\displaystyle \textbf{Question 1: }\text{If }\sin A=\frac{4}{5}\text{ and }\cos B=\frac{5}{13},\text{ where }0<A,B<\frac{\pi}{2},
\displaystyle \text{find the values of the following:}
\displaystyle \text{(i) }\sin(A+B)\qquad\text{(ii) }\cos(A+B)\qquad\text{(iii) }\sin(A-B)\qquad\text{(iv) }\cos(A-B)
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\sin A=\frac{4}{5}\text{ and }\cos B=\frac{5}{13},\text{ where }0<A,B<\frac{\pi}{2}.
\displaystyle \therefore \cos A=\sqrt{1-\sin^2A}=\sqrt{1-\left(\frac45\right)^2}=\frac35.
\displaystyle \text{Also, }\sin B=\sqrt{1-\cos^2B}=\sqrt{1-\left(\frac5{13}\right)^2}=\frac{12}{13}.
\displaystyle \text{(i) }\sin(A+B)=\sin A\cos B+\cos A\sin B
\displaystyle =\frac45\times\frac5{13}+\frac35\times\frac{12}{13}=\frac{20+36}{65}=\frac{56}{65}.
\displaystyle \text{(ii) }\cos(A+B)=\cos A\cos B-\sin A\sin B
\displaystyle =\frac35\times\frac5{13}-\frac45\times\frac{12}{13}=\frac{15-48}{65}=-\frac{33}{65}.
\displaystyle \text{(iii) }\sin(A-B)=\sin A\cos B-\cos A\sin B
\displaystyle =\frac45\times\frac5{13}-\frac35\times\frac{12}{13}=\frac{20-36}{65}=-\frac{16}{65}.
\displaystyle \text{(iv) }\cos(A-B)=\cos A\cos B+\sin A\sin B
\displaystyle =\frac35\times\frac5{13}+\frac45\times\frac{12}{13}=\frac{15+48}{65}=\frac{63}{65}.
\displaystyle \\

\displaystyle \textbf{Question 2:} \text{(a) If }\sin A=\frac{12}{13}\text{ and }\sin B=\frac45,\text{ where }\frac{\pi}{2}<A<\pi,\;0<B<\frac{\pi}{2},
\displaystyle \text{find: (i) }\sin(A+B)\qquad\text{(ii) }\cos(A+B).
\displaystyle \text{(b) If }\sin A=\frac35\text{ and }\cos B=-\frac{12}{13},\text{ where }A\text{ and }B\text{ lie in Q II,}
\displaystyle \text{find }\sin(A+B).
\displaystyle \textbf{Answer:}
\displaystyle \text{(a) Given }\sin A=\frac{12}{13},\;\sin B=\frac45,\;\frac{\pi}{2}<A<\pi,\;0<B<\frac{\pi}{2}.
\displaystyle \therefore \cos A=-\sqrt{1-\sin^2A}=-\sqrt{1-\left(\frac{12}{13}\right)^2}=-\sqrt{\frac{25}{169}}=-\frac{5}{13}.
\displaystyle \text{Also, }\cos B=\sqrt{1-\sin^2B}=\sqrt{1-\left(\frac45\right)^2}=\frac35.
\displaystyle \text{(i) }\sin(A+B)=\sin A\cos B+\cos A\sin B
\displaystyle =\frac{12}{13}\times\frac35+\left(-\frac{5}{13}\right)\times\frac45=\frac{36-20}{65}=\frac{16}{65}.
\displaystyle \text{(ii) }\cos(A+B)=\cos A\cos B-\sin A\sin B
\displaystyle =\left(-\frac{5}{13}\right)\times\frac35-\frac{12}{13}\times\frac45=\frac{-15-48}{65}=-\frac{63}{65}.
\displaystyle \text{(b) Given }\sin A=\frac35,\;\cos B=-\frac{12}{13},\text{ where }A\text{ and }B\text{ lie in Q II.}
\displaystyle \therefore \cos A=-\sqrt{1-\sin^2A}=-\sqrt{1-\left(\frac35\right)^2}=-\frac45.
\displaystyle \text{Also, }\sin B=\sqrt{1-\cos^2B}=\sqrt{1-\left(-\frac{12}{13}\right)^2}=\frac{5}{13}.
\displaystyle \sin(A+B)=\sin A\cos B+\cos A\sin B
\displaystyle =\frac35\times\left(-\frac{12}{13}\right)+\left(-\frac45\right)\times\frac5{13}=\frac{-36-20}{65}=-\frac{56}{65}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }\cos A=-\frac{24}{25}\text{ and }\cos B=\frac35,\text{ where }\pi<A<\frac{3\pi}{2}
\displaystyle \text{and }\frac{3\pi}{2}<B<2\pi,\text{ find: (i) }\sin(A+B)\qquad\text{(ii) }\cos(A+B).
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\cos A=-\frac{24}{25},\;\cos B=\frac35,\;\pi<A<\frac{3\pi}{2},\;\frac{3\pi}{2}<B<2\pi.
\displaystyle \therefore A\text{ lies in Quadrant III and }B\text{ lies in Quadrant IV.}
\displaystyle \sin A=-\sqrt{1-\cos^2A}=-\sqrt{1-\left(-\frac{24}{25}\right)^2}=-\sqrt{\frac{49}{625}}=-\frac7{25}.
\displaystyle \text{Also, }\sin B=-\sqrt{1-\cos^2B}=-\sqrt{1-\left(\frac35\right)^2}=-\sqrt{\frac{16}{25}}=-\frac45.
\displaystyle \text{(i) }\sin(A+B)=\sin A\cos B+\cos A\sin B
\displaystyle =\left(-\frac7{25}\right)\times\frac35+\left(-\frac{24}{25}\right)\times\left(-\frac45\right)
\displaystyle =\frac{-21+96}{125}=\frac{75}{125}=\frac35.
\displaystyle \text{(ii) }\cos(A+B)=\cos A\cos B-\sin A\sin B
\displaystyle =\left(-\frac{24}{25}\right)\times\frac35-\left(-\frac7{25}\right)\times\left(-\frac45\right)
\displaystyle =\frac{-72-28}{125}=-\frac{100}{125}=-\frac45.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }\tan A=\frac34\text{ and }\cos B=\frac9{41},\text{ where }\pi<A<\frac{3\pi}{2}
\displaystyle \text{and }0<B<\frac{\pi}{2},\text{ find }\tan(A+B).
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\tan A=\frac34,\;\cos B=\frac9{41},\;\pi<A<\frac{3\pi}{2},\;0<B<\frac{\pi}{2}.
\displaystyle \therefore A\text{ lies in Quadrant III and }B\text{ lies in Quadrant I.}
\displaystyle \sin B=\sqrt{1-\cos^2B}=\sqrt{1-\left(\frac9{41}\right)^2}=\sqrt{\frac{1600}{1681}}=\frac{40}{41}.
\displaystyle \therefore \tan B=\frac{\sin B}{\cos B}=\frac{\frac{40}{41}}{\frac9{41}}=\frac{40}{9}.
