\displaystyle \textbf{Question 1: }\text{Evaluate }\sin45^\circ\sin30^\circ+\cos45^\circ\cos30^\circ.
\displaystyle \text{Answer:}
\displaystyle \sin45^\circ=\frac{1}{\sqrt2},\quad\sin30^\circ=\frac12,\quad\cos45^\circ=\frac{1}{\sqrt2},\quad\cos30^\circ=\frac{\sqrt3}{2}
\displaystyle \therefore \sin45^\circ\sin30^\circ+\cos45^\circ\cos30^\circ
\displaystyle =\left(\frac{1}{\sqrt2}\right)\left(\frac12\right)+\left(\frac{1}{\sqrt2}\right)\left(\frac{\sqrt3}{2}\right)
\displaystyle =\frac{1}{2\sqrt2}+\frac{\sqrt3}{2\sqrt2}=\frac{1+\sqrt3}{2\sqrt2}
\displaystyle =\frac{\sqrt2+\sqrt6}{4}
\displaystyle \therefore \text{The required value is }\frac{\sqrt6+\sqrt2}{4}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Evaluate }\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ.
\displaystyle \text{Answer:}
\displaystyle \cos60^\circ=\frac12,\quad\cos45^\circ=\frac{1}{\sqrt2},\quad\sin60^\circ=\frac{\sqrt3}{2},\quad\sin45^\circ=\frac{1}{\sqrt2}
\displaystyle \therefore \cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ
\displaystyle =\left(\frac12\right)\left(\frac{1}{\sqrt2}\right)-\left(\frac{\sqrt3}{2}\right)\left(\frac{1}{\sqrt2}\right)
\displaystyle =\frac{1}{2\sqrt2}-\frac{\sqrt3}{2\sqrt2}=\frac{1-\sqrt3}{2\sqrt2}
\displaystyle =\frac{\sqrt2-\sqrt6}{4}
\displaystyle \therefore \text{The required value is }\frac{\sqrt2-\sqrt6}{4}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Evaluate: } \\ \sin^2 30^\circ\cos^2 45^\circ+4\tan^2 30^\circ+\frac{1}{2}\sin^2 90^\circ
\displaystyle \qquad -2\cos^2 90^\circ+\frac{1}{24}\cos^2 0^\circ.
\displaystyle \text{Answer:}
\displaystyle \sin30^\circ=\frac12,\quad\cos45^\circ=\frac{1}{\sqrt2},\quad\tan30^\circ=\frac{1}{\sqrt3}
\displaystyle \sin90^\circ=1,\quad\cos90^\circ=0,\quad\cos0^\circ=1
\displaystyle \therefore \sin^2 30^\circ\cos^2 45^\circ+4\tan^2 30^\circ+\frac12\sin^2 90^\circ
\displaystyle \qquad -2\cos^2 90^\circ+\frac{1}{24}\cos^2 0^\circ
\displaystyle =\left(\frac12\right)^2\left(\frac{1}{\sqrt2}\right)^2+4\left(\frac{1}{\sqrt3}\right)^2+\frac12(1)^2-2(0)^2+\frac{1}{24}(1)^2
\displaystyle =\frac18+\frac43+\frac12+\frac{1}{24}
\displaystyle =\frac{3+32+12+1}{24}=\frac{48}{24}=2
\displaystyle \therefore \text{The required value is }2.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Evaluate: } \\ 4(\sin^4 60^\circ+\cos^4 30^\circ)-3(\tan^2 60^\circ-\tan^2 45^\circ)
\displaystyle \qquad +5\cos^2 45^\circ.
\displaystyle \text{Answer:}
\displaystyle \sin60^\circ=\frac{\sqrt3}{2},\quad\cos30^\circ=\frac{\sqrt3}{2},\quad\tan60^\circ=\sqrt3
\displaystyle \tan45^\circ=1,\qquad\cos45^\circ=\frac{1}{\sqrt2}
\displaystyle \therefore 4(\sin^4 60^\circ+\cos^4 30^\circ)-3(\tan^2 60^\circ-\tan^2 45^\circ)+5\cos^2 45^\circ
\displaystyle =4\left[\left(\frac{\sqrt3}{2}\right)^4+\left(\frac{\sqrt3}{2}\right)^4\right]-3\left[(\sqrt3)^2-1^2\right]+5\left(\frac{1}{\sqrt2}\right)^2
\displaystyle =4\left(\frac{9}{16}+\frac{9}{16}\right)-3(3-1)+\frac52
\displaystyle =4\left(\frac98\right)-6+\frac52
\displaystyle =\frac92-6+\frac52=7-6=1
\displaystyle \therefore \text{The required value is }1.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Evaluate: }(\mathrm{cosec}^2 45^\circ\sec^2 30^\circ)(\sin^2 30^\circ+4\cot^2 45^\circ-\sec^2 60^\circ).
