\displaystyle \textbf{Question 1: }\text{Define an identity.}
\displaystyle \text{Answer:}
\displaystyle \text{An identity is an equation which is true for all values of the variable(s) for which} \\ \text{the expressions are defined.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{What is the value of }(1-\cos^2\theta)\mathrm{cosec}^2\theta?
\displaystyle \text{Answer:}
\displaystyle (1-\cos^2\theta)\mathrm{cosec}^2\theta
\displaystyle =\sin^2\theta\,\mathrm{cosec}^2\theta
\displaystyle =\sin^2\theta\times\frac{1}{\sin^2\theta}
\displaystyle =1
\displaystyle \therefore \text{The required value is }1.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{What is the value of }(1+\cot^2\theta)\sin^2\theta?
\displaystyle \text{Answer:}
\displaystyle (1+\cot^2\theta)\sin^2\theta
\displaystyle =\mathrm{cosec}^2\theta\sin^2\theta
\displaystyle =\frac{1}{\sin^2\theta}\times\sin^2\theta
\displaystyle =1
\displaystyle \therefore \text{The required value is }1.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{What is the value of }\sin^2\theta+\frac{1}{1+\tan^2\theta}?\hfill\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \sin^2\theta+\frac{1}{1+\tan^2\theta}
\displaystyle =\sin^2\theta+\frac{1}{\sec^2\theta}
\displaystyle =\sin^2\theta+\cos^2\theta
\displaystyle =1
\displaystyle \therefore \text{The required value is }1.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }\sec^2\theta(1+\sin\theta)(1-\sin\theta)=k,\text{ then find the value of }k.
\displaystyle \hfill\text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle k=\sec^2\theta(1+\sin\theta)(1-\sin\theta)
\displaystyle =\sec^2\theta(1-\sin^2\theta)
\displaystyle =\sec^2\theta\cos^2\theta
\displaystyle =\frac{1}{\cos^2\theta}\times\cos^2\theta=1
\displaystyle \therefore k=1.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }\mathrm{cosec}^2\theta(1+\cos\theta)(1-\cos\theta)=\lambda,\text{ then find the value of }\lambda.
\displaystyle \text{Answer:}
\displaystyle \lambda=\mathrm{cosec}^2\theta(1+\cos\theta)(1-\cos\theta)
\displaystyle =\mathrm{cosec}^2\theta(1-\cos^2\theta)
\displaystyle =\mathrm{cosec}^2\theta\sin^2\theta
\displaystyle =\frac{1}{\sin^2\theta}\times\sin^2\theta=1
\displaystyle \therefore \lambda=1.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Write the value of }\cot^2\theta-\frac{1}{\sin^2\theta}.
\displaystyle \text{Answer:}
\displaystyle \cot^2\theta-\frac{1}{\sin^2\theta}
\displaystyle =\cot^2\theta-\mathrm{cosec}^2\theta
\displaystyle =-1\qquad[\because \mathrm{cosec}^2\theta-\cot^2\theta=1]
\displaystyle \therefore \text{The required value is }-1.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }x=a\sin\theta\text{ and }y=b\cos\theta,\text{ what is the value of }b^2x^2+a^2y^2?
\displaystyle \text{Answer:}
\displaystyle b^2x^2+a^2y^2=b^2(a\sin\theta)^2+a^2(b\cos\theta)^2
\displaystyle =a^2b^2\sin^2\theta+a^2b^2\cos^2\theta
\displaystyle =a^2b^2(\sin^2\theta+\cos^2\theta)
\displaystyle =a^2b^2
\displaystyle \therefore \text{The required value is }a^2b^2.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }\sin\theta=\frac{4}{5},\text{ what is the value of }\cot\theta+\mathrm{cosec}\theta?
\displaystyle \text{Answer:}
\displaystyle \sin\theta=\frac{4}{5}
\displaystyle \cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}
\displaystyle \cot\theta=\frac{\cos\theta}{\sin\theta}=\frac{3/5}{4/5}=\frac{3}{4}
\displaystyle \mathrm{cosec}\theta=\frac{1}{\sin\theta}=\frac{5}{4}
\displaystyle \therefore \cot\theta+\mathrm{cosec}\theta=\frac{3}{4}+\frac{5}{4}=2
\displaystyle \therefore \text{The required value is }2.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{What is the value of }9\cot^2\theta-9\mathrm{cosec}^2\theta?
\displaystyle \text{Answer:}
\displaystyle 9\cot^2\theta-9\mathrm{cosec}^2\theta
\displaystyle =9(\cot^2\theta-\mathrm{cosec}^2\theta)
\displaystyle =9(-1)\qquad[\because \mathrm{cosec}^2\theta-\cot^2\theta=1]
\displaystyle =-9
\displaystyle \therefore \text{The required value is }-9.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{What is the value of }6\tan^2\theta-\frac{6}{\cos^2\theta}?