\displaystyle \therefore \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}
\displaystyle =\frac{\frac34+\frac{40}{9}}{1-\frac34\times\frac{40}{9}}=\frac{\frac{187}{36}}{-\frac73}
\displaystyle =\frac{187}{36}\times\left(-\frac37\right)=-\frac{187}{84}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }\sin A=\frac12\text{ and }\cos B=\frac{12}{13},\text{ where }\frac{\pi}{2}<A<\pi
\displaystyle \text{and }\frac{3\pi}{2}<B<2\pi,\text{ find }\tan(A-B).
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\sin A=\frac12,\;\cos B=\frac{12}{13},\;\frac{\pi}{2}<A<\pi,\;\frac{3\pi}{2}<B<2\pi.
\displaystyle \therefore A\text{ lies in Quadrant II and }B\text{ lies in Quadrant IV.}
\displaystyle \cos A=-\sqrt{1-\sin^2A}=-\sqrt{1-\left(\frac12\right)^2}=-\sqrt{\frac34}=-\frac{\sqrt3}{2}.
\displaystyle \text{Also, }\sin B=-\sqrt{1-\cos^2B}=-\sqrt{1-\left(\frac{12}{13}\right)^2}=-\frac5{13}.
\displaystyle \therefore \tan A=\frac{\sin A}{\cos A}=\frac{\frac12}{-\frac{\sqrt3}{2}}=-\frac1{\sqrt3}.
\displaystyle \text{Also, }\tan B=\frac{\sin B}{\cos B}=\frac{-\frac5{13}}{\frac{12}{13}}=-\frac5{12}.
\displaystyle \therefore \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}
\displaystyle =\frac{-\frac1{\sqrt3}-\left(-\frac5{12}\right)}{1+\left(-\frac1{\sqrt3}\right)\left(-\frac5{12}\right)}
\displaystyle =\frac{-12+5\sqrt3}{12\sqrt3+5}=\frac{5\sqrt3-12}{12\sqrt3+5}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }\sin A=\frac12\text{ and }\cos B=\frac{\sqrt3}{2},\text{ where }\frac{\pi}{2}<A<\pi
\displaystyle \text{and }0<B<\frac{\pi}{2},\text{ find: (i) }\tan(A+B)\qquad\text{(ii) }\tan(A-B).
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\sin A=\frac12,\;\cos B=\frac{\sqrt3}{2},\;\frac{\pi}{2}<A<\pi,\;0<B<\frac{\pi}{2}.
\displaystyle \therefore A\text{ lies in Quadrant II and }B\text{ lies in Quadrant I.}
\displaystyle \cos A=-\sqrt{1-\sin^2A}=-\sqrt{1-\left(\frac12\right)^2}=-\sqrt{\frac34}=-\frac{\sqrt3}{2}.
\displaystyle \text{Also, }\sin B=\sqrt{1-\cos^2B}=\sqrt{1-\left(\frac{\sqrt3}{2}\right)^2}=\sqrt{\frac14}=\frac12.
\displaystyle \therefore \tan A=\frac{\sin A}{\cos A}=\frac{\frac12}{-\frac{\sqrt3}{2}}=-\frac1{\sqrt3}.
\displaystyle \text{Also, }\tan B=\frac{\sin B}{\cos B}=\frac{\frac12}{\frac{\sqrt3}{2}}=\frac1{\sqrt3}.
\displaystyle \text{(i) }\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}
\displaystyle =\frac{-\frac1{\sqrt3}+\frac1{\sqrt3}}{1-\left(-\frac1{\sqrt3}\right)\left(\frac1{\sqrt3}\right)}=\frac0{1+\frac13}=0.
\displaystyle \text{(ii) }\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}
\displaystyle =\frac{-\frac1{\sqrt3}-\frac1{\sqrt3}}{1+\left(-\frac1{\sqrt3}\right)\left(\frac1{\sqrt3}\right)}=\frac{-\frac2{\sqrt3}}{\frac23}=-\sqrt3.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Evaluate the following:}
\displaystyle \text{(i) }\sin78^\circ\cos18^\circ-\cos78^\circ\sin18^\circ\qquad\text{(ii) }\cos47^\circ\cos13^\circ-\sin47^\circ\sin13^\circ
\displaystyle \text{(iii) }\sin36^\circ\cos9^\circ+\cos36^\circ\sin9^\circ\qquad\text{(iv) }\cos80^\circ\cos20^\circ+\sin80^\circ\sin20^\circ
\displaystyle \textbf{Answer:}
\displaystyle \text{(i) }\sin78^\circ\cos18^\circ-\cos78^\circ\sin18^\circ=\sin(78^\circ-18^\circ)=\sin60^\circ=\frac{\sqrt3}{2}.
\displaystyle \text{(ii) }\cos47^\circ\cos13^\circ-\sin47^\circ\sin13^\circ=\cos(47^\circ+13^\circ)=\cos60^\circ=\frac12.
\displaystyle \text{(iii) }\sin36^\circ\cos9^\circ+\cos36^\circ\sin9^\circ=\sin(36^\circ+9^\circ)=\sin45^\circ=\frac1{\sqrt2}.
\displaystyle \text{(iv) }\cos80^\circ\cos20^\circ+\sin80^\circ\sin20^\circ=\cos(80^\circ-20^\circ)=\cos60^\circ=\frac12.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }\cos A=-\frac{12}{13}\text{ and }\cot B=\frac{24}{7},\text{ where }A\text{ lies in}
\displaystyle \text{the second quadrant and }B\text{ lies in the third quadrant, find the following:}
\displaystyle \text{(i) }\sin(A+B)\qquad\text{(ii) }\cos(A+B)\qquad\text{(iii) }\tan(A+B)
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\cos A=-\frac{12}{13}\text{ and }\cot B=\frac{24}{7}.
\displaystyle \text{Since }A\text{ lies in Quadrant II, }\sin A>0.
\displaystyle \therefore \sin A=\sqrt{1-\cos^2A}=\sqrt{1-\left(-\frac{12}{13}\right)^2}=\sqrt{\frac{25}{169}}=\frac5{13}.
\displaystyle \text{Since }B\text{ lies in Quadrant III, }\sin B<0\text{ and }\cos B<0.
\displaystyle \sin B=-\frac1{\sqrt{1+\cot^2B}}=-\frac1{\sqrt{1+\left(\frac{24}{7}\right)^2}}=-\frac7{25}.
\displaystyle \cos B=\cot B\sin B=\frac{24}{7}\times\left(-\frac7{25}\right)=-\frac{24}{25}.
\displaystyle \text{(i) }\sin(A+B)=\sin A\cos B+\cos A\sin B
\displaystyle =\frac5{13}\times\left(-\frac{24}{25}\right)+\left(-\frac{12}{13}\right)\times\left(-\frac7{25}\right)
\displaystyle =\frac{-120+84}{325}=-\frac{36}{325}.
\displaystyle \text{(ii) }\cos(A+B)=\cos A\cos B-\sin A\sin B
\displaystyle =\left(-\frac{12}{13}\right)\times\left(-\frac{24}{25}\right)-\frac5{13}\times\left(-\frac7{25}\right)
\displaystyle =\frac{288+35}{325}=\frac{323}{325}.
\displaystyle \text{(iii) }\tan(A+B)=\frac{\sin(A+B)}{\cos(A+B)}
\displaystyle =\frac{-\frac{36}{325}}{\frac{323}{325}}=-\frac{36}{323}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Prove that }\cos\frac{7\pi}{12}+\cos\frac{\pi}{12}=\sin\frac{5\pi}{12}-\sin\frac{\pi}{12}.