\displaystyle \text{Answer:}
\displaystyle \mathrm{cosec}\,45^\circ=\sqrt2,\quad\sec30^\circ=\frac{2}{\sqrt3},\quad\sin30^\circ=\frac12
\displaystyle \cot45^\circ=1,\qquad\sec60^\circ=2
\displaystyle \therefore (\mathrm{cosec}^2 45^\circ\sec^2 30^\circ)(\sin^2 30^\circ+4\cot^2 45^\circ-\sec^2 60^\circ)
\displaystyle =\left[(\sqrt2)^2\left(\frac{2}{\sqrt3}\right)^2\right]\left[\left(\frac12\right)^2+4(1)^2-(2)^2\right]
\displaystyle =\left(2\times\frac43\right)\left(\frac14+4-4\right)
\displaystyle =\frac83\times\frac14=\frac23
\displaystyle \therefore \text{The required value is }\frac23.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Evaluate: }\mathrm{cosec}^3 30^\circ\cos60^\circ\tan^3 45^\circ\sin^2 90^\circ\sec^2 45^\circ\cot30^\circ.
\displaystyle \text{Answer:}
\displaystyle \mathrm{cosec}\,30^\circ=2,\quad\cos60^\circ=\frac12,\quad\tan45^\circ=1
\displaystyle \sin90^\circ=1,\quad\sec45^\circ=\sqrt2,\quad\cot30^\circ=\sqrt3
\displaystyle \therefore \mathrm{cosec}^3 30^\circ\cos60^\circ\tan^3 45^\circ\sin^2 90^\circ\sec^2 45^\circ\cot30^\circ
\displaystyle =(2)^3\left(\frac12\right)(1)^3(1)^2(\sqrt2)^2(\sqrt3)
\displaystyle =8\times\frac12\times1\times1\times2\times\sqrt3
\displaystyle =8\sqrt3
\displaystyle \therefore \text{The required value is }8\sqrt3.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Evaluate: }\cot^2 30^\circ-2\cos^2 60^\circ-\frac34\sec^2 45^\circ-4\sec^2 30^\circ.
\displaystyle \text{Answer:}
\displaystyle \cot30^\circ=\sqrt3,\quad\cos60^\circ=\frac12,\quad\sec45^\circ=\sqrt2,\quad\sec30^\circ=\frac{2}{\sqrt3}
\displaystyle \therefore \cot^2 30^\circ-2\cos^2 60^\circ-\frac34\sec^2 45^\circ-4\sec^2 30^\circ
\displaystyle =(\sqrt3)^2-2\left(\frac12\right)^2-\frac34(\sqrt2)^2-4\left(\frac{2}{\sqrt3}\right)^2
\displaystyle =3-\frac12-\frac32-\frac{16}{3}
\displaystyle =1-\frac{16}{3}=-\frac{13}{3}
\displaystyle \therefore \text{The required value is }-\frac{13}{3}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Evaluate: }\frac{\tan^2 60^\circ+4\cos^2 45^\circ+3\sec^2 30^\circ+5\cos^2 90^\circ}{\mathrm{cosec}\,30^\circ+\sec60^\circ-\cot^2 30^\circ}.
