\displaystyle \textbf{Question 1: }\text{Find the values of the following trigonometric ratios:}
\displaystyle \text{(i) }\sin\left(\frac{5\pi}{3}\right)\qquad\text{(ii) }\sin17\pi\qquad\text{(iii) }\tan\left(\frac{11\pi}{6}\right)
\displaystyle \text{(iv) }\cos\left(-\frac{25\pi}{4}\right)\qquad\text{(v) }\tan\left(\frac{7\pi}{4}\right)\qquad\text{(vi) }\sin\left(\frac{17\pi}{6}\right)
\displaystyle \text{(vii) }\cos\left(\frac{19\pi}{6}\right)\qquad\text{(viii) }\sin\left(-\frac{11\pi}{6}\right)
\displaystyle \text{(ix) }\mathrm{cosec}\left(-\frac{20\pi}{3}\right)\qquad\text{(x) }\tan\left(-\frac{13\pi}{4}\right)
\displaystyle \text{(xi) }\cos\left(\frac{19\pi}{4}\right)\qquad\text{(xii) }\sin\left(\frac{41\pi}{4}\right)
\displaystyle \text{(xiii) }\cos\left(\frac{39\pi}{4}\right)\qquad\text{(xiv) }\sin\left(\frac{151\pi}{6}\right)
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\sin\left(\frac{5\pi}{3}\right)=\sin\left(2\pi-\frac{\pi}{3}\right)
\displaystyle =-\sin\frac{\pi}{3}=-\frac{\sqrt3}{2}

\displaystyle \text{(ii) }\sin17\pi=0\qquad[\because\sin n\pi=0,\ n\in\mathbb Z]

\displaystyle \text{(iii) }\tan\left(\frac{11\pi}{6}\right)=\tan\left(2\pi-\frac{\pi}{6}\right)
\displaystyle =-\tan\frac{\pi}{6}=-\frac{1}{\sqrt3}=-\frac{\sqrt3}{3}

\displaystyle \text{(iv) }\cos\left(-\frac{25\pi}{4}\right)=\cos\left(\frac{25\pi}{4}\right)
\displaystyle =\cos\left(6\pi+\frac{\pi}{4}\right)=\cos\frac{\pi}{4}=\frac{\sqrt2}{2}

\displaystyle \text{(v) }\tan\left(\frac{7\pi}{4}\right)=\tan\left(2\pi-\frac{\pi}{4}\right)
\displaystyle =-\tan\frac{\pi}{4}=-1

\displaystyle \text{(vi) }\sin\left(\frac{17\pi}{6}\right)=\sin\left(3\pi-\frac{\pi}{6}\right)
\displaystyle =\sin\frac{\pi}{6}=\frac12

\displaystyle \text{(vii) }\cos\left(\frac{19\pi}{6}\right)=\cos\left(3\pi+\frac{\pi}{6}\right)
\displaystyle =-\cos\frac{\pi}{6}=-\frac{\sqrt3}{2}

\displaystyle \text{(viii) }\sin\left(-\frac{11\pi}{6}\right)=\sin\left(-\frac{11\pi}{6}+2\pi\right)
\displaystyle =\sin\frac{\pi}{6}=\frac12

\displaystyle \text{(ix) }\mathrm{cosec}\left(-\frac{20\pi}{3}\right)
\displaystyle =\mathrm{cosec}\left(-\frac{20\pi}{3}+6\pi\right)
\displaystyle =\mathrm{cosec}\left(-\frac{2\pi}{3}\right)
\displaystyle =-\mathrm{cosec}\frac{2\pi}{3}=-\frac{2}{\sqrt3}=-\frac{2\sqrt3}{3}

\displaystyle \text{(x) }\tan\left(-\frac{13\pi}{4}\right)=\tan\left(-\frac{13\pi}{4}+4\pi\right)
\displaystyle =\tan\frac{3\pi}{4}=-1