\displaystyle \textbf{Answer:}
\displaystyle \text{LHS}=\cos\frac{7\pi}{12}+\cos\frac{\pi}{12}
\displaystyle =\cos\left(\frac{\pi}{2}+\frac{\pi}{12}\right)+\cos\left(\frac{\pi}{2}-\frac{5\pi}{12}\right)
\displaystyle =-\sin\frac{\pi}{12}+\sin\frac{5\pi}{12}
\displaystyle =\sin\frac{5\pi}{12}-\sin\frac{\pi}{12}=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Prove that }\frac{\tan A+\tan B}{\tan A-\tan B}=\frac{\sin(A+B)}{\sin(A-B)}.
\displaystyle \textbf{Answer:}
\displaystyle \text{LHS}=\frac{\tan A+\tan B}{\tan A-\tan B}
\displaystyle =\frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}}
\displaystyle =\frac{\sin A\cos B+\cos A\sin B}{\sin A\cos B-\cos A\sin B}
\displaystyle =\frac{\sin(A+B)}{\sin(A-B)}=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Prove the following:}
\displaystyle \text{(i) }\frac{\cos11^\circ+\sin11^\circ}{\cos11^\circ-\sin11^\circ}=\tan56^\circ
\displaystyle \text{(ii) }\frac{\cos9^\circ+\sin9^\circ}{\cos9^\circ-\sin9^\circ}=\tan54^\circ
\displaystyle \text{(iii) }\frac{\cos8^\circ-\sin8^\circ}{\cos8^\circ+\sin8^\circ}=\tan37^\circ
\displaystyle \textbf{Answer:}
\displaystyle \text{(i) }\frac{\cos11^\circ+\sin11^\circ}{\cos11^\circ-\sin11^\circ}
\displaystyle =\frac{1+\tan11^\circ}{1-\tan11^\circ}
\displaystyle =\frac{\tan45^\circ+\tan11^\circ}{1-\tan45^\circ\tan11^\circ}
\displaystyle =\tan(45^\circ+11^\circ)=\tan56^\circ.
\displaystyle \text{Hence proved.}
\displaystyle \text{(ii) }\frac{\cos9^\circ+\sin9^\circ}{\cos9^\circ-\sin9^\circ}
\displaystyle =\frac{1+\tan9^\circ}{1-\tan9^\circ}
\displaystyle =\frac{\tan45^\circ+\tan9^\circ}{1-\tan45^\circ\tan9^\circ}
\displaystyle =\tan(45^\circ+9^\circ)=\tan54^\circ.
\displaystyle \text{Hence proved.}
\displaystyle \text{(iii) }\frac{\cos8^\circ-\sin8^\circ}{\cos8^\circ+\sin8^\circ}
\displaystyle =\frac{1-\tan8^\circ}{1+\tan8^\circ}
\displaystyle =\frac{\tan45^\circ-\tan8^\circ}{1+\tan45^\circ\tan8^\circ}
\displaystyle =\tan(45^\circ-8^\circ)=\tan37^\circ.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Prove the following:}
\displaystyle \text{(i) }\sin\left(\frac{\pi}{3}-x\right)\cos\left(\frac{\pi}{6}+x\right)+\cos\left(\frac{\pi}{3}-x\right)\sin\left(\frac{\pi}{6}+x\right)=1
\displaystyle \text{(ii) }\sin\left(\frac{4\pi}{9}+7\right)\cos\left(\frac{\pi}{9}+7\right)-\cos\left(\frac{4\pi}{9}+7\right)\sin\left(\frac{\pi}{9}+7\right)=\frac{\sqrt3}{2}
\displaystyle \text{(iii) }\sin\left(\frac{3\pi}{8}-5\right)\cos\left(\frac{\pi}{8}+5\right)+\cos\left(\frac{3\pi}{8}-5\right)\sin\left(\frac{\pi}{8}+5\right)=1
\displaystyle \textbf{Answer:}
\displaystyle \text{(i) LHS}=\sin\left(\frac{\pi}{3}-x\right)\cos\left(\frac{\pi}{6}+x\right)+\cos\left(\frac{\pi}{3}-x\right)\sin\left(\frac{\pi}{6}+x\right)
\displaystyle =\sin\left[\left(\frac{\pi}{3}-x\right)+\left(\frac{\pi}{6}+x\right)\right]
\displaystyle =\sin\left(\frac{\pi}{3}+\frac{\pi}{6}\right)=\sin\frac{\pi}{2}=1=\text{RHS. Hence proved.}
\displaystyle \text{(ii) LHS}=\sin\left(\frac{4\pi}{9}+7\right)\cos\left(\frac{\pi}{9}+7\right)-\cos\left(\frac{4\pi}{9}+7\right)\sin\left(\frac{\pi}{9}+7\right)
\displaystyle =\sin\left[\left(\frac{4\pi}{9}+7\right)-\left(\frac{\pi}{9}+7\right)\right]
\displaystyle =\sin\left(\frac{4\pi}{9}-\frac{\pi}{9}\right)=\sin\frac{3\pi}{9}
\displaystyle =\sin\frac{\pi}{3}=\frac{\sqrt3}{2}=\text{RHS. Hence proved.}
\displaystyle \text{(iii) LHS}=\sin\left(\frac{3\pi}{8}-5\right)\cos\left(\frac{\pi}{8}+5\right)+\cos\left(\frac{3\pi}{8}-5\right)\sin\left(\frac{\pi}{8}+5\right)
\displaystyle =\sin\left[\left(\frac{3\pi}{8}-5\right)+\left(\frac{\pi}{8}+5\right)\right]
\displaystyle =\sin\left(\frac{3\pi}{8}+\frac{\pi}{8}\right)=\sin\frac{\pi}{2}=1=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Prove that }\frac{\tan69^\circ+\tan66^\circ}{1-\tan69^\circ\tan66^\circ}=-1.
\displaystyle \textbf{Answer:}
\displaystyle \text{LHS}=\frac{\tan69^\circ+\tan66^\circ}{1-\tan69^\circ\tan66^\circ}
\displaystyle =\tan(69^\circ+66^\circ)
\displaystyle =\tan135^\circ=-1=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{(i) If }\tan A=\frac56\text{ and }\tan B=\frac1{11},\text{ where }A\text{ and }B\text{ are acute angles,}
\displaystyle \text{prove that }A+B=\frac{\pi}{4}.
\displaystyle \text{(ii) If }\tan A=\frac{m}{m-1}\text{ and }\tan B=\frac1{2m-1},\text{ where }m\ne1,\frac12
\displaystyle \text{and }-\frac{\pi}{2}<A-B<\frac{\pi}{2},\text{ prove that }A-B=\frac{\pi}{4}.
\displaystyle \textbf{Answer:}
\displaystyle \text{(i) }\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}
\displaystyle =\frac{\frac56+\frac1{11}}{1-\frac56\times\frac1{11}}
\displaystyle =\frac{55+6}{66-5}=\frac{61}{61}=1=\tan\frac{\pi}{4}.
\displaystyle \text{Since }A\text{ and }B\text{ are acute and }\tan(A+B)>0,\text{ we have }0<A+B<\frac{\pi}{2}.
\displaystyle \therefore A+B=\frac{\pi}{4}.