\displaystyle \text{Answer:}
\displaystyle \tan60^\circ=\sqrt3,\quad\cos45^\circ=\frac{1}{\sqrt2},\quad\sec30^\circ=\frac{2}{\sqrt3}
\displaystyle \cos90^\circ=0,\quad\mathrm{cosec}\,30^\circ=2,\quad\sec60^\circ=2,\quad\cot30^\circ=\sqrt3
\displaystyle \therefore \frac{\tan^2 60^\circ+4\cos^2 45^\circ+3\sec^2 30^\circ+5\cos^2 90^\circ}{\mathrm{cosec}\,30^\circ+\sec60^\circ-\cot^2 30^\circ}
\displaystyle =\frac{(\sqrt3)^2+4\left(\frac{1}{\sqrt2}\right)^2+3\left(\frac{2}{\sqrt3}\right)^2+5(0)^2}{2+2-(\sqrt3)^2}
\displaystyle =\frac{3+2+4+0}{4-3}
\displaystyle =\frac{9}{1}=9
\displaystyle \therefore \text{The required value is }9.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Evaluate:}
\displaystyle \text{(i) }\frac{\tan45^\circ}{\mathrm{cosec}\,30^\circ}+\frac{\sec60^\circ}{\cot45^\circ}-\frac{5\sin90^\circ}{2\cos0^\circ}

\displaystyle \text{Answer:}
\displaystyle \tan45^\circ=1,\quad\mathrm{cosec}\,30^\circ=2,\quad\sec60^\circ=2
\displaystyle \cot45^\circ=1,\quad\sin90^\circ=1,\quad\cos0^\circ=1
\displaystyle \therefore \frac{\tan45^\circ}{\mathrm{cosec}\,30^\circ}+\frac{\sec60^\circ}{\cot45^\circ}-\frac{5\sin90^\circ}{2\cos0^\circ}
\displaystyle =\frac12+\frac21-\frac{5(1)}{2(1)}
\displaystyle =\frac12+2-\frac52
\displaystyle =\frac{1+4-5}{2}=0
\displaystyle \therefore \text{The required value is }0.

\displaystyle \text{(ii) }\frac{5\cos^2 60^\circ+4\sec^2 30^\circ-\tan^2 45^\circ}{\sin^2 30^\circ+\cos^2 30^\circ}   \hfill\text{[CBSE 2023]}

\displaystyle \text{Answer:}
\displaystyle \cos60^\circ=\frac12,\quad\sec30^\circ=\frac{2}{\sqrt3},\quad\tan45^\circ=1
\displaystyle \sin30^\circ=\frac12,\qquad\cos30^\circ=\frac{\sqrt3}{2}
\displaystyle \therefore \frac{5\cos^2 60^\circ+4\sec^2 30^\circ-\tan^2 45^\circ}{\sin^2 30^\circ+\cos^2 30^\circ}
\displaystyle =\frac{5\left(\frac12\right)^2+4\left(\frac{2}{\sqrt3}\right)^2-(1)^2}{\left(\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}
\displaystyle =\frac{\frac54+\frac{16}{3}-1}{\frac14+\frac34}
\displaystyle =\frac{\frac14+\frac{16}{3}}{1}
\displaystyle =\frac{3+64}{12}=\frac{67}{12}
\displaystyle \therefore \text{The required value is }\frac{67}{12}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the value of }x\text{ in each of the following:}
\displaystyle \text{(i) }2\sin\frac{x}{2}=1\qquad\text{(ii) }\sqrt3\sin x=\cos x
\displaystyle \text{(iii) }\sqrt3\tan2x=\cos60^\circ+\sin45^\circ\cos45^\circ
\displaystyle \text{Answer:}
\displaystyle \text{(i) }2\sin\frac{x}{2}=1
\displaystyle \sin\frac{x}{2}=\frac12
\displaystyle \sin\frac{x}{2}=\sin30^\circ
\displaystyle \therefore \frac{x}{2}=30^\circ
\displaystyle \therefore x=60^\circ.

\displaystyle \text{(ii) }\sqrt3\sin x=\cos x
\displaystyle \frac{\sin x}{\cos x}=\frac{1}{\sqrt3}
\displaystyle \tan x=\frac{1}{\sqrt3}=\tan30^\circ
\displaystyle \therefore x=30^\circ.