\displaystyle \text{Answer:}
\displaystyle 6\tan^2\theta-\frac{6}{\cos^2\theta}
\displaystyle =6\tan^2\theta-6\sec^2\theta
\displaystyle =6(\tan^2\theta-\sec^2\theta)
\displaystyle =6(-1)\qquad[\because \sec^2\theta-\tan^2\theta=1]
\displaystyle =-6
\displaystyle \therefore \text{The required value is }-6.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{What is the value of }\frac{\tan^2\theta-\sec^2\theta}{\cot^2\theta-\mathrm{cosec}^2\theta}?
\displaystyle \text{Answer:}
\displaystyle \frac{\tan^2\theta-\sec^2\theta}{\cot^2\theta-\mathrm{cosec}^2\theta}
\displaystyle =\frac{-1}{-1}
\displaystyle \qquad[\because \sec^2\theta-\tan^2\theta=1\text{ and }\mathrm{cosec}^2\theta-\cot^2\theta=1]
\displaystyle =1
\displaystyle \therefore \text{The required value is }1.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{What is the value of }(1+\tan^2\theta)(1-\sin\theta)(1+\sin\theta)?\hfill\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle (1+\tan^2\theta)(1-\sin\theta)(1+\sin\theta)
\displaystyle =\sec^2\theta(1-\sin^2\theta)
\displaystyle =\sec^2\theta\cos^2\theta
\displaystyle =\frac{1}{\cos^2\theta}\times\cos^2\theta
\displaystyle =1
\displaystyle \therefore \text{The required value is }1.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }\cos A=\frac{7}{25},\text{ find the value of }\tan A+\cot A.\hfill\text{[CBSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \cos A=\frac{7}{25}
\displaystyle \sin A=\sqrt{1-\cos^2 A}=\sqrt{1-\frac{49}{625}}=\frac{24}{25}
\displaystyle \tan A=\frac{\sin A}{\cos A}=\frac{24}{7}
\displaystyle \cot A=\frac{1}{\tan A}=\frac{7}{24}
\displaystyle \therefore \tan A+\cot A=\frac{24}{7}+\frac{7}{24}
\displaystyle =\frac{576+49}{168}=\frac{625}{168}
\displaystyle \therefore \text{The required value is }\frac{625}{168}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }\sin\theta=\frac{1}{3},\text{ then find the value of }2\cot^2\theta+2.\hfill\text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle 2\cot^2\theta+2=2(\cot^2\theta+1)
\displaystyle =2\mathrm{cosec}^2\theta
\displaystyle =2\left(\frac{1}{\sin^2\theta}\right)
\displaystyle =2\left(\frac{1}{(1/3)^2}\right)
\displaystyle =2\times9=18
\displaystyle \therefore \text{The required value is }18.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }\cos\theta=\frac{3}{4},\text{ then find the value of }9\tan^2\theta+9.
\displaystyle \text{Answer:}
\displaystyle 9\tan^2\theta+9=9(\tan^2\theta+1)
\displaystyle =9\sec^2\theta
\displaystyle =9\left(\frac{1}{\cos^2\theta}\right)
\displaystyle =9\left(\frac{1}{(3/4)^2}\right)
\displaystyle =9\times\frac{16}{9}=16
\displaystyle \therefore \text{The required value is }16.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }\sec\theta+\tan\theta=x,\text{ write the value of }\sec\theta-\tan\theta\text{ in terms of }x.
\displaystyle \text{Answer:}
\displaystyle (\sec\theta+\tan\theta)(\sec\theta-\tan\theta)
\displaystyle =\sec^2\theta-\tan^2\theta=1
\displaystyle \text{Given, }\sec\theta+\tan\theta=x.
\displaystyle \therefore x(\sec\theta-\tan\theta)=1
\displaystyle \therefore \sec\theta-\tan\theta=\frac{1}{x}.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }\mathrm{cosec}\theta-\cot\theta=a,\text{ write the value of }\mathrm{cosec}\theta+\cot\theta.
\displaystyle \text{Answer:}
\displaystyle (\mathrm{cosec}\theta-\cot\theta)(\mathrm{cosec}\theta+\cot\theta)
\displaystyle =\mathrm{cosec}^2\theta-\cot^2\theta=1
\displaystyle \text{Given, }\mathrm{cosec}\theta-\cot\theta=a.
\displaystyle \therefore a(\mathrm{cosec}\theta+\cot\theta)=1
\displaystyle \therefore \mathrm{cosec}\theta+\cot\theta=\frac{1}{a}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }\sin^2\theta\cos^2\theta(1+\tan^2\theta)(1+\cot^2\theta)=\lambda,
\displaystyle \text{then find the value of }\lambda.