\displaystyle \text{(ii) }\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}
\displaystyle =\frac{\frac{m}{m-1}-\frac1{2m-1}}{1+\frac{m}{m-1}\times\frac1{2m-1}}
\displaystyle =\frac{m(2m-1)-(m-1)}{(m-1)(2m-1)+m}
\displaystyle =\frac{2m^2-2m+1}{2m^2-2m+1}=1=\tan\frac{\pi}{4}.
\displaystyle \text{Since }-\frac{\pi}{2}<A-B<\frac{\pi}{2},
\displaystyle \therefore A-B=\frac{\pi}{4}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Prove that:}
\displaystyle \text{(i) }\cos^2\frac{\pi}{4}-\sin^2\frac{\pi}{12}=\frac{\sqrt3}{4}
\displaystyle \text{(ii) }\sin^2(n+1)A-\sin^2nA=\sin(2n+1)A\sin A
\displaystyle \textbf{Answer:}
\displaystyle \text{(i) LHS}=\cos^2\frac{\pi}{4}-\sin^2\frac{\pi}{12}
\displaystyle =\left(\frac1{\sqrt2}\right)^2-\sin^215^\circ
\displaystyle \sin15^\circ=\sin(45^\circ-30^\circ)
\displaystyle =\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ
\displaystyle =\frac1{\sqrt2}\times\frac{\sqrt3}{2}-\frac1{\sqrt2}\times\frac12
\displaystyle =\frac{\sqrt3-1}{2\sqrt2}.
\displaystyle \therefore \sin^215^\circ=\left(\frac{\sqrt3-1}{2\sqrt2}\right)^2
\displaystyle =\frac{(\sqrt3-1)^2}{8}=\frac{4-2\sqrt3}{8}=\frac{2-\sqrt3}{4}.
\displaystyle \therefore \text{LHS}=\frac12-\frac{2-\sqrt3}{4}
\displaystyle =\frac{2-2+\sqrt3}{4}=\frac{\sqrt3}{4}=\text{RHS. Hence proved.}
\displaystyle \text{(ii) LHS}=\sin^2(n+1)A-\sin^2nA
\displaystyle \text{Using }\sin^2x-\sin^2y=\sin(x+y)\sin(x-y),
\displaystyle \text{LHS}=\sin\left[(n+1)A+nA\right]\sin\left[(n+1)A-nA\right]
\displaystyle =\sin(2n+1)A\sin A
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Prove that:}
\displaystyle \text{(i) }\frac{\sin(A+B)+\sin(A-B)}{\cos(A+B)+\cos(A-B)}=\tan A
\displaystyle \text{(ii) }\frac{\sin(A-B)}{\cos A\cos B}+\frac{\sin(B-C)}{\cos B\cos C}+\frac{\sin(C-A)}{\cos C\cos A}=0
\displaystyle \text{(iii) }\frac{\sin(A-B)}{\sin A\sin B}+\frac{\sin(B-C)}{\sin B\sin C}+\frac{\sin(C-A)}{\sin C\sin A}=0
\displaystyle \text{(iv) }\sin^2B=\sin^2A+\sin^2(A-B)-2\sin A\cos B\sin(A-B)
\displaystyle \text{(v) }\cos^2A+\cos^2B-2\cos A\cos B\cos(A+B)=\sin^2(A+B)
\displaystyle \text{(vi) }\frac{\tan(A+B)}{\cot(A-B)}=\frac{\tan^2A-\tan^2B}{1-\tan^2A\tan^2B}
\displaystyle \textbf{Answer:}
\displaystyle \text{(i) LHS}=\frac{\sin(A+B)+\sin(A-B)}{\cos(A+B)+\cos(A-B)}
\displaystyle =\frac{2\sin A\cos B}{2\cos A\cos B}
\displaystyle =\tan A=\text{RHS. Hence proved.}
\displaystyle \text{(ii) LHS}=\frac{\sin(A-B)}{\cos A\cos B}+\frac{\sin(B-C)}{\cos B\cos C}+\frac{\sin(C-A)}{\cos C\cos A}
\displaystyle =\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B}
\displaystyle \quad+\frac{\sin B\cos C-\cos B\sin C}{\cos B\cos C}+\frac{\sin C\cos A-\cos C\sin A}{\cos C\cos A}
\displaystyle =\tan A-\tan B+\tan B-\tan C+\tan C-\tan A
\displaystyle =0=\text{RHS. Hence proved.}
\displaystyle \text{(iii) LHS}=\frac{\sin(A-B)}{\sin A\sin B}+\frac{\sin(B-C)}{\sin B\sin C}+\frac{\sin(C-A)}{\sin C\sin A}
\displaystyle =\frac{\sin A\cos B-\cos A\sin B}{\sin A\sin B}
\displaystyle \quad+\frac{\sin B\cos C-\cos B\sin C}{\sin B\sin C}+\frac{\sin C\cos A-\cos C\sin A}{\sin C\sin A}
\displaystyle =\cot B-\cot A+\cot C-\cot B+\cot A-\cot C
\displaystyle =0=\text{RHS. Hence proved.}
\displaystyle \text{(iv) RHS}=\sin^2A+\sin^2(A-B)-2\sin A\cos B\sin(A-B)
\displaystyle =\sin^2A+\sin(A-B)\left[\sin(A-B)-2\sin A\cos B\right]
\displaystyle =\sin^2A+\sin(A-B)\left[\sin A\cos B-\cos A\sin B-2\sin A\cos B\right]
\displaystyle =\sin^2A-\sin(A-B)\left[\sin A\cos B+\cos A\sin B\right]
\displaystyle =\sin^2A-\sin(A-B)\sin(A+B)
\displaystyle =\sin^2A-\left(\sin^2A-\sin^2B\right)
\displaystyle =\sin^2B=\text{LHS. Hence proved.}
\displaystyle \text{(v) LHS}=\cos^2A+\cos^2B-2\cos A\cos B\cos(A+B)
\displaystyle =\cos^2A+1-\sin^2B-2\cos A\cos B\cos(A+B)
\displaystyle =1+\left(\cos^2A-\sin^2B\right)-2\cos A\cos B\cos(A+B)
\displaystyle =1+\cos(A+B)\cos(A-B)-2\cos A\cos B\cos(A+B)
\displaystyle =1+\cos(A+B)\left[\cos(A-B)-2\cos A\cos B\right]
\displaystyle =1+\cos(A+B)\left[\cos A\cos B+\sin A\sin B-2\cos A\cos B\right]
\displaystyle =1-\cos(A+B)\left[\cos A\cos B-\sin A\sin B\right]
\displaystyle =1-\cos^2(A+B)
\displaystyle =\sin^2(A+B)=\text{RHS. Hence proved.}
\displaystyle \text{(vi) LHS}=\frac{\tan(A+B)}{\cot(A-B)}
\displaystyle =\tan(A+B)\tan(A-B)
\displaystyle =\left(\frac{\tan A+\tan B}{1-\tan A\tan B}\right)\left(\frac{\tan A-\tan B}{1+\tan A\tan B}\right)
\displaystyle =\frac{(\tan A+\tan B)(\tan A-\tan B)}{(1-\tan A\tan B)(1+\tan A\tan B)}
\displaystyle =\frac{\tan^2A-\tan^2B}{1-\tan^2A\tan^2B}
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Prove that:}
\displaystyle \text{(i) }\tan8x-\tan6x-\tan2x=\tan8x\tan6x\tan2x
\displaystyle \text{(ii) }\tan\frac{\pi}{12}+\tan\frac{\pi}{6}+\tan\frac{\pi}{12}\tan\frac{\pi}{6}=1
\displaystyle \text{(iii) }\tan36^\circ+\tan9^\circ+\tan36^\circ\tan9^\circ=1
\displaystyle \text{(iv) }\tan13x-\tan9x-\tan4x=\tan13x\tan9x\tan4x
\displaystyle \textbf{Answer:}
\displaystyle \text{(i) Since }8x=6x+2x,
\displaystyle \tan8x=\tan(6x+2x)
\displaystyle =\frac{\tan6x+\tan2x}{1-\tan6x\tan2x}.