\displaystyle \text{(iii) }\sqrt3\tan2x=\cos60^\circ+\sin45^\circ\cos45^\circ
\displaystyle \sqrt3\tan2x=\frac12+\left(\frac{1}{\sqrt2}\right)\left(\frac{1}{\sqrt2}\right)
\displaystyle \sqrt3\tan2x=\frac12+\frac12=1
\displaystyle \tan2x=\frac{1}{\sqrt3}=\tan30^\circ
\displaystyle \therefore 2x=30^\circ
\displaystyle \therefore x=15^\circ.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }\theta=30^\circ,\text{ verify that:}
\displaystyle \text{(i) }\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}\qquad\text{(ii) }\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}
\displaystyle \text{Answer:}
\displaystyle \text{(i) For }\theta=30^\circ,\quad 2\theta=60^\circ
\displaystyle \text{LHS}=\sin2\theta=\sin60^\circ=\frac{\sqrt3}{2}
\displaystyle \text{RHS}=\frac{2\tan\theta}{1+\tan^2\theta}
\displaystyle =\frac{2\tan30^\circ}{1+\tan^2 30^\circ}
\displaystyle =\frac{2\left(\frac{1}{\sqrt3}\right)}{1+\left(\frac{1}{\sqrt3}\right)^2}
\displaystyle =\frac{\frac{2}{\sqrt3}}{1+\frac13}=\frac{\frac{2}{\sqrt3}}{\frac43}
\displaystyle =\frac{2}{\sqrt3}\times\frac34=\frac{\sqrt3}{2}
\displaystyle \therefore \text{LHS}=\text{RHS}=\frac{\sqrt3}{2}.
\displaystyle \text{Hence verified.}

\displaystyle \text{(ii) For }\theta=30^\circ,\quad 2\theta=60^\circ
\displaystyle \text{LHS}=\cos2\theta=\cos60^\circ=\frac12
\displaystyle \text{RHS}=\frac{1-\tan^2\theta}{1+\tan^2\theta}
\displaystyle =\frac{1-\tan^2 30^\circ}{1+\tan^2 30^\circ}
\displaystyle =\frac{1-\left(\frac{1}{\sqrt3}\right)^2}{1+\left(\frac{1}{\sqrt3}\right)^2}
\displaystyle =\frac{1-\frac13}{1+\frac13}=\frac{\frac23}{\frac43}=\frac12
\displaystyle \therefore \text{LHS}=\text{RHS}=\frac12.
\displaystyle \text{Hence verified.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }A=B=60^\circ,\text{ verify that:}
\displaystyle \text{(i) }\cos(A-B)=\cos A\cos B+\sin A\sin B
\displaystyle \text{(ii) }\sin(A-B)=\sin A\cos B-\cos A\sin B
\displaystyle \text{(iii) }\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=B=60^\circ.
\displaystyle \text{(i) LHS}=\cos(A-B)
\displaystyle =\cos(60^\circ-60^\circ)=\cos0^\circ=1
\displaystyle \text{RHS}=\cos A\cos B+\sin A\sin B
\displaystyle =\cos60^\circ\cos60^\circ+\sin60^\circ\sin60^\circ
\displaystyle =\left(\frac12\right)\left(\frac12\right)+\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt3}{2}\right)
\displaystyle =\frac14+\frac34=1
\displaystyle \therefore \text{LHS}=\text{RHS}=1.
\displaystyle \text{Hence verified.}

\displaystyle \text{(ii) LHS}=\sin(A-B)
\displaystyle =\sin(60^\circ-60^\circ)=\sin0^\circ=0
\displaystyle \text{RHS}=\sin A\cos B-\cos A\sin B
\displaystyle =\sin60^\circ\cos60^\circ-\cos60^\circ\sin60^\circ
\displaystyle =\left(\frac{\sqrt3}{2}\right)\left(\frac12\right)-\left(\frac12\right)\left(\frac{\sqrt3}{2}\right)
\displaystyle =\frac{\sqrt3}{4}-\frac{\sqrt3}{4}=0
\displaystyle \therefore \text{LHS}=\text{RHS}=0.
\displaystyle \text{Hence verified.}

\displaystyle \text{(iii) LHS}=\tan(A-B)
\displaystyle =\tan(60^\circ-60^\circ)=\tan0^\circ=0
\displaystyle \text{RHS}=\frac{\tan A-\tan B}{1+\tan A\tan B}
\displaystyle =\frac{\tan60^\circ-\tan60^\circ}{1+\tan60^\circ\tan60^\circ}
\displaystyle =\frac{\sqrt3-\sqrt3}{1+(\sqrt3)(\sqrt3)}
\displaystyle =\frac{0}{4}=0
\displaystyle \therefore \text{LHS}=\text{RHS}=0.
