\displaystyle \textbf{Question 1: }\text{In a right-angled triangle }ABC,\text{ right-angled at }B,\text{ if }
\displaystyle \sin A=\frac{3}{5}, \text{find all the six trigonometric ratios of }\angle C.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\sin A=\frac{3}{5}
\displaystyle \Rightarrow \frac{BC}{AC}=\frac{3}{5}
\displaystyle \text{Let }BC=3\text{ and }AC=5.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle AB^2+BC^2=AC^2
\displaystyle \Rightarrow AB^2=AC^2-BC^2
\displaystyle \Rightarrow AB^2=5^2-3^2=25-9=16
\displaystyle \Rightarrow AB=4
\displaystyle \therefore \sin C=\frac{AB}{AC}=\frac{4}{5},\qquad \mathrm{cosec}\,C=\frac{AC}{AB}=\frac{5}{4}
\displaystyle \cos C=\frac{BC}{AC}=\frac{3}{5},\qquad \sec C=\frac{AC}{BC}=\frac{5}{3}
\displaystyle \tan C=\frac{AB}{BC}=\frac{4}{3},\qquad \cot C=\frac{BC}{AB}=\frac{3}{4}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }\sin A=\frac{a^2-b^2}{a^2+b^2},\text{ find the value of the other five}
\displaystyle \text{trigonometric ratios.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\sin A=\frac{a^2-b^2}{a^2+b^2}
\displaystyle \therefore \frac{BC}{AC}=\frac{a^2-b^2}{a^2+b^2}
\displaystyle \text{Let }BC=a^2-b^2\text{ and }AC=a^2+b^2.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle AB^2=AC^2-BC^2
\displaystyle =(a^2+b^2)^2-(a^2-b^2)^2
\displaystyle =a^4+2a^2b^2+b^4-(a^4-2a^2b^2+b^4)
\displaystyle =4a^2b^2
\displaystyle \therefore AB=2ab
\displaystyle \cos A=\frac{AB}{AC}=\frac{2ab}{a^2+b^2},\qquad \sec A=\frac{a^2+b^2}{2ab}
\displaystyle \tan A=\frac{BC}{AB}=\frac{a^2-b^2}{2ab},\qquad \cot A=\frac{2ab}{a^2-b^2}
\displaystyle \mathrm{cosec}\,A=\frac{AC}{BC}=\frac{a^2+b^2}{a^2-b^2}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }\mathrm{cosec}\,A=2,\text{ find the value of }\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\mathrm{cosec}\,A=2
\displaystyle \therefore \sin A=\frac12
\displaystyle \text{Let the perpendicular }=1\text{ and hypotenuse }=2.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Base}=\sqrt{2^2-1^2}=\sqrt3
\displaystyle \therefore \tan A=\frac1{\sqrt3},\qquad \cos A=\frac{\sqrt3}{2}
\displaystyle \frac{1}{\tan A}+\frac{\sin A}{1+\cos A}
\displaystyle =\sqrt3+\frac{\frac12}{1+\frac{\sqrt3}{2}}
\displaystyle =\sqrt3+\frac{1}{2+\sqrt3}
\displaystyle =\sqrt3+\frac{2-\sqrt3}{(2+\sqrt3)(2-\sqrt3)}
\displaystyle =\sqrt3+2-\sqrt3
\displaystyle =2
\displaystyle \therefore \frac{1}{\tan A}+\frac{\sin A}{1+\cos A}=2
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }\tan A=\sqrt2-1,\text{ show that }\sin A\cos A=\frac{\sqrt2}{4}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\tan A=\sqrt2-1
\displaystyle \therefore \frac{\text{Perpendicular}}{\text{Base}}=\frac{\sqrt2-1}{1}
\displaystyle \text{Let the perpendicular }=\sqrt2-1\text{ and the base }=1.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Hypotenuse}=\sqrt{(\sqrt2-1)^2+1^2}
\displaystyle =\sqrt{3-2\sqrt2+1}=\sqrt{4-2\sqrt2}
\displaystyle \therefore \sin A=\frac{\sqrt2-1}{\sqrt{4-2\sqrt2}},\qquad \cos A=\frac{1}{\sqrt{4-2\sqrt2}}
\displaystyle \therefore \sin A\cos A=\frac{\sqrt2-1}{4-2\sqrt2}
\displaystyle =\frac{\sqrt2-1}{2\sqrt2(\sqrt2-1)}
\displaystyle =\frac{1}{2\sqrt2}=\frac{\sqrt2}{4}
\displaystyle \therefore \sin A\cos A=\frac{\sqrt2}{4}.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In }\triangle ABC,\text{ right-angled at }C,\text{ if }\tan A=\frac{1}{\sqrt3}\text{ and }\tan B=\sqrt3,
\displaystyle \text{show that }\sin A\cos B+\cos A\sin B=1.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\tan A=\frac{1}{\sqrt3}=\tan30^\circ
\displaystyle \therefore A=30^\circ
\displaystyle \text{Since }\angle C=90^\circ,\quad A+B=90^\circ.