\displaystyle \text{Answer:}
\displaystyle \lambda=\sin^2\theta\cos^2\theta(1+\tan^2\theta)(1+\cot^2\theta)
\displaystyle =\sin^2\theta\cos^2\theta\,\sec^2\theta\,\mathrm{cosec}^2\theta
\displaystyle =\sin^2\theta\cos^2\theta\times\frac{1}{\cos^2\theta}\times\frac{1}{\sin^2\theta}
\displaystyle =1
\displaystyle \therefore \lambda=1.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }5x=\sec\theta\text{ and }\frac{5}{x}=\tan\theta,\text{ find the value of}
\displaystyle 5\left(x^2-\frac{1}{x^2}\right).\hfill\text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \sec\theta=5x,\qquad \tan\theta=\frac{5}{x}
\displaystyle \sec^2\theta-\tan^2\theta=1
\displaystyle (5x)^2-\left(\frac{5}{x}\right)^2=1
\displaystyle 25x^2-\frac{25}{x^2}=1
\displaystyle 25\left(x^2-\frac{1}{x^2}\right)=1
\displaystyle \therefore 5\left(x^2-\frac{1}{x^2}\right)=\frac{1}{5}.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Evaluate: }\frac{3\sin30^\circ-4\sin^3 30^\circ}{2\sin^2 60^\circ+2\cos^2 60^\circ}.\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \frac{3\sin30^\circ-4\sin^3 30^\circ}{2\sin^2 60^\circ+2\cos^2 60^\circ}
\displaystyle =\frac{3\left(\frac{1}{2}\right)-4\left(\frac{1}{2}\right)^3}{2(\sin^2 60^\circ+\cos^2 60^\circ)}
\displaystyle =\frac{\frac{3}{2}-\frac{1}{2}}{2(1)}
\displaystyle =\frac{1}{2}
\displaystyle \therefore \text{The required value is }\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Find the value of }x\text{ for which }(\sin A+\mathrm{cosec} A)^2+(\cos A+\sec A)^2
\displaystyle =x+\tan^2 A+\cot^2 A.\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle (\sin A+\mathrm{cosec} A)^2+(\cos A+\sec A)^2
\displaystyle =\sin^2 A+\mathrm{cosec}^2 A+2+\cos^2 A+\sec^2 A+2
\displaystyle =1+\mathrm{cosec}^2 A+\sec^2 A+4
\displaystyle =5+(1+\cot^2 A)+(1+\tan^2 A)
\displaystyle =7+\tan^2 A+\cot^2 A
\displaystyle \therefore x+\tan^2 A+\cot^2 A=7+\tan^2 A+\cot^2 A
\displaystyle \therefore x=7.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If }\sin A=y,\text{ then express }\cos A\text{ and }\tan A\text{ in terms of }y.
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \sin A=y
\displaystyle \cos A=\sqrt{1-\sin^2 A}
\displaystyle =\sqrt{1-y^2}
\displaystyle \tan A=\frac{\sin A}{\cos A}
\displaystyle =\frac{y}{\sqrt{1-y^2}}
\displaystyle \therefore \cos A=\sqrt{1-y^2}\text{ and }\tan A=\frac{y}{\sqrt{1-y^2}}.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If }\mathrm{cosec}\theta=2x\text{ and }\cot\theta=\frac{2}{x},\text{ find the value of}
\displaystyle 2\left(x^2-\frac{1}{x^2}\right).\hfill\text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \mathrm{cosec}^2\theta-\cot^2\theta=1
\displaystyle (2x)^2-\left(\frac{2}{x}\right)^2=1
\displaystyle 4x^2-\frac{4}{x^2}=1
\displaystyle 4\left(x^2-\frac{1}{x^2}\right)=1
\displaystyle \therefore 2\left(x^2-\frac{1}{x^2}\right)=\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Write `True' or `False' and justify your answer in each of the following:}
\displaystyle \text{(i) The value of }\sin\theta\text{ is }x+\frac{1}{x},\text{ where }x\text{ is a positive real number.}
\displaystyle \text{Answer: False.}
\displaystyle \text{Since }x>0,\quad \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\geq0
\displaystyle \therefore x+\frac{1}{x}-2\geq0
\displaystyle \therefore x+\frac{1}{x}\geq2
\displaystyle \text{But }-1\leq\sin\theta\leq1.
\displaystyle \therefore \sin\theta\text{ cannot have the value }x+\frac{1}{x}.

\displaystyle \text{(ii) }\cos\theta=\frac{a^2+b^2}{2ab},\text{ where }a\text{ and }b\text{ are two distinct numbers such that }ab>0.
\displaystyle \text{Answer: False.}
\displaystyle \text{Since }a\neq b,\quad (a-b)^2>0
\displaystyle \therefore a^2+b^2>2ab
\displaystyle \text{Since }ab>0,\quad \frac{a^2+b^2}{2ab}>1
\displaystyle \text{But }-1\leq\cos\theta\leq1.
\displaystyle \therefore \cos\theta\text{ cannot be equal to }\frac{a^2+b^2}{2ab}.

\displaystyle \text{(iii) The value of }\sin\theta+\cos\theta\text{ is always greater than }1.
\displaystyle \text{Answer: False.}
\displaystyle \text{For }\theta=0^\circ,
\displaystyle \sin0^\circ+\cos0^\circ=0+1=1
\displaystyle \therefore \sin\theta+\cos\theta\text{ is not always greater than }1.
\displaystyle \\

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