\displaystyle \text{Hence verified.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Prove that }(\sqrt3+1)(3-\cot30^\circ)=\tan^3 60^\circ-2\sin60^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(\sqrt3+1)(3-\cot30^\circ)
\displaystyle =(\sqrt3+1)(3-\sqrt3)
\displaystyle =3\sqrt3-3+3-\sqrt3
\displaystyle =2\sqrt3
\displaystyle \text{RHS}=\tan^3 60^\circ-2\sin60^\circ
\displaystyle =(\sqrt3)^3-2\left(\frac{\sqrt3}{2}\right)
\displaystyle =3\sqrt3-\sqrt3=2\sqrt3
\displaystyle \therefore \text{LHS}=\text{RHS}=2\sqrt3.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{(i) If }\tan(A-B)=\frac{1}{\sqrt3}\text{ and }\tan(A+B)=\sqrt3,\ 
\displaystyle 0^\circ<A+B\leq90^\circ, \ \text{and }A>B,\text{ find }A\text{ and }B.
\displaystyle \text{Answer:}
\displaystyle \tan(A-B)=\frac{1}{\sqrt3}=\tan30^\circ
\displaystyle \therefore A-B=30^\circ
\displaystyle \tan(A+B)=\sqrt3=\tan60^\circ
\displaystyle \therefore A+B=60^\circ
\displaystyle \text{Adding the two equations,}
\displaystyle 2A=90^\circ
\displaystyle \therefore A=45^\circ
\displaystyle \text{Substituting }A=45^\circ\text{ in }A+B=60^\circ,
\displaystyle 45^\circ+B=60^\circ
\displaystyle \therefore B=15^\circ
\displaystyle \therefore A=45^\circ,\qquad B=15^\circ.

\displaystyle \text{(ii) If }\tan(A+B)=1\text{ and }\tan(A-B)=\frac{1}{\sqrt3},\ 0^\circ<A+B<90^\circ,\ A>B,
\displaystyle \text{then find the values of }A\text{ and }B.\hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \tan(A+B)=1=\tan45^\circ
\displaystyle \therefore A+B=45^\circ
\displaystyle \tan(A-B)=\frac{1}{\sqrt3}=\tan30^\circ
\displaystyle \therefore A-B=30^\circ
\displaystyle \text{Adding the two equations,}
\displaystyle 2A=75^\circ
\displaystyle \therefore A=37.5^\circ=37^\circ30'
\displaystyle \text{Substituting }A=37.5^\circ\text{ in }A+B=45^\circ,
\displaystyle 37.5^\circ+B=45^\circ
\displaystyle \therefore B=7.5^\circ=7^\circ30'
\displaystyle \therefore A=37^\circ30',\qquad B=7^\circ30'.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }\sin(A-B)=\frac12\text{ and }\cos(A+B)=\frac12,\ 
\displaystyle 0^\circ<A+B\leq90^\circ,\ A>B, \ \text{find }A\text{ and }B.
\displaystyle \text{Answer:}
\displaystyle \sin(A-B)=\frac12=\sin30^\circ
\displaystyle \therefore A-B=30^\circ
\displaystyle \cos(A+B)=\frac12=\cos60^\circ
\displaystyle \therefore A+B=60^\circ
\displaystyle \text{Adding the two equations,}
\displaystyle 2A=90^\circ
\displaystyle \therefore A=45^\circ
\displaystyle \text{Substituting }A=45^\circ\text{ in }A+B=60^\circ,
\displaystyle 45^\circ+B=60^\circ
\displaystyle \therefore B=15^\circ
\displaystyle \therefore A=45^\circ,\qquad B=15^\circ.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In a }\triangle ABC\text{ right angled at }B,\ \angle A=\angle C.\text{ Find the values of}
\displaystyle \text{(i) }\sin A\cos C+\cos A\sin C\qquad\text{(ii) }\sin A\sin B+\cos A\cos B.
\displaystyle \text{Answer:}
\displaystyle \angle A+\angle B+\angle C=180^\circ
\displaystyle \angle A+\angle C=180^\circ-90^\circ=90^\circ
\displaystyle \text{Since }\angle A=\angle C,\quad A=C=45^\circ.
\displaystyle \text{Also, }B=90^\circ.
\displaystyle \text{(i) }\sin A\cos C+\cos A\sin C
\displaystyle =\sin45^\circ\cos45^\circ+\cos45^\circ\sin45^\circ
\displaystyle =\left(\frac{1}{\sqrt2}\right)\left(\frac{1}{\sqrt2}\right)+\left(\frac{1}{\sqrt2}\right)\left(\frac{1}{\sqrt2}\right)
\displaystyle =\frac12+\frac12=1
\displaystyle \therefore \text{The required value is }1.