\displaystyle \text{(xi) }\cos\left(\frac{19\pi}{4}\right)=\cos\left(\frac{19\pi}{4}-4\pi\right)
\displaystyle =\cos\frac{3\pi}{4}=-\cos\frac{\pi}{4}=-\frac{\sqrt2}{2}

\displaystyle \text{(xii) }\sin\left(\frac{41\pi}{4}\right)=\sin\left(\frac{41\pi}{4}-10\pi\right)
\displaystyle =\sin\frac{\pi}{4}=\frac{\sqrt2}{2}

\displaystyle \text{(xiii) }\cos\left(\frac{39\pi}{4}\right)=\cos\left(\frac{39\pi}{4}-8\pi\right)
\displaystyle =\cos\frac{7\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt2}{2}

\displaystyle \text{(xiv) }\sin\left(\frac{151\pi}{6}\right)=\sin\left(\frac{151\pi}{6}-24\pi\right)
\displaystyle =\sin\frac{7\pi}{6}=-\sin\frac{\pi}{6}=-\frac12
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Prove that:}
\displaystyle \text{(i) }\tan225^\circ\cot405^\circ+\tan765^\circ\cot675^\circ=0
\displaystyle \text{(ii) }\sin\frac{8\pi}{3}\cos\frac{23\pi}{6}+\cos\frac{13\pi}{3}\sin\frac{35\pi}{6}=\frac12
\displaystyle \text{(iii) }\cos24^\circ+\cos55^\circ+\cos125^\circ+\cos204^\circ+\cos300^\circ=\frac12
\displaystyle \text{(iv) }\tan(-225^\circ)\cot(-405^\circ)-\tan(-765^\circ)\cot675^\circ=0
\displaystyle \text{(v) }\cos570^\circ\sin510^\circ+\sin(-300^\circ)\cos(-390^\circ)=\frac{3-\sqrt3}{4}
\displaystyle \text{(vi) }\tan\frac{11\pi}{3}-2\sin\frac{4\pi}{6}-\frac34\mathrm{cosec}^2\frac{\pi}{4}
\displaystyle \qquad+4\cos^2\frac{17\pi}{6}=\frac{3-4\sqrt3}{2}
\displaystyle \text{(vii) }3\sin\frac{\pi}{6}\sec\frac{\pi}{3}-4\sin\frac{5\pi}{6}\cot\frac{\pi}{4}=1
\displaystyle \text{Answer:}
\displaystyle \text{(i) LHS}=\tan225^\circ\cot405^\circ+\tan765^\circ\cot675^\circ
\displaystyle =\tan(180^\circ+45^\circ)\cot(360^\circ+45^\circ)
\displaystyle \qquad+\tan(720^\circ+45^\circ)\cot(720^\circ-45^\circ)
\displaystyle =\tan45^\circ\cot45^\circ+\tan45^\circ\cot(-45^\circ)
\displaystyle =(1)(1)+(1)(-1)
\displaystyle =0=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \text{(ii) LHS}=\sin\frac{8\pi}{3}\cos\frac{23\pi}{6}+\cos\frac{13\pi}{3}\sin\frac{35\pi}{6}
\displaystyle =\sin\left(2\pi+\frac{2\pi}{3}\right)\cos\left(4\pi-\frac{\pi}{6}\right)
\displaystyle \qquad+\cos\left(4\pi+\frac{\pi}{3}\right)\sin\left(6\pi-\frac{\pi}{6}\right)
\displaystyle =\sin\frac{2\pi}{3}\cos\frac{\pi}{6}+\cos\frac{\pi}{3}\sin\left(-\frac{\pi}{6}\right)
\displaystyle =\frac{\sqrt3}{2}\times\frac{\sqrt3}{2}+\frac12\left(-\frac12\right)
\displaystyle =\frac34-\frac14=\frac12=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \text{(iii) LHS}=\cos24^\circ+\cos55^\circ+\cos125^\circ+\cos204^\circ+\cos300^\circ
\displaystyle =(\cos24^\circ+\cos204^\circ)+(\cos55^\circ+\cos125^\circ)+\cos300^\circ
\displaystyle =[\cos24^\circ+\cos(180^\circ+24^\circ)]
\displaystyle \qquad+[\cos55^\circ+\cos(180^\circ-55^\circ)]+\cos(360^\circ-60^\circ)
\displaystyle =\cos24^\circ-\cos24^\circ+\cos55^\circ-\cos55^\circ+\cos60^\circ