\displaystyle \therefore B=90^\circ-30^\circ=60^\circ
\displaystyle \text{L.H.S.}=\sin A\cos B+\cos A\sin B
\displaystyle =\sin30^\circ\cos60^\circ+\cos30^\circ\sin60^\circ
\displaystyle =\frac12\times\frac12+\frac{\sqrt3}{2}\times\frac{\sqrt3}{2}
\displaystyle =\frac14+\frac34=1
\displaystyle =\text{R.H.S.}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }\cot B=\frac{12}{5},\text{ prove that }\tan^2 B-\sin^2 B=\sin^4 B\sec^2 B.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\cot B=\frac{12}{5}
\displaystyle \therefore \frac{\text{Adjacent side}}{\text{Opposite side}}=\frac{12}{5}
\displaystyle \text{Let the adjacent side }=12\text{ and the opposite side }=5.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Hypotenuse}=\sqrt{12^2+5^2}=\sqrt{169}=13
\displaystyle \therefore \tan B=\frac{5}{12},\qquad \sin B=\frac{5}{13},\qquad \sec B=\frac{13}{12}
\displaystyle \text{L.H.S.}=\tan^2 B-\sin^2 B
\displaystyle =\left(\frac{5}{12}\right)^2-\left(\frac{5}{13}\right)^2
\displaystyle =\frac{25}{144}-\frac{25}{169}
\displaystyle =\frac{25(169-144)}{144\times169}
\displaystyle =\frac{625}{12^2\times13^2}=\frac{5^4}{12^2\times13^2}
\displaystyle \text{R.H.S.}=\sin^4 B\sec^2 B
\displaystyle =\left(\frac{5}{13}\right)^4\left(\frac{13}{12}\right)^2
\displaystyle =\frac{5^4}{13^4}\times\frac{13^2}{12^2}
\displaystyle =\frac{5^4}{12^2\times13^2}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In the given figure, }AD=DB\text{ and }\angle B=90^\circ.\text{ Determine:}
\displaystyle \text{(i) }\sin\theta\qquad\text{(ii) }\cos\theta\qquad\text{(iii) }\tan\theta\qquad\text{(iv) }\sin^2\theta+\cos^2\theta
\displaystyle \text{Answer:}
\displaystyle \text{From the figure, }AB=a,\quad AC=b\text{ and }AD=DB.