\displaystyle \therefore \tan8x\left(1-\tan6x\tan2x\right)=\tan6x+\tan2x
\displaystyle \Rightarrow \tan8x-\tan8x\tan6x\tan2x=\tan6x+\tan2x
\displaystyle \Rightarrow \tan8x-\tan6x-\tan2x=\tan8x\tan6x\tan2x.
\displaystyle \text{Hence proved.}
\displaystyle \text{(ii) Since }\frac{\pi}{4}=\frac{\pi}{6}+\frac{\pi}{12},
\displaystyle \tan\frac{\pi}{4}=\tan\left(\frac{\pi}{6}+\frac{\pi}{12}\right)
\displaystyle \Rightarrow 1=\frac{\tan\frac{\pi}{6}+\tan\frac{\pi}{12}}{1-\tan\frac{\pi}{6}\tan\frac{\pi}{12}}
\displaystyle \Rightarrow 1-\tan\frac{\pi}{6}\tan\frac{\pi}{12}=\tan\frac{\pi}{6}+\tan\frac{\pi}{12}
\displaystyle \Rightarrow \tan\frac{\pi}{12}+\tan\frac{\pi}{6}+\tan\frac{\pi}{12}\tan\frac{\pi}{6}=1.
\displaystyle \text{Hence proved.}
\displaystyle \text{(iii) Since }45^\circ=36^\circ+9^\circ,
\displaystyle \tan45^\circ=\tan(36^\circ+9^\circ)
\displaystyle \Rightarrow 1=\frac{\tan36^\circ+\tan9^\circ}{1-\tan36^\circ\tan9^\circ}
\displaystyle \Rightarrow 1-\tan36^\circ\tan9^\circ=\tan36^\circ+\tan9^\circ
\displaystyle \Rightarrow \tan36^\circ+\tan9^\circ+\tan36^\circ\tan9^\circ=1.
\displaystyle \text{Hence proved.}
\displaystyle \text{(iv) Since }13x=9x+4x,
\displaystyle \tan13x=\tan(9x+4x)
\displaystyle =\frac{\tan9x+\tan4x}{1-\tan9x\tan4x}.
\displaystyle \therefore \tan13x\left(1-\tan9x\tan4x\right)=\tan9x+\tan4x
\displaystyle \Rightarrow \tan13x-\tan13x\tan9x\tan4x=\tan9x+\tan4x
\displaystyle \Rightarrow \tan13x-\tan9x-\tan4x=\tan13x\tan9x\tan4x.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Prove that }\frac{\tan^22x-\tan^2x}{1-\tan^22x\tan^2x}=\tan3x\tan x.
\displaystyle \textbf{Answer:}
\displaystyle \text{RHS}=\tan3x\tan x
\displaystyle =\tan(2x+x)\tan(2x-x)
\displaystyle =\left(\frac{\tan2x+\tan x}{1-\tan2x\tan x}\right)\left(\frac{\tan2x-\tan x}{1+\tan2x\tan x}\right)
\displaystyle =\frac{(\tan2x+\tan x)(\tan2x-\tan x)}{(1-\tan2x\tan x)(1+\tan2x\tan x)}
\displaystyle =\frac{\tan^22x-\tan^2x}{1-\tan^22x\tan^2x}
\displaystyle =\text{LHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }\frac{\sin(x+y)}{\sin(x-y)}=\frac{a+b}{a-b},\text{ show that }\frac{\tan x}{\tan y}=\frac{a}{b}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\frac{\sin(x+y)}{\sin(x-y)}=\frac{a+b}{a-b}.
\displaystyle \Rightarrow \frac{\sin x\cos y+\cos x\sin y}{\sin x\cos y-\cos x\sin y}=\frac{a+b}{a-b}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{(\sin x\cos y+\cos x\sin y)+(\sin x\cos y-\cos x\sin y)}{(\sin x\cos y+\cos x\sin y)-(\sin x\cos y-\cos x\sin y)}
\displaystyle =\frac{(a+b)+(a-b)}{(a+b)-(a-b)}
\displaystyle \Rightarrow \frac{2\sin x\cos y}{2\cos x\sin y}=\frac{2a}{2b}
\displaystyle \Rightarrow \frac{\sin x\cos y}{\cos x\sin y}=\frac{a}{b}
\displaystyle \Rightarrow \frac{\tan x}{\tan y}=\frac{a}{b}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }\tan A=x\tan B,\text{ prove that }\frac{\sin(A-B)}{\sin(A+B)}=\frac{x-1}{x+1}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\tan A=x\tan B.
\displaystyle \Rightarrow \frac{\sin A}{\cos A}=x\frac{\sin B}{\cos B}
\displaystyle \Rightarrow \sin A\cos B=x\cos A\sin B.
\displaystyle \text{Now, }\frac{\sin(A-B)}{\sin(A+B)}
\displaystyle =\frac{\sin A\cos B-\cos A\sin B}{\sin A\cos B+\cos A\sin B}
\displaystyle =\frac{x\cos A\sin B-\cos A\sin B}{x\cos A\sin B+\cos A\sin B}
\displaystyle =\frac{(x-1)\cos A\sin B}{(x+1)\cos A\sin B}
\displaystyle =\frac{x-1}{x+1}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }\tan(A+B)=x\text{ and }\tan(A-B)=y,\text{ find }\tan2A\text{ and }\tan2B.
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\tan(A+B)=x\text{ and }\tan(A-B)=y.
\displaystyle \tan2A=\tan\left[(A+B)+(A-B)\right]
\displaystyle =\frac{\tan(A+B)+\tan(A-B)}{1-\tan(A+B)\tan(A-B)}
\displaystyle =\frac{x+y}{1-xy}.
\displaystyle \text{Similarly,}
\displaystyle \tan2B=\tan\left[(A+B)-(A-B)\right]
\displaystyle =\frac{\tan(A+B)-\tan(A-B)}{1+\tan(A+B)\tan(A-B)}
\displaystyle =\frac{x-y}{1+xy}.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If }\cos A+\sin B=m\text{ and }\sin A+\cos B=n,\text{ prove that}
\displaystyle 2\sin(A+B)=m^2+n^2-2.
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\cos A+\sin B=m\text{ and }\sin A+\cos B=n.