\displaystyle =\frac12=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \text{(iv) LHS}=\tan(-225^\circ)\cot(-405^\circ)-\tan(-765^\circ)\cot675^\circ
\displaystyle =[-\tan225^\circ][-\cot405^\circ]+\tan765^\circ\cot675^\circ
\displaystyle =\tan45^\circ\cot45^\circ+\tan45^\circ\cot135^\circ
\displaystyle =(1)(1)+(1)(-1)
\displaystyle =0=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \text{(v) LHS}=\cos570^\circ\sin510^\circ+\sin(-300^\circ)\cos(-390^\circ)
\displaystyle =\cos210^\circ\sin150^\circ+\sin60^\circ\cos30^\circ
\displaystyle =\left(-\frac{\sqrt3}{2}\right)\left(\frac12\right)
\displaystyle \qquad+\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt3}{2}\right)
\displaystyle =-\frac{\sqrt3}{4}+\frac34
\displaystyle =\frac{3-\sqrt3}{4}=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \text{(vi) LHS}=\tan\frac{11\pi}{3}-2\sin\frac{4\pi}{6}-\frac34\mathrm{cosec}^2\frac{\pi}{4}
\displaystyle \qquad+4\cos^2\frac{17\pi}{6}
\displaystyle =\tan\left(4\pi-\frac{\pi}{3}\right)-2\sin\frac{2\pi}{3}
\displaystyle \qquad-\frac34\mathrm{cosec}^2\frac{\pi}{4}+4\cos^2\frac{5\pi}{6}
\displaystyle =-\tan\frac{\pi}{3}-2\sin\frac{\pi}{3}-\frac34(\sqrt2)^2
\displaystyle \qquad+4\left(-\frac{\sqrt3}{2}\right)^2
\displaystyle =-\sqrt3-\sqrt3-\frac32+3
\displaystyle =\frac32-2\sqrt3
\displaystyle =\frac{3-4\sqrt3}{2}=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \text{(vii) LHS}=3\sin\frac{\pi}{6}\sec\frac{\pi}{3}-4\sin\frac{5\pi}{6}\cot\frac{\pi}{4}
\displaystyle =3\left(\frac12\right)(2)-4\sin\left(\pi-\frac{\pi}{6}\right)(1)
\displaystyle =3-4\sin\frac{\pi}{6}
\displaystyle =3-4\left(\frac12\right)
\displaystyle =1=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Prove that:}
\displaystyle \text{(i) }\frac{\cos(2\pi+x)\mathrm{cosec}(2\pi+x)\tan\left(\frac{\pi}{2}+x\right)}{\sec\left(\frac{\pi}{2}+x\right)\cos x\cot(\pi+x)}=1
\displaystyle \text{(ii) }\frac{\mathrm{cosec}(90^\circ+x)+\cot(450^\circ+x)}{\mathrm{cosec}(90^\circ-x)+\tan(180^\circ-x)}
\displaystyle \qquad+\frac{\tan(180^\circ+x)+\sec(180^\circ-x)}{\tan(360^\circ+x)-\sec(-x)}=2
\displaystyle \text{(iii) }\frac{\sin(\pi+x)\cos\left(\frac{\pi}{2}+x\right)\tan\left(\frac{3\pi}{2}-x\right)\cot(2\pi-x)}{\sin(2\pi-x)\cos(2\pi+x)\mathrm{cosec}(-x)\sin\left(\frac{3\pi}{2}-x\right)}=1
\displaystyle \text{(iv) }\left\{1+\cot x-\sec\left(\frac{\pi}{2}+x\right)\right\}\left\{1+\cot x+\sec\left(\frac{\pi}{2}+x\right)\right\}=2\cot x
\displaystyle \text{(v) }\frac{\tan\left(\frac{\pi}{2}-x\right)\sec(\pi-x)\sin(-x)}{\sin(\pi+x)\cot(2\pi-x)\mathrm{cosec}\left(\frac{\pi}{2}-x\right)}=1
\displaystyle \text{Answer:}
\displaystyle \text{(i) LHS}=\frac{\cos(2\pi+x)\mathrm{cosec}(2\pi+x)\tan\left(\frac{\pi}{2}+x\right)}{\sec\left(\frac{\pi}{2}+x\right)\cos x\cot(\pi+x)}
\displaystyle =\frac{\cos x\cdot\mathrm{cosec}\,x\cdot(-\cot x)}{(-\mathrm{cosec}\,x)\cdot\cos x\cdot\cot x}
\displaystyle =1=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \text{(ii) LHS}=\frac{\mathrm{cosec}(90^\circ+x)+\cot(450^\circ+x)}{\mathrm{cosec}(90^\circ-x)+\tan(180^\circ-x)}