\displaystyle \therefore AD=DB=\frac{a}{2}
\displaystyle \text{In right-angled }\triangle ABC,\text{ by Pythagoras theorem,}
\displaystyle BC^2=AC^2-AB^2
\displaystyle =b^2-a^2
\displaystyle \therefore BC=\sqrt{b^2-a^2}
\displaystyle \text{In right-angled }\triangle BCD,\text{ by Pythagoras theorem,}
\displaystyle CD^2=BC^2+BD^2
\displaystyle =b^2-a^2+\left(\frac{a}{2}\right)^2
\displaystyle =b^2-\frac{3a^2}{4}=\frac{4b^2-3a^2}{4}
\displaystyle \therefore CD=\frac{\sqrt{4b^2-3a^2}}{2}
\displaystyle \text{(i) }\sin\theta=\frac{BD}{CD}
\displaystyle =\frac{\frac{a}{2}}{\frac{\sqrt{4b^2-3a^2}}{2}}=\frac{a}{\sqrt{4b^2-3a^2}}

\displaystyle \text{(ii) }\cos\theta=\frac{BC}{CD}
\displaystyle =\frac{\sqrt{b^2-a^2}}{\frac{\sqrt{4b^2-3a^2}}{2}}
\displaystyle =\frac{2\sqrt{b^2-a^2}}{\sqrt{4b^2-3a^2}}

\displaystyle \text{(iii) }\tan\theta=\frac{BD}{BC}
\displaystyle =\frac{\frac{a}{2}}{\sqrt{b^2-a^2}}=\frac{a}{2\sqrt{b^2-a^2}}

\displaystyle \text{(iv) }\sin^2\theta+\cos^2\theta
\displaystyle =\left(\frac{a}{\sqrt{4b^2-3a^2}}\right)^2+\left(\frac{2\sqrt{b^2-a^2}}{\sqrt{4b^2-3a^2}}\right)^2
\displaystyle =\frac{a^2}{4b^2-3a^2}+\frac{4(b^2-a^2)}{4b^2-3a^2}
\displaystyle =\frac{a^2+4b^2-4a^2}{4b^2-3a^2}
\displaystyle =\frac{4b^2-3a^2}{4b^2-3a^2}=1
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In }\triangle ABC,\text{ right-angled at }C\text{ and }\angle A=\angle B,\text{ determine whether:}
\displaystyle \text{(i) }\cos A=\cos B\qquad\text{(ii) }\tan A=\tan B
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\angle C=90^\circ\text{ and }\angle A=\angle B.
\displaystyle \text{Since }\angle A+\angle B+\angle C=180^\circ,
\displaystyle \angle A+\angle B=90^\circ
\displaystyle \therefore \angle A=\angle B=45^\circ
\displaystyle \text{(i) }\cos A=\cos B=\cos45^\circ=\frac{1}{\sqrt2}
\displaystyle \therefore \cos A=\cos B

\displaystyle \text{(ii) }\tan A=\tan B=\tan45^\circ=1
\displaystyle \therefore \tan A=\tan B
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }5\tan A=4,\text{ show that }\frac{5\sin A-3\cos A}{5\sin A+2\cos A}=\frac{1}{6}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }5\tan A=4
\displaystyle \therefore \tan A=\frac45
\displaystyle \text{L.H.S.}=\frac{5\sin A-3\cos A}{5\sin A+2\cos A}
\displaystyle \text{Dividing the numerator and denominator by }\cos A,
\displaystyle \text{L.H.S.}=\frac{5\tan A-3}{5\tan A+2}
\displaystyle =\frac{5\times\frac45-3}{5\times\frac45+2}
\displaystyle =\frac{4-3}{4+2}=\frac16
\displaystyle =\text{R.H.S.}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }\tan A+\frac{1}{\tan A}=2,\text{ find the value of }\tan^2 A+\frac{1}{\tan^2 A}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\tan A+\frac{1}{\tan A}=2
\displaystyle \text{Squaring both sides,}
\displaystyle \left(\tan A+\frac{1}{\tan A}\right)^2=4
\displaystyle \Rightarrow \tan^2 A+\frac{1}{\tan^2 A}+2=4
\displaystyle \therefore \tan^2 A+\frac{1}{\tan^2 A}=2
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }\cot A=\frac{7}{8},\text{ evaluate:}
\displaystyle \text{(i) }\frac{(1+\sin A)(1-\sin A)}{(1+\cos A)(1-\cos A)}\qquad\text{(ii) }\cot^2 A
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\frac{(1+\sin A)(1-\sin A)}{(1+\cos A)(1-\cos A)}
\displaystyle =\frac{1-\sin^2 A}{1-\cos^2 A}
\displaystyle =\frac{\cos^2 A}{\sin^2 A}=\cot^2 A
\displaystyle =\left(\frac{7}{8}\right)^2=\frac{49}{64}

\displaystyle \text{(ii) }\cot^2 A=\left(\frac{7}{8}\right)^2=\frac{49}{64}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }3\cot A=4,\text{ check if }\frac{1-\tan^2 A}{1+\tan^2 A}=\cos^2 A-\sin^2 A.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }3\cot A=4
\displaystyle \Rightarrow \cot A=\frac43
\displaystyle \therefore \tan A=\frac34
\displaystyle \text{Hence, }\sin A=\frac35\text{ and }\cos A=\frac45.