\displaystyle m^2+n^2-2
\displaystyle =(\cos A+\sin B)^2+(\sin A+\cos B)^2-2
\displaystyle =\cos^2A+\sin^2B+2\cos A\sin B+\sin^2A+\cos^2B+2\sin A\cos B-2
\displaystyle =(\cos^2A+\sin^2A)+(\cos^2B+\sin^2B)
\displaystyle \quad+2(\cos A\sin B+\sin A\cos B)-2
\displaystyle =2+2(\cos A\sin B+\sin A\cos B)-2
\displaystyle =2(\sin A\cos B+\cos A\sin B)
\displaystyle =2\sin(A+B).
\displaystyle \therefore 2\sin(A+B)=m^2+n^2-2.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If }\tan A+\tan B=a\text{ and }\cot A+\cot B=b,\text{ prove that}
\displaystyle \cot(A+B)=\frac1a-\frac1b.
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\tan A+\tan B=a\text{ and }\cot A+\cot B=b.
\displaystyle \cot A+\cot B=b
\displaystyle \Rightarrow \frac1{\tan A}+\frac1{\tan B}=b
\displaystyle \Rightarrow \frac{\tan A+\tan B}{\tan A\tan B}=b
\displaystyle \Rightarrow \frac{a}{\tan A\tan B}=b
\displaystyle \Rightarrow \tan A\tan B=\frac ab.
\displaystyle \text{Now, }\cot(A+B)=\frac1{\tan(A+B)}
\displaystyle =\frac{1-\tan A\tan B}{\tan A+\tan B}
\displaystyle =\frac{1-\frac ab}{a}
\displaystyle =\frac{b-a}{ab}
\displaystyle =\frac1a-\frac1b.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If }x\text{ lies in the first quadrant and }\cos x=\frac8{17},\text{ prove that:}
\displaystyle \cos\left(\frac{\pi}{6}+x\right)+\cos\left(\frac{\pi}{4}-x\right)+\cos\left(\frac{2\pi}{3}-x\right)
\displaystyle =\left(\frac{\sqrt3-1}{2}+\frac1{\sqrt2}\right)\frac{23}{17}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\cos x=\frac8{17}.
\displaystyle \text{Since }x\text{ lies in the first quadrant,}
\displaystyle \sin x=\sqrt{1-\cos^2x}=\sqrt{1-\left(\frac8{17}\right)^2}=\sqrt{\frac{225}{289}}=\frac{15}{17}.
\displaystyle \text{LHS}=\cos\left(\frac{\pi}{6}+x\right)+\cos\left(\frac{\pi}{4}-x\right)+\cos\left(\frac{2\pi}{3}-x\right)
\displaystyle =\left(\cos\frac{\pi}{6}\cos x-\sin\frac{\pi}{6}\sin x\right)
\displaystyle \quad+\left(\cos\frac{\pi}{4}\cos x+\sin\frac{\pi}{4}\sin x\right)
\displaystyle \quad+\left(\cos\frac{2\pi}{3}\cos x+\sin\frac{2\pi}{3}\sin x\right)
\displaystyle =\left(\cos\frac{\pi}{6}+\cos\frac{\pi}{4}+\cos\frac{2\pi}{3}\right)\cos x
\displaystyle \quad+\left(-\sin\frac{\pi}{6}+\sin\frac{\pi}{4}+\sin\frac{2\pi}{3}\right)\sin x
\displaystyle =\left(\frac{\sqrt3}{2}+\frac1{\sqrt2}-\frac12\right)\frac8{17}
\displaystyle \quad+\left(-\frac12+\frac1{\sqrt2}+\frac{\sqrt3}{2}\right)\frac{15}{17}
\displaystyle =\left(\frac{\sqrt3-1}{2}+\frac1{\sqrt2}\right)\left(\frac8{17}+\frac{15}{17}\right)
\displaystyle =\left(\frac{\sqrt3-1}{2}+\frac1{\sqrt2}\right)\frac{23}{17}
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If }\tan x+\tan\left(x+\frac{\pi}{3}\right)+\tan\left(x+\frac{2\pi}{3}\right)=3,
\displaystyle \text{prove that }\frac{3\tan x-\tan^3x}{1-3\tan^2x}=1.
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\tan x+\tan\left(x+\frac{\pi}{3}\right)+\tan\left(x+\frac{2\pi}{3}\right)=3.
\displaystyle \text{Let }\tan x=t.
\displaystyle \therefore t+\frac{t+\tan\frac{\pi}{3}}{1-t\tan\frac{\pi}{3}}+\frac{t+\tan\frac{2\pi}{3}}{1-t\tan\frac{2\pi}{3}}=3
\displaystyle \Rightarrow t+\frac{t+\sqrt3}{1-\sqrt3t}+\frac{t-\sqrt3}{1+\sqrt3t}=3
\displaystyle \Rightarrow t+\frac{(t+\sqrt3)(1+\sqrt3t)+(t-\sqrt3)(1-\sqrt3t)}{(1-\sqrt3t)(1+\sqrt3t)}=3
\displaystyle \Rightarrow t+\frac{8t}{1-3t^2}=3
\displaystyle \Rightarrow \frac{t(1-3t^2)+8t}{1-3t^2}=3
\displaystyle \Rightarrow \frac{9t-3t^3}{1-3t^2}=3
\displaystyle \Rightarrow \frac{3t-t^3}{1-3t^2}=1
\displaystyle \Rightarrow \frac{3\tan x-\tan^3x}{1-3\tan^2x}=1.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{If }\sin(\alpha+\beta)=1\text{ and }\sin(\alpha-\beta)=\frac12,\text{ where}
\displaystyle 0\leq\alpha,\beta\leq\frac{\pi}{2},\text{ find }\tan(\alpha+2\beta)\text{ and }\tan(2\alpha+\beta).
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\sin(\alpha+\beta)=1.
\displaystyle \text{Since }0\leq\alpha+\beta\leq\pi,
\displaystyle \alpha+\beta=\frac{\pi}{2}.\qquad\text{...(i)}
\displaystyle \text{Also, }\sin(\alpha-\beta)=\frac12.
\displaystyle \text{Since }-\frac{\pi}{2}\leq\alpha-\beta\leq\frac{\pi}{2},
\displaystyle \alpha-\beta=\frac{\pi}{6}.\qquad\text{...(ii)}
\displaystyle \text{Adding (i) and (ii),}
\displaystyle 2\alpha=\frac{\pi}{2}+\frac{\pi}{6}=\frac{2\pi}{3}
\displaystyle \Rightarrow \alpha=\frac{\pi}{3}.
\displaystyle \text{From (i), }\beta=\frac{\pi}{2}-\frac{\pi}{3}=\frac{\pi}{6}.
\displaystyle \therefore \tan(\alpha+2\beta)=\tan\left(\frac{\pi}{3}+2\times\frac{\pi}{6}\right)
\displaystyle =\tan\frac{2\pi}{3}=\tan\left(\frac{\pi}{2}+\frac{\pi}{6}\right)
\displaystyle =-\cot\frac{\pi}{6}=-\sqrt3.
\displaystyle \text{Also, }\tan(2\alpha+\beta)=\tan\left(2\times\frac{\pi}{3}+\frac{\pi}{6}\right)
\displaystyle =\tan\frac{5\pi}{6}=\tan\left(\frac{\pi}{2}+\frac{\pi}{3}\right)
\displaystyle =-\cot\frac{\pi}{3}=-\frac1{\sqrt3}.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{If }\alpha\text{ and }\beta\text{ are two different values of }x\text{ lying between }0\text{ and }2\pi
\displaystyle \text{which satisfy }6\cos x+8\sin x=9,\text{ find }\sin(\alpha+\beta).