\displaystyle \qquad+\frac{\tan(180^\circ+x)+\sec(180^\circ-x)}{\tan(360^\circ+x)-\sec(-x)}
\displaystyle =\frac{\sec x+\cot(90^\circ+x)}{\sec x-\tan x}+\frac{\tan x-\sec x}{\tan x-\sec x}
\displaystyle =\frac{\sec x-\tan x}{\sec x-\tan x}+1
\displaystyle =1+1=2=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \text{(iii) LHS}=\frac{\sin(\pi+x)\cos\left(\frac{\pi}{2}+x\right)\tan\left(\frac{3\pi}{2}-x\right)\cot(2\pi-x)}{\sin(2\pi-x)\cos(2\pi+x)\mathrm{cosec}(-x)\sin\left(\frac{3\pi}{2}-x\right)}
\displaystyle =\frac{(-\sin x)(-\sin x)(\cot x)(-\cot x)}{(-\sin x)(\cos x)(-\mathrm{cosec}\,x)(-\cos x)}
\displaystyle =\frac{-\sin^2x\cot^2x}{-\cos^2x}
\displaystyle =\frac{\sin^2x}{\cos^2x}\cot^2x
\displaystyle =\tan^2x\cot^2x
\displaystyle =1=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \text{(iv) LHS}=\left\{1+\cot x-\sec\left(\frac{\pi}{2}+x\right)\right\}
\displaystyle \qquad\times\left\{1+\cot x+\sec\left(\frac{\pi}{2}+x\right)\right\}
\displaystyle =\left(1+\cot x+\mathrm{cosec}\,x\right)\left(1+\cot x-\mathrm{cosec}\,x\right)
\displaystyle =(1+\cot x)^2-\mathrm{cosec}^2x
\displaystyle =1+\cot^2x+2\cot x-\mathrm{cosec}^2x
\displaystyle =\mathrm{cosec}^2x+2\cot x-\mathrm{cosec}^2x
\displaystyle =2\cot x=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \text{(v) LHS}=\frac{\tan\left(\frac{\pi}{2}-x\right)\sec(\pi-x)\sin(-x)}{\sin(\pi+x)\cot(2\pi-x)\mathrm{cosec}\left(\frac{\pi}{2}-x\right)}
\displaystyle =\frac{\cot x\cdot(-\sec x)\cdot(-\sin x)}{(-\sin x)\cdot(-\cot x)\cdot\sec x}
\displaystyle =1=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Prove that}
\displaystyle \sin^2\frac{\pi}{18}+\sin^2\frac{\pi}{9}+\sin^2\frac{7\pi}{18}+\sin^2\frac{4\pi}{9}=2.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sin^2\frac{\pi}{18}+\sin^2\frac{\pi}{9}+\sin^2\frac{7\pi}{18}+\sin^2\frac{4\pi}{9}
\displaystyle =\sin^2\left(\frac{\pi}{2}-\frac{4\pi}{9}\right)+\sin^2\frac{\pi}{9}
\displaystyle \qquad+\sin^2\left(\frac{\pi}{2}-\frac{\pi}{9}\right)+\sin^2\frac{4\pi}{9}
\displaystyle =\cos^2\frac{4\pi}{9}+\sin^2\frac{4\pi}{9}
\displaystyle \qquad+\sin^2\frac{\pi}{9}+\cos^2\frac{\pi}{9}
\displaystyle =1+1
\displaystyle =2=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Prove that}
\displaystyle \sec\left(\frac{3\pi}{2}-x\right)\sec\left(x-\frac{5\pi}{2}\right)
\displaystyle \qquad+\tan\left(\frac{5\pi}{2}+x\right)\tan\left(x-\frac{3\pi}{2}\right)=-1.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sec\left(\frac{3\pi}{2}-x\right)\sec\left(x-\frac{5\pi}{2}\right)
\displaystyle \qquad+\tan\left(\frac{5\pi}{2}+x\right)\tan\left(x-\frac{3\pi}{2}\right)
\displaystyle =(-\mathrm{cosec}\,x)(\mathrm{cosec}\,x)+(-\cot x)(-\cot x)
\displaystyle =-\mathrm{cosec}^2x+\cot^2x
\displaystyle =-(1+\cot^2x)+\cot^2x
\displaystyle =-1
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In a }\triangle ABC,\text{ prove that:}