\displaystyle \text{L.H.S.}=\frac{1-\tan^2 A}{1+\tan^2 A}
\displaystyle =\frac{1-\left(\frac34\right)^2}{1+\left(\frac34\right)^2}
\displaystyle =\frac{1-\frac9{16}}{1+\frac9{16}}
\displaystyle =\frac{\frac7{16}}{\frac{25}{16}}=\frac7{25}
\displaystyle \text{R.H.S.}=\cos^2 A-\sin^2 A
\displaystyle =\left(\frac45\right)^2-\left(\frac35\right)^2
\displaystyle =\frac{16}{25}-\frac9{25}=\frac7{25}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence verified.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }\tan A=\frac{a}{b},\text{ find the value of }\frac{\cos A+\sin A}{\cos A-\sin A}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\tan A=\frac{a}{b}
\displaystyle \frac{\cos A+\sin A}{\cos A-\sin A}
\displaystyle \text{Dividing the numerator and denominator by }\cos A,
\displaystyle =\frac{1+\tan A}{1-\tan A}
\displaystyle =\frac{1+\frac{a}{b}}{1-\frac{a}{b}}
\displaystyle =\frac{\frac{a+b}{b}}{\frac{b-a}{b}}
\displaystyle =\frac{a+b}{b-a}
\displaystyle \therefore \frac{\cos A+\sin A}{\cos A-\sin A}=\frac{a+b}{b-a}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }\cos A=\frac{12}{13},\text{ find the value of }\sin A(1-\tan A).
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\cos A=\frac{12}{13}
\displaystyle \text{Let the base }=12\text{ and the hypotenuse }=13.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Perpendicular}=\sqrt{13^2-12^2}
\displaystyle =\sqrt{169-144}=5
\displaystyle \therefore \sin A=\frac{5}{13},\qquad \tan A=\frac{5}{12}
\displaystyle \sin A(1-\tan A)
\displaystyle =\frac{5}{13}\left(1-\frac{5}{12}\right)
\displaystyle =\frac{5}{13}\times\frac{7}{12}
\displaystyle =\frac{35}{156}
\displaystyle \therefore \sin A(1-\tan A)=\frac{35}{156}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }\sec A=\frac{5}{4},\text{ find the value of }\frac{\sin A-2\cos A}{\tan A-\cot A}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\sec A=\frac{5}{4}
\displaystyle \therefore \cos A=\frac{4}{5}
\displaystyle \text{Let the base }=4\text{ and the hypotenuse }=5.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Perpendicular}=\sqrt{5^2-4^2}=\sqrt9=3
\displaystyle \therefore \sin A=\frac35,\quad \cos A=\frac45,\quad \tan A=\frac34,\quad \cot A=\frac43
\displaystyle \frac{\sin A-2\cos A}{\tan A-\cot A}
\displaystyle =\frac{\frac35-2\times\frac45}{\frac34-\frac43}
\displaystyle =\frac{-1}{-\frac7{12}}
\displaystyle =\frac{12}{7}
\displaystyle \therefore \frac{\sin A-2\cos A}{\tan A-\cot A}=\frac{12}{7}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }\sin A=\frac{3}{5},\text{ evaluate }\frac{\cos A-\frac{1}{\tan A}}{2\cot A}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\sin A=\frac35
\displaystyle \text{Let the perpendicular }=3\text{ and the hypotenuse }=5.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Base}=\sqrt{5^2-3^2}=\sqrt{25-9}=4
\displaystyle \therefore \cos A=\frac45,\qquad \tan A=\frac34,\qquad \cot A=\frac43
\displaystyle \frac{\cos A-\frac{1}{\tan A}}{2\cot A}
\displaystyle =\frac{\frac45-\frac{1}{\frac34}}{2\times\frac43}
\displaystyle =\frac{\frac45-\frac43}{\frac83}
\displaystyle =\frac{-\frac8{15}}{\frac83}
