\displaystyle \textbf{Question 1: }\text{Write the maximum and minimum values of }\sin\theta.
\displaystyle \text{Answer:}
\displaystyle -1\leq\sin\theta\leq1
\displaystyle \therefore \text{Maximum value of }\sin\theta=1.
\displaystyle \therefore \text{Minimum value of }\sin\theta=-1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Write the maximum and minimum values of }\cos\theta.
\displaystyle \text{Answer:}
\displaystyle -1\leq\cos\theta\leq1
\displaystyle \therefore \text{Maximum value of }\cos\theta=1.
\displaystyle \therefore \text{Minimum value of }\cos\theta=-1.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{What is the maximum value of }\frac{1}{\sec\theta}\text{?}
\displaystyle \text{Answer:}
\displaystyle \frac{1}{\sec\theta}=\cos\theta
\displaystyle \text{Maximum value of }\cos\theta=1.
\displaystyle \therefore \text{The maximum value of }\frac{1}{\sec\theta}\text{ is }1.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{What is the maximum value of }\frac{1}{\mathrm{cosec}\,\theta}\text{?}
\displaystyle \text{Answer:}
\displaystyle \frac{1}{\mathrm{cosec}\,\theta}=\sin\theta
\displaystyle \text{Maximum value of }\sin\theta=1.
\displaystyle \therefore \text{The maximum value of }\frac{1}{\mathrm{cosec}\,\theta}\text{ is }1.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }\tan\theta=\frac45,\text{ find the value of }\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}.
\displaystyle \text{Answer:}
\displaystyle \frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}
\displaystyle =\frac{1-\frac{\sin\theta}{\cos\theta}}{1+\frac{\sin\theta}{\cos\theta}}
\displaystyle =\frac{1-\tan\theta}{1+\tan\theta}
\displaystyle =\frac{1-\frac45}{1+\frac45}
\displaystyle =\frac{\frac15}{\frac95}=\frac19
\displaystyle \therefore \text{The required value is }\frac19.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }\cos\theta=\frac23,\text{ find the value of }\frac{\sec\theta-1}{\sec\theta+1}.
\displaystyle \text{Answer:}
\displaystyle \cos\theta=\frac23\quad\therefore\quad\sec\theta=\frac32
\displaystyle \frac{\sec\theta-1}{\sec\theta+1}=\frac{\frac32-1}{\frac32+1}
\displaystyle =\frac{\frac12}{\frac52}=\frac15
\displaystyle \therefore \text{The required value is }\frac15.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }3\cot\theta=4,\text{ find the value of }\frac{4\cos\theta-\sin\theta}{2\cos\theta+\sin\theta}.
\displaystyle \text{Answer:}
\displaystyle 3\cot\theta=4\quad\therefore\quad\cot\theta=\frac43
\displaystyle \therefore \cos\theta=\frac45,\qquad\sin\theta=\frac35
\displaystyle \frac{4\cos\theta-\sin\theta}{2\cos\theta+\sin\theta}
\displaystyle =\frac{4\left(\frac45\right)-\frac35}{2\left(\frac45\right)+\frac35}
\displaystyle =\frac{\frac{13}{5}}{\frac{11}{5}}=\frac{13}{11}
\displaystyle \therefore \text{The required value is }\frac{13}{11}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Given }\tan\theta=\frac{1}{\sqrt5},\text{ what is the value of}
\displaystyle \frac{\mathrm{cosec}^2\theta-\sec^2\theta}{\mathrm{cosec}^2\theta+\sec^2\theta}\text{?}
\displaystyle \text{Answer:}
\displaystyle \tan\theta=\frac{1}{\sqrt5}=\frac{\text{Perpendicular}}{\text{Base}}
\displaystyle \text{Let perpendicular}=1,\quad\text{base}=\sqrt5,\quad\text{hypotenuse}=\sqrt6.