\displaystyle \text{(i) }\cos(A+B)+\cos C=0
\displaystyle \text{(ii) }\cos\left(\frac{A+B}{2}\right)=\sin\left(\frac{C}{2}\right)
\displaystyle \text{(iii) }\tan\left(\frac{A+B}{2}\right)=\cot\left(\frac{C}{2}\right)
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABC,\ A+B+C=180^\circ.
\displaystyle \therefore A+B=180^\circ-C.
\displaystyle \text{(i) LHS}=\cos(A+B)+\cos C
\displaystyle =\cos(180^\circ-C)+\cos C
\displaystyle =-\cos C+\cos C
\displaystyle =0=\text{RHS. Hence proved.}
\displaystyle \text{(ii) LHS}=\cos\left(\frac{A+B}{2}\right)
\displaystyle =\cos\left(\frac{180^\circ-C}{2}\right)
\displaystyle =\cos\left(90^\circ-\frac{C}{2}\right)
\displaystyle =\sin\frac{C}{2}=\text{RHS. Hence proved.}
\displaystyle \text{(iii) LHS}=\tan\left(\frac{A+B}{2}\right)
\displaystyle =\tan\left(\frac{180^\circ-C}{2}\right)
\displaystyle =\tan\left(90^\circ-\frac{C}{2}\right)
\displaystyle =\cot\frac{C}{2}=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }A,\ B,\ C,\ D\text{ are the angles of a cyclic quadrilateral, taken in order,}
\displaystyle \text{prove that }\cos(180^\circ-A)+\cos(180^\circ+B)+\cos(180^\circ+C)
\displaystyle \qquad-\sin(90^\circ+D)=0.
\displaystyle \text{Answer:}
\displaystyle \text{In a cyclic quadrilateral, opposite angles are supplementary.}
\displaystyle \therefore A+C=180^\circ\text{ and }B+D=180^\circ.
\displaystyle \therefore 180^\circ-A=C\text{ and }180^\circ-B=D.
\displaystyle \text{LHS}=\cos(180^\circ-A)+\cos(180^\circ+B)
\displaystyle \qquad+\cos(180^\circ+C)-\sin(90^\circ+D)
\displaystyle =\cos C-\cos B-\cos C-\cos D
\displaystyle =\cos C+\cos D-\cos C-\cos D
\displaystyle \qquad[\because\cos D=\cos(180^\circ-B)=-\cos B]
\displaystyle =0=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find }x\text{ from the following equations:}
\displaystyle \text{(i) }\mathrm{cosec}\left(\frac{\pi}{2}+\theta\right)+x\cos\theta\cot\left(\frac{\pi}{2}+\theta\right)
\displaystyle \qquad=\sin\left(\frac{\pi}{2}+\theta\right)
\displaystyle \text{(ii) }x\cot\left(\frac{\pi}{2}+\theta\right)+\tan\left(\frac{\pi}{2}+\theta\right)\sin\theta
\displaystyle \qquad+\mathrm{cosec}\left(\frac{\pi}{2}+\theta\right)=0
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }\mathrm{cosec}\left(\frac{\pi}{2}+\theta\right)+x\cos\theta\cot\left(\frac{\pi}{2}+\theta\right)
\displaystyle \qquad=\sin\left(\frac{\pi}{2}+\theta\right)
\displaystyle \sec\theta+x\cos\theta(-\tan\theta)=\cos\theta
\displaystyle \frac{1}{\cos\theta}-x\sin\theta=\cos\theta
\displaystyle \frac{1-\cos^2\theta}{\cos\theta}=x\sin\theta
\displaystyle \frac{\sin^2\theta}{\cos\theta}=x\sin\theta
\displaystyle \sin\theta(\tan\theta-x)=0
\displaystyle \therefore \sin\theta=0\text{ or }x=\tan\theta.
\displaystyle \text{If }\sin\theta\neq0,\text{ then }x=\tan\theta.
\displaystyle \text{If }\sin\theta=0,\text{ the equation is true for every real value of }x.
\displaystyle \\