\displaystyle =-\frac8{15}\times\frac38=-\frac15
\displaystyle \therefore \frac{\cos A-\frac{1}{\tan A}}{2\cot A}=-\frac15
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }\sec A=\frac54,\text{ verify } \\ \\ \frac{3\sin A-4\sin^3 A}{4\cos^3 A-3\cos A}  =\frac{3\tan A-\tan^3 A}{1-3\tan^2 A}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\sec A=\frac54
\displaystyle \therefore \cos A=\frac45
\displaystyle \text{Let the base }=4\text{ and the hypotenuse }=5.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Perpendicular}=\sqrt{5^2-4^2}=\sqrt9=3
\displaystyle \therefore \sin A=\frac35,\qquad \cos A=\frac45,\qquad \tan A=\frac34
\displaystyle \text{L.H.S.}=\frac{3\sin A-4\sin^3 A}{4\cos^3 A-3\cos A}
\displaystyle =\frac{3\left(\frac35\right)-4\left(\frac35\right)^3}{4\left(\frac45\right)^3-3\left(\frac45\right)}
\displaystyle =\frac{\frac95-\frac{108}{125}}{\frac{256}{125}-\frac{12}{5}}
\displaystyle =\frac{\frac{225-108}{125}}{\frac{256-300}{125}}
\displaystyle =\frac{117}{-44}=-\frac{117}{44}
\displaystyle \text{R.H.S.}=\frac{3\tan A-\tan^3 A}{1-3\tan^2 A}
\displaystyle =\frac{3\left(\frac34\right)-\left(\frac34\right)^3}{1-3\left(\frac34\right)^2}
\displaystyle =\frac{\frac94-\frac{27}{64}}{1-\frac{27}{16}}
\displaystyle =\frac{\frac{144-27}{64}}{\frac{16-27}{16}}
\displaystyle =\frac{\frac{117}{64}}{-\frac{11}{16}}
\displaystyle =\frac{117}{64}\times\frac{16}{-11}=-\frac{117}{44}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence verified.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }\sin A=\frac{3}{4},\text{ prove that }\sqrt{\frac{\mathrm{cosec}^2 A-\cot^2 A}{\sec^2 A-1}}=\frac{\sqrt7}{3}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\sin A=\frac34
\displaystyle \text{Let the perpendicular }=3\text{ and the hypotenuse }=4.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Base}=\sqrt{4^2-3^2}=\sqrt7
\displaystyle \therefore \cos A=\frac{\sqrt7}{4},\quad \sec A=\frac4{\sqrt7},\quad \cot A=\frac{\sqrt7}{3},\quad \mathrm{cosec}\,A=\frac43
\displaystyle \text{L.H.S.}=\sqrt{\frac{\mathrm{cosec}^2 A-\cot^2 A}{\sec^2 A-1}}
\displaystyle =\sqrt{\frac{\left(\frac43\right)^2-\left(\frac{\sqrt7}{3}\right)^2}{\left(\frac4{\sqrt7}\right)^2-1}}
\displaystyle =\sqrt{\frac{\frac{16}{9}-\frac79}{\frac{16}{7}-1}}
\displaystyle =\sqrt{\frac{1}{\frac97}}=\sqrt{\frac79}
\displaystyle =\frac{\sqrt7}{3}=\text{R.H.S.}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }8\tan A=15,\text{ find }\sin A-\cos A.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }8\tan A=15
\displaystyle \Rightarrow \tan A=\frac{15}{8}
\displaystyle \text{Let the perpendicular }=15\text{ and the base }=8.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Hypotenuse}=\sqrt{15^2+8^2}
\displaystyle =\sqrt{225+64}=\sqrt{289}=17
\displaystyle \therefore \sin A=\frac{15}{17}\text{ and }\cos A=\frac8{17}
\displaystyle \sin A-\cos A=\frac{15}{17}-\frac8{17}
\displaystyle =\frac7{17}
\displaystyle \therefore \sin A-\cos A=\frac7{17}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }3\cos A-4\sin A=2\cos A+\sin A,\text{ find }\tan A.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }3\cos A-4\sin A=2\cos A+\sin A
