Class 10: Trigonometry – CBSE Board Problems

$\displaystyle \textbf{Question 1. }\text{Find the value of }x\text{ for which}$
$\displaystyle (\sin A+\mathrm{cosec}\,A)^{2}+(\cos A+\sec A)^{2}=x+\tan^{2}A+\cot^{2}A \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have to find value of }x\text{ for which}$
$\displaystyle (\sin A+\mathrm{cosec} A)^2+(\cos A+\sec A)^2=x+\tan^2A+\cot^2A$
$\displaystyle \text{Taking LHS, we have}$
$\displaystyle =\sin^2A+\mathrm{cosec}^2A+2+\cos^2A+\sec^2A+2$
$\displaystyle =\sin^2A+\cos^2A+\mathrm{cosec}^2A+\sec^2A+4$
$\displaystyle =1+(1+\cot^2A)+(1+\tan^2A)+4$
$\displaystyle =7+\cot^2A+\tan^2A$
$\displaystyle \text{Comparing with RHS,}$
$\displaystyle x=7$
$\\$

$\displaystyle \textbf{Question 2. }\text{Prove that }\frac{\cos A+\sin A-1}{\cos A-\sin A+1}=\mathrm{cosec}\,A-\cot A. \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{To prove: }\frac{\cos A+\sin A-1}{\cos A-\sin A+1}=\mathrm{cosec} A-\cot A$
$\displaystyle \text{Taking LHS and dividing numerator and denominator by }\sin A,$
$\displaystyle =\frac{\frac{\cos A}{\sin A}+\frac{\sin A}{\sin A}-\frac{1}{\sin A}}{\frac{\cos A}{\sin A}-\frac{\sin A}{\sin A}+\frac{1}{\sin A}}$
$\displaystyle =\frac{\cot A+1-\mathrm{cosec} A}{\cot A-1+\mathrm{cosec} A}$
$\displaystyle \text{As, }1+\cot^2A=\mathrm{cosec}^2A$
$\displaystyle \Rightarrow 1=\mathrm{cosec}^2A-\cot^2A$
$\displaystyle =(\mathrm{cosec} A-\cot A)(\mathrm{cosec} A+\cot A)$
$\displaystyle \text{Substituting in numerator,}$
$\displaystyle =\frac{(\cot A-\mathrm{cosec} A)+(\mathrm{cosec}^2A-\cot^2A)}{\cot A-1+\mathrm{cosec} A}$
$\displaystyle =\frac{(\cot A-\mathrm{cosec} A)+(\mathrm{cosec} A-\cot A)(\mathrm{cosec} A+\cot A)}{\cot A-1+\mathrm{cosec} A}$
$\displaystyle =(\mathrm{cosec} A-\cot A)\frac{-1+\mathrm{cosec} A+\cot A}{\cot A-1+\mathrm{cosec} A}$
$\displaystyle =\mathrm{cosec} A-\cot A$
$\displaystyle \therefore \text{LHS = RHS}$
$\\$

$\displaystyle \textbf{Question 3. }\text{If }\cot\theta+\cos\theta=p\text{ and }\cot\theta-\cos\theta=q,\text{ prove that } \\ p^{2}-q^{2}=4\sqrt{pq}. \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given } \cot\theta+\cos\theta=p,\qquad \cot\theta-\cos\theta=q$
$\displaystyle \text{To prove: }p^2-q^2=4\sqrt{pq}$
$\displaystyle \text{Taking LHS,}$
$\displaystyle (\cot\theta+\cos\theta)^2-(\cot\theta-\cos\theta)^2$
$\displaystyle =4\cot\theta\cos\theta \qquad \cdots (i)$
$\displaystyle \text{Now, taking RHS,}$
$\displaystyle 4\sqrt{pq}=4\sqrt{(\cot\theta+\cos\theta)(\cot\theta-\cos\theta)}$
$\displaystyle =4\sqrt{\cot^2\theta-\cos^2\theta}$
$\displaystyle =4\sqrt{\frac{\cos^2\theta}{\sin^2\theta}-\cos^2\theta}$
$\displaystyle =4\sqrt{\frac{\cos^2\theta(1-\sin^2\theta)}{\sin^2\theta}}$
$\displaystyle =4\sqrt{\frac{\cos^4\theta}{\sin^2\theta}}$
$\displaystyle =4\frac{\cos^2\theta}{\sin\theta}$
$\displaystyle =4\cos\theta\cot\theta \qquad \cdots (ii)$
$\displaystyle \text{From (i) and (ii),}$
$\displaystyle p^2-q^2=4\sqrt{pq}$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 4. }\tan 2A=3\tan A\text{ is true, when the measure of }\angle A\text{ is:} \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{(a) }90^{\circ} \qquad \text{(b) }60^{\circ} \qquad \text{(c) }45^{\circ} \qquad \text{(d) }30^{\circ}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(d)}$
$\displaystyle \text{If }\tan 2A=3\tan A$
$\displaystyle \text{For }A=30^\circ,\ \tan 2\times30^\circ=3\tan30^\circ$
$\displaystyle \tan60^\circ=3\times\frac{1}{\sqrt{3}}$
$\displaystyle \sqrt{3}=\sqrt{3}$
$\\$