\displaystyle \text{(ii) Given, }x\cot\left(\frac{\pi}{2}+\theta\right)+\tan\left(\frac{\pi}{2}+\theta\right)\sin\theta
\displaystyle \qquad+\mathrm{cosec}\left(\frac{\pi}{2}+\theta\right)=0
\displaystyle -x\tan\theta-\cot\theta\sin\theta+\sec\theta=0
\displaystyle -x\tan\theta-\cos\theta+\sec\theta=0
\displaystyle x\tan\theta=\sec\theta-\cos\theta
\displaystyle =\frac{1}{\cos\theta}-\cos\theta
\displaystyle =\frac{1-\cos^2\theta}{\cos\theta}
\displaystyle =\frac{\sin^2\theta}{\cos\theta}
\displaystyle \therefore x=\frac{\sin^2\theta}{\cos\theta}\times\frac{\cos\theta}{\sin\theta}
\displaystyle =\sin\theta
\displaystyle \therefore x=\sin\theta.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Prove that:}
\displaystyle \text{(i) }\tan4\pi-\cos\frac{3\pi}{2}-\sin\frac{5\pi}{6}\cos\frac{2\pi}{3}=\frac14
\displaystyle \text{(ii) }\sin\frac{13\pi}{3}\sin\frac{8\pi}{3}+\cos\frac{2\pi}{3}\sin\frac{5\pi}{6}=\frac12
\displaystyle \text{(iii) }\sin\frac{13\pi}{3}\sin\frac{2\pi}{3}+\cos\frac{4\pi}{3}\sin\frac{13\pi}{6}=\frac12
\displaystyle \text{(iv) }\sin\frac{10\pi}{3}\cos\frac{13\pi}{6}+\cos\frac{8\pi}{3}\sin\frac{5\pi}{6}=-1
\displaystyle \text{(v) }\tan\frac{5\pi}{4}\cot\frac{9\pi}{4}+\tan\frac{17\pi}{4}\cot\frac{15\pi}{4}=0
\displaystyle \text{Answer:}
\displaystyle \text{(i) LHS}=\tan4\pi-\cos\frac{3\pi}{2}-\sin\frac{5\pi}{6}\cos\frac{2\pi}{3}
\displaystyle =0-0-\sin\left(\pi-\frac{\pi}{6}\right)\cos\left(\pi-\frac{\pi}{3}\right)
\displaystyle =-\sin\frac{\pi}{6}\left(-\cos\frac{\pi}{3}\right)
\displaystyle =\frac12\times\frac12
\displaystyle =\frac14=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \text{(ii) LHS}=\sin\frac{13\pi}{3}\sin\frac{8\pi}{3}+\cos\frac{2\pi}{3}\sin\frac{5\pi}{6}
\displaystyle =\sin\left(4\pi+\frac{\pi}{3}\right)\sin\left(2\pi+\frac{2\pi}{3}\right)
\displaystyle \qquad+\cos\left(\pi-\frac{\pi}{3}\right)\sin\left(\pi-\frac{\pi}{6}\right)