\displaystyle \therefore \mathrm{cosec}^2\theta=6,\qquad\sec^2\theta=\frac65
\displaystyle \frac{\mathrm{cosec}^2\theta-\sec^2\theta}{\mathrm{cosec}^2\theta+\sec^2\theta}
\displaystyle =\frac{6-\frac65}{6+\frac65}=\frac{\frac{24}{5}}{\frac{36}{5}}=\frac23
\displaystyle \therefore \text{The required value is }\frac23.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }\cot\theta=\frac{1}{\sqrt3},\text{ write the value of }\frac{1-\cos^2\theta}{2-\sin^2\theta}.
\displaystyle \text{Answer:}
\displaystyle \cot\theta=\frac{1}{\sqrt3}=\cot60^\circ
\displaystyle \therefore \theta=60^\circ
\displaystyle \frac{1-\cos^2\theta}{2-\sin^2\theta}
\displaystyle =\frac{1-\cos^2 60^\circ}{2-\sin^2 60^\circ}
\displaystyle =\frac{1-\frac14}{2-\frac34}=\frac{\frac34}{\frac54}=\frac35
\displaystyle \therefore \text{The required value is }\frac35.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Write the acute angle }\theta\text{ satisfying }\sqrt3\sin\theta=\cos\theta.
\displaystyle \text{Answer:}
\displaystyle \sqrt3\sin\theta=\cos\theta
\displaystyle \frac{\sin\theta}{\cos\theta}=\frac{1}{\sqrt3}
\displaystyle \tan\theta=\frac{1}{\sqrt3}=\tan30^\circ
\displaystyle \therefore \theta=30^\circ.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Write the value of }\cos1^\circ\cos2^\circ\cos3^\circ\ldots\cos179^\circ\cos180^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{The given product contains the factor }\cos90^\circ.
\displaystyle \cos90^\circ=0
\displaystyle \therefore \cos1^\circ\cos2^\circ\cos3^\circ\ldots\cos179^\circ\cos180^\circ=0.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Evaluate: }\sin^2 60^\circ+2\tan45^\circ-\cos^2 30^\circ.   \hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \sin60^\circ=\frac{\sqrt3}{2},\quad\tan45^\circ=1,\quad\cos30^\circ=\frac{\sqrt3}{2}
\displaystyle \sin^2 60^\circ+2\tan45^\circ-\cos^2 30^\circ
\displaystyle =\left(\frac{\sqrt3}{2}\right)^2+2(1)-\left(\frac{\sqrt3}{2}\right)^2
\displaystyle =\frac34+2-\frac34=2
\displaystyle \therefore \text{The required value is }2.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }\sin A=\frac34,\text{ calculate }\sec A.   \hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \sin^2 A+\cos^2 A=1
\displaystyle \left(\frac34\right)^2+\cos^2 A=1
\displaystyle \cos^2 A=1-\frac9{16}=\frac7{16}
\displaystyle \therefore \cos A=\frac{\sqrt7}{4}
\displaystyle \sec A=\frac{1}{\cos A}=\frac{4}{\sqrt7}=\frac{4\sqrt7}{7}
\displaystyle \therefore \sec A=\frac{4\sqrt7}{7}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }\tan A=\frac{5}{12},\text{ find the value of }(\sin A+\cos A)\sec A.
\displaystyle \hfill\text{[CBSE 2008]}
\displaystyle \text{Answer:}
\displaystyle (\sin A+\cos A)\sec A
\displaystyle =(\sin A+\cos A)\frac{1}{\cos A}
\displaystyle =\frac{\sin A}{\cos A}+1
\displaystyle =\tan A+1
\displaystyle =\frac5{12}+1=\frac{17}{12}
\displaystyle \therefore \text{The required value is }\frac{17}{12}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In Fig. 10.46, }PS=3\text{ cm},\ QS=4\text{ cm},\ \angle PSQ=90^\circ,\ 
\displaystyle PQ\perp RQ \ \text{and }RQ=9\text{ cm. If }\angle PRQ=\theta,\text{ evaluate }\tan\theta.\hfill\text{[CBSE 2019]}

\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle PSQ,
\displaystyle PQ=\sqrt{PS^2+QS^2}
\displaystyle =\sqrt{3^2+4^2}=\sqrt{25}=5\text{ cm}.