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }6\cos x+8\sin x=9.
\displaystyle \Rightarrow 10\left(\frac35\cos x+\frac45\sin x\right)=9
\displaystyle \text{Let }\cos\theta=\frac45\text{ and }\sin\theta=\frac35,\text{ where }0<\theta<\frac{\pi}{2}.
\displaystyle \therefore 10(\sin x\cos\theta+\cos x\sin\theta)=9
\displaystyle \Rightarrow \sin(x+\theta)=\frac9{10}.
\displaystyle \text{Since }\alpha\text{ and }\beta\text{ are the two different solutions between }0\text{ and }2\pi,
\displaystyle (\alpha+\theta)+(\beta+\theta)=\pi+2\pi k,\text{ for some integer }k.
\displaystyle \therefore \alpha+\beta=\pi-2\theta+2\pi k.
\displaystyle \sin(\alpha+\beta)=\sin(\pi-2\theta+2\pi k)=\sin2\theta
\displaystyle =2\sin\theta\cos\theta
\displaystyle =2\times\frac35\times\frac45=\frac{24}{25}.
\displaystyle \therefore \sin(\alpha+\beta)=\frac{24}{25}.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{If }\sin\alpha+\sin\beta=a\text{ and }\cos\alpha+\cos\beta=b,\text{ show that:}
\displaystyle \text{(i) }\sin(\alpha+\beta)=\frac{2ab}{a^2+b^2}
\displaystyle \text{(ii) }\cos(\alpha+\beta)=\frac{b^2-a^2}{b^2+a^2}
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\sin\alpha+\sin\beta=a\text{ and }\cos\alpha+\cos\beta=b.
\displaystyle \text{Using the sum-to-product identities,}
\displaystyle a=2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)
\displaystyle b=2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right).
\displaystyle \therefore ab=4\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha+\beta}{2}\right)\cos^2\left(\frac{\alpha-\beta}{2}\right)
\displaystyle =2\sin(\alpha+\beta)\cos^2\left(\frac{\alpha-\beta}{2}\right).
\displaystyle \text{Also, }a^2+b^2
\displaystyle =4\cos^2\left(\frac{\alpha-\beta}{2}\right)\left[\sin^2\left(\frac{\alpha+\beta}{2}\right)+\cos^2\left(\frac{\alpha+\beta}{2}\right)\right]
\displaystyle =4\cos^2\left(\frac{\alpha-\beta}{2}\right).
\displaystyle \text{(i) }\frac{2ab}{a^2+b^2}
\displaystyle =\frac{4\sin(\alpha+\beta)\cos^2\left(\frac{\alpha-\beta}{2}\right)}{4\cos^2\left(\frac{\alpha-\beta}{2}\right)}
\displaystyle =\sin(\alpha+\beta).
\displaystyle \therefore \sin(\alpha+\beta)=\frac{2ab}{a^2+b^2}.
\displaystyle \text{(ii) }b^2-a^2
\displaystyle =4\cos^2\left(\frac{\alpha-\beta}{2}\right)\left[\cos^2\left(\frac{\alpha+\beta}{2}\right)-\sin^2\left(\frac{\alpha+\beta}{2}\right)\right]
\displaystyle =4\cos^2\left(\frac{\alpha-\beta}{2}\right)\cos(\alpha+\beta).
\displaystyle \therefore \frac{b^2-a^2}{b^2+a^2}
\displaystyle =\frac{4\cos^2\left(\frac{\alpha-\beta}{2}\right)\cos(\alpha+\beta)}{4\cos^2\left(\frac{\alpha-\beta}{2}\right)}
\displaystyle =\cos(\alpha+\beta).
\displaystyle \therefore \cos(\alpha+\beta)=\frac{b^2-a^2}{b^2+a^2}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Prove that:}
\displaystyle \text{(i) }\frac{1}{\sin(x-a)\sin(x-b)}=\frac{\cot(x-a)-\cot(x-b)}{\sin(a-b)}
\displaystyle \text{(ii) }\frac{1}{\sin(x-a)\cos(x-b)}=\frac{\cot(x-a)+\tan(x-b)}{\cos(a-b)}
\displaystyle \text{(iii) }\frac{1}{\cos(x-a)\cos(x-b)}=\frac{\tan(x-b)-\tan(x-a)}{\sin(a-b)}
\displaystyle \text{Answer:}
\displaystyle \text{The identities hold for all values of }x,\ a\text{ and }b\text{ for which both sides are defined.}
\displaystyle \text{(i) LHS}=\frac{1}{\sin(x-a)\sin(x-b)}
\displaystyle =\frac{1}{\sin(a-b)}\left[\frac{\sin(a-b)}{\sin(x-a)\sin(x-b)}\right]
\displaystyle =\frac{1}{\sin(a-b)}\left[\frac{\sin\{(x-b)-(x-a)\}}{\sin(x-a)\sin(x-b)}\right]
\displaystyle =\frac{1}{\sin(a-b)}\left[\frac{\sin(x-b)\cos(x-a)-\cos(x-b)\sin(x-a)}{\sin(x-a)\sin(x-b)}\right]
\displaystyle =\frac{1}{\sin(a-b)}\left[\cot(x-a)-\cot(x-b)\right]
\displaystyle =\frac{\cot(x-a)-\cot(x-b)}{\sin(a-b)}=\text{RHS.}
\displaystyle \therefore \text{The identity is proved.}
\displaystyle \text{(ii) LHS}=\frac{1}{\sin(x-a)\cos(x-b)}
\displaystyle =\frac{1}{\cos(a-b)}\left[\frac{\cos(a-b)}{\sin(x-a)\cos(x-b)}\right]
\displaystyle =\frac{1}{\cos(a-b)}\left[\frac{\cos\{(x-b)-(x-a)\}}{\sin(x-a)\cos(x-b)}\right]
\displaystyle =\frac{1}{\cos(a-b)}\left[\frac{\cos(x-b)\cos(x-a)+\sin(x-b)\sin(x-a)}{\sin(x-a)\cos(x-b)}\right]
\displaystyle =\frac{1}{\cos(a-b)}\left[\cot(x-a)+\tan(x-b)\right]
\displaystyle =\frac{\cot(x-a)+\tan(x-b)}{\cos(a-b)}=\text{RHS.}
\displaystyle \therefore \text{The identity is proved.}
\displaystyle \text{(iii) LHS}=\frac{1}{\cos(x-a)\cos(x-b)}
\displaystyle =\frac{1}{\sin(a-b)}\left[\frac{\sin(a-b)}{\cos(x-a)\cos(x-b)}\right]
\displaystyle =\frac{1}{\sin(a-b)}\left[\frac{\sin\{(x-b)-(x-a)\}}{\cos(x-a)\cos(x-b)}\right]
\displaystyle =\frac{1}{\sin(a-b)}\left[\frac{\sin(x-b)\cos(x-a)-\cos(x-b)\sin(x-a)}{\cos(x-a)\cos(x-b)}\right]
\displaystyle =\frac{1}{\sin(a-b)}\left[\tan(x-b)-\tan(x-a)\right]
\displaystyle =\frac{\tan(x-b)-\tan(x-a)}{\sin(a-b)}=\text{RHS.}
\displaystyle \therefore \text{The identity is proved.}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{If }\sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0,\text{ prove that}
\displaystyle 1+\cot\alpha\tan\beta=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0.