\displaystyle \text{(ii) }\sin A\sin B+\cos A\cos B
\displaystyle =\sin45^\circ\sin90^\circ+\cos45^\circ\cos90^\circ
\displaystyle =\left(\frac{1}{\sqrt2}\right)(1)+\left(\frac{1}{\sqrt2}\right)(0)
\displaystyle =\frac{1}{\sqrt2}
\displaystyle \therefore \text{The required value is }\frac{1}{\sqrt2}.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }4\cot^2 45^\circ-\sec^2 60^\circ+\sin^2 60^\circ+p=\frac34,\text{ find the value of }p.
\displaystyle \hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle 4\cot^2 45^\circ-\sec^2 60^\circ+\sin^2 60^\circ+p=\frac34
\displaystyle 4(1)^2-(2)^2+\left(\frac{\sqrt3}{2}\right)^2+p=\frac34
\displaystyle 4-4+\frac34+p=\frac34
\displaystyle \frac34+p=\frac34
\displaystyle \therefore p=0.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }A\text{ and }B\text{ are acute angles such that }\sin(A-B)=0\text{ and}
\displaystyle 2\cos(A+B)-1=0,\text{ then find angles }A\text{ and }B.\hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \sin(A-B)=0
\displaystyle \sin(A-B)=\sin0^\circ
\displaystyle \therefore A-B=0^\circ
\displaystyle \therefore A=B
\displaystyle \text{Also, }2\cos(A+B)-1=0
\displaystyle 2\cos(A+B)=1
\displaystyle \cos(A+B)=\frac12=\cos60^\circ
\displaystyle \therefore A+B=60^\circ
\displaystyle \text{Since }A=B,
\displaystyle 2A=60^\circ
\displaystyle \therefore A=30^\circ
\displaystyle \therefore B=30^\circ
\displaystyle \therefore A=30^\circ,\qquad B=30^\circ.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Find acute angles }A\text{ and }B,\text{ if }\sin(A+2B)=\frac{\sqrt3}{2}\text{ and}
\displaystyle \cos(A+4B)=0,\ A>B.
\displaystyle \text{Answer:}
\displaystyle \sin(A+2B)=\frac{\sqrt3}{2}=\sin60^\circ
\displaystyle \therefore A+2B=60^\circ
\displaystyle \cos(A+4B)=0=\cos90^\circ
\displaystyle \therefore A+4B=90^\circ
\displaystyle \text{Subtracting }A+2B=60^\circ\text{ from }A+4B=90^\circ,
\displaystyle 2B=30^\circ
\displaystyle \therefore B=15^\circ
\displaystyle \text{Substituting }B=15^\circ\text{ in }A+2B=60^\circ,
\displaystyle A+30^\circ=60^\circ
\displaystyle \therefore A=30^\circ
\displaystyle \therefore A=30^\circ,\qquad B=15^\circ.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{In }\triangle PQR,\text{ right-angled at }Q,\ PQ=3\text{ cm and }PR=6\text{ cm. Determine}
\displaystyle \angle P\text{ and }\angle R.
\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle PQR,\quad PR=6\text{ cm is the hypotenuse.}
\displaystyle \cos P=\frac{PQ}{PR}=\frac{3}{6}=\frac12
\displaystyle \cos P=\cos60^\circ
\displaystyle \therefore \angle P=60^\circ
\displaystyle \angle P+\angle Q+\angle R=180^\circ
\displaystyle 60^\circ+90^\circ+\angle R=180^\circ
\displaystyle \therefore \angle R=30^\circ
\displaystyle \therefore \angle P=60^\circ,\qquad\angle R=30^\circ.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }\sin(A-B)=\sin A\cos B-\cos A\sin B\text{ and}
\displaystyle \cos(A-B)=\cos A\cos B+\sin A\sin B,\text{ find the values of }\sin15^\circ\text{ and }\cos15^\circ.
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle 15^\circ=45^\circ-30^\circ
\displaystyle \sin15^\circ=\sin(45^\circ-30^\circ)
\displaystyle =\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ
\displaystyle =\left(\frac{1}{\sqrt2}\right)\left(\frac{\sqrt3}{2}\right)-\left(\frac{1}{\sqrt2}\right)\left(\frac12\right)
\displaystyle =\frac{\sqrt3-1}{2\sqrt2}
\displaystyle =\frac{\sqrt6-\sqrt2}{4}
\displaystyle \therefore \sin15^\circ=\frac{\sqrt6-\sqrt2}{4}.