\displaystyle \text{In right-angled }\triangle PRQ,\text{ for }\angle PRQ=\theta,
\displaystyle \tan\theta=\frac{PQ}{RQ}=\frac{5}{9}
\displaystyle \therefore \text{The required value is }\frac{5}{9}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Evaluate: }2\sin^2 30^\circ\sec60^\circ+\tan^2 60^\circ.   \hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \sin30^\circ=\frac12,\qquad\sec60^\circ=2,\qquad\tan60^\circ=\sqrt3
\displaystyle 2\sin^2 30^\circ\sec60^\circ+\tan^2 60^\circ
\displaystyle =2\left(\frac12\right)^2(2)+(\sqrt3)^2
\displaystyle =1+3=4
\displaystyle \therefore \text{The required value is }4.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }2\sin(A+B)=\sqrt3\text{ and }\cos(A-B)=1,\text{ then find the measures of}
\displaystyle \text{angles }A\text{ and }B,\text{ where }0\leq A,B\text{ and }(A+B)\leq90^\circ.\hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle 2\sin(A+B)=\sqrt3
\displaystyle \sin(A+B)=\frac{\sqrt3}{2}=\sin60^\circ
\displaystyle \therefore A+B=60^\circ
\displaystyle \cos(A-B)=1=\cos0^\circ
\displaystyle \therefore A-B=0^\circ
\displaystyle \therefore A=B
\displaystyle \text{Adding }A+B=60^\circ\text{ and }A-B=0^\circ,
\displaystyle 2A=60^\circ
\displaystyle \therefore A=30^\circ,\qquad B=30^\circ.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Evaluate: }2\sqrt2\cos45^\circ\sin30^\circ+2\sqrt3\cos30^\circ.   \hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \cos45^\circ=\frac{1}{\sqrt2},\qquad\sin30^\circ=\frac12,\qquad\cos30^\circ=\frac{\sqrt3}{2}
\displaystyle 2\sqrt2\cos45^\circ\sin30^\circ+2\sqrt3\cos30^\circ
\displaystyle =2\sqrt2\left(\frac{1}{\sqrt2}\right)\left(\frac12\right)+2\sqrt3\left(\frac{\sqrt3}{2}\right)
\displaystyle =1+3=4
\displaystyle \therefore \text{The required value is }4.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }A=60^\circ\text{ and }B=30^\circ,\text{ verify that }
\displaystyle \sin(A+B)=\sin A\cos B+\cos A\sin B. \ \hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sin(A+B)
\displaystyle =\sin(60^\circ+30^\circ)=\sin90^\circ=1
\displaystyle \text{RHS}=\sin A\cos B+\cos A\sin B
\displaystyle =\sin60^\circ\cos30^\circ+\cos60^\circ\sin30^\circ
\displaystyle =\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt3}{2}\right)+\left(\frac12\right)\left(\frac12\right)
\displaystyle =\frac34+\frac14=1
\displaystyle \therefore \text{LHS}=\text{RHS}=1.