\displaystyle =\sin\frac{\pi}{3}\sin\frac{2\pi}{3}-\cos\frac{\pi}{3}\sin\frac{\pi}{6}
\displaystyle =\frac{\sqrt3}{2}\times\frac{\sqrt3}{2}-\frac12\times\frac12
\displaystyle =\frac34-\frac14
\displaystyle =\frac12=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \text{(iii) LHS}=\sin\frac{13\pi}{3}\sin\frac{2\pi}{3}+\cos\frac{4\pi}{3}\sin\frac{13\pi}{6}
\displaystyle =\sin\left(4\pi+\frac{\pi}{3}\right)\sin\left(\pi-\frac{\pi}{3}\right)
\displaystyle \qquad+\cos\left(\pi+\frac{\pi}{3}\right)\sin\left(2\pi+\frac{\pi}{6}\right)
\displaystyle =\sin\frac{\pi}{3}\sin\frac{\pi}{3}-\cos\frac{\pi}{3}\sin\frac{\pi}{6}
\displaystyle =\left(\frac{\sqrt3}{2}\right)^2-\frac12\times\frac12
\displaystyle =\frac34-\frac14
\displaystyle =\frac12=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \text{(iv) LHS}=\sin\frac{10\pi}{3}\cos\frac{13\pi}{6}+\cos\frac{8\pi}{3}\sin\frac{5\pi}{6}
\displaystyle =\sin\left(2\pi+\frac{4\pi}{3}\right)\cos\left(2\pi+\frac{\pi}{6}\right)
\displaystyle \qquad+\cos\left(2\pi+\frac{2\pi}{3}\right)\sin\left(\pi-\frac{\pi}{6}\right)
\displaystyle =\sin\frac{4\pi}{3}\cos\frac{\pi}{6}+\cos\frac{2\pi}{3}\sin\frac{\pi}{6}
\displaystyle =\left(-\frac{\sqrt3}{2}\right)\left(\frac{\sqrt3}{2}\right)+\left(-\frac12\right)\left(\frac12\right)
\displaystyle =-\frac34-\frac14
\displaystyle =-1=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \text{(v) LHS}=\tan\frac{5\pi}{4}\cot\frac{9\pi}{4}+\tan\frac{17\pi}{4}\cot\frac{15\pi}{4}
\displaystyle =\tan\left(\pi+\frac{\pi}{4}\right)\cot\left(2\pi+\frac{\pi}{4}\right)
\displaystyle \qquad+\tan\left(4\pi+\frac{\pi}{4}\right)\cot\left(4\pi-\frac{\pi}{4}\right)
\displaystyle =\tan\frac{\pi}{4}\cot\frac{\pi}{4}+\tan\frac{\pi}{4}\cot\left(-\frac{\pi}{4}\right)
\displaystyle =(1)(1)+(1)(-1)
\displaystyle =0=\text{RHS. Hence proved.}
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.