\displaystyle \cos15^\circ=\cos(45^\circ-30^\circ)
\displaystyle =\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ
\displaystyle =\left(\frac{1}{\sqrt2}\right)\left(\frac{\sqrt3}{2}\right)+\left(\frac{1}{\sqrt2}\right)\left(\frac12\right)
\displaystyle =\frac{\sqrt3+1}{2\sqrt2}
\displaystyle =\frac{\sqrt6+\sqrt2}{4}
\displaystyle \therefore \cos15^\circ=\frac{\sqrt6+\sqrt2}{4}.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{In a right triangle }ABC,\text{ right angled at }C,\text{ if }\angle B=60^\circ\text{ and }
\displaystyle AB=15\text{ units.} \ \text{Find the remaining angles and sides.}
\displaystyle \text{Answer:}
\displaystyle \angle A+\angle B+\angle C=180^\circ
\displaystyle \angle A+60^\circ+90^\circ=180^\circ
\displaystyle \therefore \angle A=30^\circ
\displaystyle \sin30^\circ=\frac{BC}{AB}
\displaystyle \frac12=\frac{BC}{15}
\displaystyle \therefore BC=\frac{15}{2}\text{ units}.
\displaystyle \cos30^\circ=\frac{AC}{AB}
\displaystyle \frac{\sqrt3}{2}=\frac{AC}{15}
\displaystyle \therefore AC=\frac{15\sqrt3}{2}\text{ units}.
\displaystyle \therefore \angle A=30^\circ,\quad BC=\frac{15}{2}\text{ units},\quad AC=\frac{15\sqrt3}{2}\text{ units}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If }\triangle ABC\text{ is a right triangle such that }\angle C=90^\circ,\ \angle A=45^\circ\text{ and}
\displaystyle BC=7\text{ units. Find }\angle B,\ AB\text{ and }AC.
\displaystyle \text{Answer:}
\displaystyle \angle A+\angle B+\angle C=180^\circ
\displaystyle 45^\circ+\angle B+90^\circ=180^\circ
\displaystyle \therefore \angle B=45^\circ
\displaystyle \tan45^\circ=\frac{BC}{AC}
\displaystyle 1=\frac{7}{AC}
\displaystyle \therefore AC=7\text{ units}.
\displaystyle \sin45^\circ=\frac{BC}{AB}
\displaystyle \frac{1}{\sqrt2}=\frac{7}{AB}
\displaystyle \therefore AB=7\sqrt2\text{ units}.
\displaystyle \therefore \angle B=45^\circ,\quad AB=7\sqrt2\text{ units},\quad AC=7\text{ units}.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If }A\text{ and }B\text{ are acute angles such that }\tan A=\frac12,\ 
\displaystyle \tan B=\frac13\text{ and} \ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B},\text{ find }A+B.
\displaystyle \text{Answer:}
\displaystyle \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}
\displaystyle =\frac{\frac12+\frac13}{1-\left(\frac12\right)\left(\frac13\right)}
\displaystyle =\frac{\frac56}{1-\frac16}=\frac{\frac56}{\frac56}=1
\displaystyle \therefore \tan(A+B)=1=\tan45^\circ
\displaystyle \text{Since }A\text{ and }B\text{ are acute angles, }0^\circ<A+B<180^\circ.
\displaystyle \therefore A+B=45^\circ.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If }\tan A=\sqrt3,\text{ where }A\text{ is an acute angle, then find the value of}
\displaystyle \frac{\sin^2 A}{1+\cos^2 A}.\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \tan A=\sqrt3=\tan60^\circ
\displaystyle \therefore A=60^\circ
\displaystyle \frac{\sin^2 A}{1+\cos^2 A}
\displaystyle =\frac{\sin^2 60^\circ}{1+\cos^2 60^\circ}
\displaystyle =\frac{\left(\frac{\sqrt3}{2}\right)^2}{1+\left(\frac12\right)^2}
\displaystyle =\frac{\frac34}{1+\frac14}=\frac{\frac34}{\frac54}
\displaystyle =\frac35
\displaystyle \therefore \text{The required value is }\frac35.
\displaystyle \\

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