\displaystyle \text{Hence verified.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Evaluate: }\frac{5\cos^2 60^\circ+4\sec^2 30^\circ-\tan^2 45^\circ}{\sin^2 30^\circ+\sin^2 60^\circ}.   \hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \cos60^\circ=\frac12,\quad\sec30^\circ=\frac{2}{\sqrt3},\quad\tan45^\circ=1
\displaystyle \sin30^\circ=\frac12,\qquad\sin60^\circ=\frac{\sqrt3}{2}
\displaystyle \frac{5\cos^2 60^\circ+4\sec^2 30^\circ-\tan^2 45^\circ}{\sin^2 30^\circ+\sin^2 60^\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}=\frac{3+64}{12}=\frac{67}{12}
\displaystyle \therefore \text{The required value is }\frac{67}{12}.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }\sin(A-B)=\frac12,\ \cos(A+B)=\frac12,\ 0^\circ<A+B\leq90^\circ,\
\displaystyle A>B,  \ \text{find }\angle A\text{ and }\angle B.\hfill\text{[CBSE 2024]}
\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 B=15^\circ
\displaystyle \therefore \angle A=45^\circ,\qquad\angle B=15^\circ.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Evaluate: }\frac{2\tan30^\circ\sec60^\circ\tan45^\circ}{1-\sin^2 60^\circ}.   \hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \tan30^\circ=\frac{1}{\sqrt3},\quad\sec60^\circ=2,\quad\tan45^\circ=1
\displaystyle \sin60^\circ=\frac{\sqrt3}{2}
\displaystyle \frac{2\tan30^\circ\sec60^\circ\tan45^\circ}{1-\sin^2 60^\circ}
\displaystyle =\frac{2\left(\frac{1}{\sqrt3}\right)(2)(1)}{1-\left(\frac{\sqrt3}{2}\right)^2}
\displaystyle =\frac{\frac{4}{\sqrt3}}{1-\frac34}
\displaystyle =\frac{\frac{4}{\sqrt3}}{\frac14}=\frac{16}{\sqrt3}=\frac{16\sqrt3}{3}
\displaystyle \therefore \text{The required value is }\frac{16\sqrt3}{3}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Evaluate: }\frac{5\tan60^\circ}{(\sin^2 60^\circ+\cos^2 60^\circ)\tan30^\circ}.   \hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \sin^2 60^\circ+\cos^2 60^\circ=1
\displaystyle \therefore \frac{5\tan60^\circ}{(\sin^2 60^\circ+\cos^2 60^\circ)\tan30^\circ}
\displaystyle =\frac{5(\sqrt3)}{(1)\left(\frac{1}{\sqrt3}\right)}
\displaystyle =5\sqrt3\times\sqrt3=15
\displaystyle \therefore \text{The required value is }15.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If }\tan A=\sqrt3,\text{ then find the value of }\cos^2 A-\sin^2 A.   \hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \tan A=\sqrt3=\tan60^\circ
\displaystyle \therefore A=60^\circ
\displaystyle \cos^2 A-\sin^2 A=\cos^2 60^\circ-\sin^2 60^\circ
\displaystyle =\left(\frac12\right)^2-\left(\frac{\sqrt3}{2}\right)^2
\displaystyle =\frac14-\frac34=-\frac12
\displaystyle \therefore \text{The required value is }-\frac12.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If }x\sin60^\circ+\cos30^\circ-\tan45^\circ=\frac{\sqrt3}{2},\text{ find the value of }x.   \hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle x\left(\frac{\sqrt3}{2}\right)+\frac{\sqrt3}{2}-1=\frac{\sqrt3}{2}
\displaystyle \frac{x\sqrt3}{2}=1
\displaystyle x=\frac{2}{\sqrt3}=\frac{2\sqrt3}{3}
\displaystyle \therefore \text{The required value of }x\text{ is }\frac{2\sqrt3}{3}.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{If }4k=\tan^2 60^\circ-2\mathrm{cosec}^2 30^\circ-2\tan^2 30^\circ,\text{ then find the value of }k.
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle 4k=(\sqrt3)^2-2(2)^2-2\left(\frac{1}{\sqrt3}\right)^2
\displaystyle 4k=3-8-\frac23
\displaystyle 4k=\frac{9-24-2}{3}=-\frac{17}{3}
\displaystyle k=-\frac{17}{12}
\displaystyle \therefore \text{The required value of }k\text{ is }-\frac{17}{12}.
\displaystyle \\

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