\displaystyle \Rightarrow \cos\alpha\cos\beta-\sin\alpha\sin\beta=1
\displaystyle \Rightarrow \cos(\alpha+\beta)=1
\displaystyle \therefore \sin(\alpha+\beta)=0.
\displaystyle 1+\cot\alpha\tan\beta
\displaystyle =1+\frac{\cos\alpha}{\sin\alpha}\times\frac{\sin\beta}{\cos\beta}
\displaystyle =\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\sin\alpha\cos\beta}
\displaystyle =\frac{\sin(\alpha+\beta)}{\sin\alpha\cos\beta}
\displaystyle =\frac{0}{\sin\alpha\cos\beta}=0
\displaystyle \therefore 1+\cot\alpha\tan\beta=0.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{If }\tan\alpha=x+1\text{ and }\tan\beta=x-1,\text{ show that}
\displaystyle 2\cot(\alpha-\beta)=x^2.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\tan\alpha=x+1\text{ and }\tan\beta=x-1.
\displaystyle \cot(\alpha-\beta)=\frac{1+\tan\alpha\tan\beta}{\tan\alpha-\tan\beta}
\displaystyle \therefore 2\cot(\alpha-\beta)=2\left[\frac{1+\tan\alpha\tan\beta}{\tan\alpha-\tan\beta}\right]
\displaystyle =2\left[\frac{1+(x+1)(x-1)}{(x+1)-(x-1)}\right]
\displaystyle =2\left[\frac{1+x^2-1}{2}\right]
\displaystyle =2\left(\frac{x^2}{2}\right)=x^2
\displaystyle \therefore 2\cot(\alpha-\beta)=x^2.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{An angle }\theta\text{ is divided into two parts such that the tangent of one part is}
\displaystyle \lambda\text{ times the tangent of the other. If }\phi\text{ is the difference between the two parts, show that}
\displaystyle \sin\phi=\frac{\lambda-1}{\lambda+1}\sin\theta.
\displaystyle \text{Answer:}
\displaystyle \text{Let }\alpha\text{ and }\beta\text{ be the two parts of the angle.}
\displaystyle \therefore \theta=\alpha+\beta\text{ and }\phi=\alpha-\beta.
\displaystyle \text{Given, }\tan\alpha=\lambda\tan\beta.
\displaystyle \Rightarrow \frac{\tan\alpha}{\tan\beta}=\frac{\lambda}{1}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{\tan\alpha+\tan\beta}{\tan\alpha-\tan\beta}=\frac{\lambda+1}{\lambda-1}
\displaystyle \Rightarrow \frac{\frac{\sin\alpha}{\cos\alpha}+\frac{\sin\beta}{\cos\beta}}{\frac{\sin\alpha}{\cos\alpha}-\frac{\sin\beta}{\cos\beta}}=\frac{\lambda+1}{\lambda-1}
\displaystyle \Rightarrow \frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\sin\alpha\cos\beta-\cos\alpha\sin\beta}=\frac{\lambda+1}{\lambda-1}
\displaystyle \Rightarrow \frac{\sin(\alpha+\beta)}{\sin(\alpha-\beta)}=\frac{\lambda+1}{\lambda-1}
\displaystyle \Rightarrow \frac{\sin\theta}{\sin\phi}=\frac{\lambda+1}{\lambda-1}
\displaystyle \Rightarrow \frac{\sin\phi}{\sin\theta}=\frac{\lambda-1}{\lambda+1}
\displaystyle \therefore \sin\phi=\frac{\lambda-1}{\lambda+1}\sin\theta.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{If }\tan x=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha},\text{ then show that}
\displaystyle \sin\alpha+\cos\alpha=\sqrt{2}\cos x.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\tan x=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}.
\displaystyle \text{Dividing the numerator and denominator by }\cos\alpha,
\displaystyle \tan x=\frac{\tan\alpha-1}{\tan\alpha+1}
\displaystyle =\frac{\tan\alpha-\tan\frac{\pi}{4}}{1+\tan\alpha\tan\frac{\pi}{4}}
\displaystyle =\tan\left(\alpha-\frac{\pi}{4}\right)
\displaystyle \therefore x=\alpha-\frac{\pi}{4},\text{ taking }x\text{ and }\alpha-\frac{\pi}{4}\text{ as the same principal angle.}
\displaystyle \Rightarrow \cos x=\cos\left(\alpha-\frac{\pi}{4}\right)
\displaystyle =\cos\alpha\cos\frac{\pi}{4}+\sin\alpha\sin\frac{\pi}{4}
\displaystyle =\frac{\cos\alpha}{\sqrt{2}}+\frac{\sin\alpha}{\sqrt{2}}
\displaystyle =\frac{\sin\alpha+\cos\alpha}{\sqrt{2}}
\displaystyle \therefore \sin\alpha+\cos\alpha=\sqrt{2}\cos x.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{If }\alpha\text{ and }\beta\text{ are two solutions of the equation}
\displaystyle a\tan x+b\sec x=c,\text{ find }\sin(\alpha+\beta)\text{ and }\cos(\alpha+\beta).
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a\tan x+b\sec x=c.
\displaystyle \Rightarrow \frac{a\sin x}{\cos x}+\frac{b}{\cos x}=c
\displaystyle \Rightarrow a\sin x-c\cos x=-b
\displaystyle \text{Let }R=\sqrt{a^2+c^2}\text{ and choose }\phi\text{ such that}
\displaystyle \cos\phi=\frac{a}{R}\text{ and }\sin\phi=\frac{c}{R}.
\displaystyle \therefore a=R\cos\phi\text{ and }c=R\sin\phi.
\displaystyle a\sin x-c\cos x=-b
\displaystyle \Rightarrow R(\sin x\cos\phi-\cos x\sin\phi)=-b
\displaystyle \Rightarrow R\sin(x-\phi)=-b
\displaystyle \Rightarrow \sin(x-\phi)=-\frac{b}{R}
\displaystyle \text{Since }\alpha\text{ and }\beta\text{ are the two solutions,}
\displaystyle (\alpha-\phi)+(\beta-\phi)=\pi
\displaystyle \Rightarrow \alpha+\beta=\pi+2\phi
\displaystyle \therefore \sin(\alpha+\beta)=\sin(\pi+2\phi)
\displaystyle =-\sin2\phi
\displaystyle =-2\sin\phi\cos\phi
\displaystyle =-2\left(\frac{c}{R}\right)\left(\frac{a}{R}\right)
\displaystyle =-\frac{2ac}{a^2+c^2}
\displaystyle \therefore \sin(\alpha+\beta)=-\frac{2ac}{a^2+c^2}.
\displaystyle \cos(\alpha+\beta)=\cos(\pi+2\phi)
\displaystyle =-\cos2\phi
\displaystyle =-(\cos^2\phi-\sin^2\phi)
\displaystyle =-\left(\frac{a^2}{R^2}-\frac{c^2}{R^2}\right)
\displaystyle =\frac{c^2-a^2}{a^2+c^2}
\displaystyle \therefore \cos(\alpha+\beta)=\frac{c^2-a^2}{a^2+c^2}.
\displaystyle \\

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