\displaystyle \Rightarrow \cos A=5\sin A
\displaystyle \Rightarrow \frac{\sin A}{\cos A}=\frac15
\displaystyle \therefore \tan A=\frac15
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }\angle A\text{ and }\angle B\text{ are acute angles such that }\cos A=\cos B,
\displaystyle \text{show that }\angle A=\angle B.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\cos A=\cos B
\displaystyle \text{For acute angles, }\cos\theta\text{ decreases as }\theta\text{ increases.}
\displaystyle \therefore \cos A=\cos B\Rightarrow A=B
\displaystyle \therefore \angle A=\angle B.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{In a triangle, }\angle A\text{ and }\angle B\text{ are acute angles such that }\tan A=\tan B,
\displaystyle \text{show that }\angle A=\angle B.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\tan A=\tan B
\displaystyle \text{For acute angles, }\tan\theta\text{ increases as }\theta\text{ increases.}
\displaystyle \therefore \tan A=\tan B\Rightarrow A=B
\displaystyle \therefore \angle A=\angle B.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If }A\text{ is an acute angle such that }3\sin A=4\cos A,\text{ find the value of}
\displaystyle 4\sin^2 A-3\cos^2 A+2.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }3\sin A=4\cos A
\displaystyle \Rightarrow \frac{\sin A}{\cos A}=\frac43
\displaystyle \Rightarrow \tan A=\frac43
\displaystyle \text{Let the perpendicular }=4\text{ and the base }=3.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Hypotenuse}=\sqrt{4^2+3^2}=\sqrt{25}=5
\displaystyle \therefore \sin A=\frac45\text{ and }\cos A=\frac35
\displaystyle 4\sin^2 A-3\cos^2 A+2
\displaystyle =4\left(\frac45\right)^2-3\left(\frac35\right)^2+2
\displaystyle =\frac{64}{25}-\frac{27}{25}+2
\displaystyle =\frac{64-27+50}{25}
\displaystyle =\frac{87}{25}
\displaystyle \therefore 4\sin^2 A-3\cos^2 A+2=\frac{87}{25}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{In the adjoining figure, }\triangle ABC\text{ is right-angled at }B\text{ and }D\text{ is the}
\displaystyle \text{midpoint of }BC.\ AC=5\text{ cm},\ BC=4\text{ cm and }\angle BAD=\theta.\text{ Find:}
\displaystyle \text{(i) }\tan\theta\qquad\text{(ii) }\sin\theta\qquad\text{(iii) }\sin^2\theta+\cos^2\theta
\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ABC,\text{ by Pythagoras theorem,}
\displaystyle AB=\sqrt{AC^2-BC^2}
\displaystyle =\sqrt{5^2-4^2}=\sqrt{25-16}=3\text{ cm}
\displaystyle \text{Since }D\text{ is the midpoint of }BC,
\displaystyle BD=\frac{BC}{2}=\frac42=2\text{ cm}
\displaystyle \text{(i) }\tan\theta=\frac{BD}{AB}=\frac23

\displaystyle \text{(ii) In right-angled }\triangle ABD,\text{ by Pythagoras theorem,}
\displaystyle AD=\sqrt{AB^2+BD^2}
\displaystyle =\sqrt{3^2+2^2}=\sqrt{13}\text{ cm}
\displaystyle \therefore \sin\theta=\frac{BD}{AD}=\frac{2}{\sqrt{13}}

\displaystyle \text{(iii) }\sin^2\theta+\cos^2\theta
\displaystyle =\left(\frac{2}{\sqrt{13}}\right)^2+\left(\frac{3}{\sqrt{13}}\right)^2
\displaystyle =\frac4{13}+\frac9{13}
\displaystyle =\frac{13}{13}=1
\displaystyle \\

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