{4\pi}{3}\right)}.$
$\displaystyle \therefore xy+yz+zx$
$\displaystyle =k^2\frac{\cos\theta+\cos\left(\theta+\frac{2\pi}{3}\right)+\cos\left(\theta+\frac{4\pi}{3}\right)}{\cos\theta\cos\left(\theta+\frac{2\pi}{3}\right)\cos\left(\theta+\frac{4\pi}{3}\right)}$
$\displaystyle \text{Now, }\cos\left(\theta+\frac{2\pi}{3}\right)+\cos\left(\theta+\frac{4\pi}{3}\right)=2\cos(\theta+\pi)\cos\left(\frac{\pi}{3}\right)$
$\displaystyle =2(-\cos\theta)\times\frac{1}{2}=-\cos\theta.$
$\displaystyle \therefore \cos\theta+\cos\left(\theta+\frac{2\pi}{3}\right)+\cos\left(\theta+\frac{4\pi}{3}\right)=0.$
$\displaystyle \therefore xy+yz+zx=0.$

$\displaystyle \text{Case II: }k=0$
$\displaystyle \text{If none of the cosine factors is zero, then }x=y=z=0.$
$\displaystyle \therefore xy+yz+zx=0.$
$\displaystyle \text{If one cosine factor is zero, the other two are non-zero, since their sum is zero.}$
$\displaystyle \text{Hence the corresponding two variables are zero, and again }xy+yz+zx=0.$
$\displaystyle \therefore xy+yz+zx=0\text{ in all cases. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{If }m\sin\theta=n\sin(\theta+2\alpha),\text{ prove that}$
$\displaystyle \tan(\theta+\alpha)\cot\alpha=\frac{m+n}{m-n}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }m\sin\theta=n\sin(\theta+2\alpha).$
$\displaystyle \Rightarrow \frac{m}{n}=\frac{\sin(\theta+2\alpha)}{\sin\theta}$
$\displaystyle \text{Applying componendo and dividendo,}$
$\displaystyle \frac{m+n}{m-n}=\frac{\sin(\theta+2\alpha)+\sin\theta}{\sin(\theta+2\alpha)-\sin\theta}$
$\displaystyle =\frac{2\sin\left(\frac{\theta+2\alpha+\theta}{2}\right)\cos\left(\frac{\theta+2\alpha-\theta}{2}\right)}{2\cos\left(\frac{\theta+2\alpha+\theta}{2}\right)\sin\left(\frac{\theta+2\alpha-\theta}{2}\right)}$
$\displaystyle =\frac{2\sin(\theta+\alpha)\cos\alpha}{2\cos(\theta+\alpha)\sin\alpha}$
$\displaystyle =\frac{\tan(\theta+\alpha)}{\tan\alpha}$
$\displaystyle =\tan(\theta+\alpha)\cot\alpha.$
$\displaystyle \therefore \tan(\theta+\alpha)\cot\alpha=\frac{m+n}{m-n}.\text{ Hence proved.}$
$\displaystyle \\$