$\displaystyle \textbf{Question 5. }\text{Which of the following statements is true?} \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{(a) }\sin 20^{\circ}>\sin 70^{\circ} \qquad \text{(b) }\sin 20^{\circ}>\cos 20^{\circ}$
$\displaystyle \text{(c) }\cos 20^{\circ}>\cos 70^{\circ} \qquad \text{(d) }\tan 20^{\circ}>\tan 70^{\circ}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(c)}$
$\displaystyle \text{In }\sin\theta,\text{ when }\theta\text{ increases, the value of }\sin\theta\text{ also increases.}$
$\displaystyle \text{In }\cos\theta,\text{ when }\theta\text{ increases, the value of }\cos\theta\text{ decreases.}$
$\displaystyle \text{In option }(c),\ \cos20^\circ>\cos70^\circ,\text{ it satisfies the condition.}$
$\\$

$\displaystyle \textbf{Question 6. }\text{Evaluate the following:}$
$\displaystyle \frac{3\sin30^{\circ}-4\sin^{3}30^{\circ}}{2\sin^{2}50^{\circ}+2\cos^{2}50^{\circ}} \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{3\sin30^\circ-4\sin^{3}30^\circ}{2\sin^{2}50^\circ+2\cos^{2}50^\circ}$
$\displaystyle \text{Put values }\sin30^\circ=\frac{1}{2}$
$\displaystyle =\frac{3\left(\frac{1}{2}\right)-4\left(\frac{1}{2}\right)^{3}}{2(\sin^{2}50^\circ+\cos^{2}50^\circ)}$
$\displaystyle =\frac{\frac{3}{2}-\frac{4}{8}}{2(1)}$
$\displaystyle =\frac{\frac{3-1}{2}}{2}$
$\displaystyle =\frac{1}{2}$
$\\$

$\displaystyle \textbf{Question 7. }\text{If }\sec\theta-\tan\theta=m,\text{ then the value of }\sec\theta+\tan\theta\text{ is:} \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{(a) }1-\frac{1}{m} \qquad \text{(b) }m^{2}-1$
$\displaystyle \text{(c) }\frac{1}{m} \qquad \text{(d) }-m$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(c)}$
$\displaystyle \text{We know}$
$\displaystyle \sec^2\theta-\tan^2\theta=1$
$\displaystyle \Rightarrow (\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1$
$\displaystyle \Rightarrow m(\sec\theta+\tan\theta)=1$
$\displaystyle \Rightarrow \sec\theta+\tan\theta=\frac{1}{m}$
$\\$

$\displaystyle \textbf{Question 8. }\text{Prove that: }\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}=1+\sec\theta\,\mathrm{cosec}\,\theta. \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}$
$\displaystyle =\frac{\tan\theta}{1-\frac{1}{\tan\theta}}+\frac{1}{\tan\theta(1-\tan\theta)}$
$\displaystyle =\frac{\tan^2\theta}{\tan\theta-1}-\frac{1}{\tan\theta(\tan\theta-1)}$
$\displaystyle =\frac{\tan^3\theta-1}{\tan\theta(\tan\theta-1)}$
$\displaystyle =\frac{(\tan\theta-1)(\tan^2\theta+\tan\theta+1)}{\tan\theta(\tan\theta-1)}$
$\displaystyle =\frac{\tan^2\theta+\tan\theta+1}{\tan\theta}$
$\displaystyle =\frac{\tan^2\theta}{\tan\theta}+\frac{\tan\theta}{\tan\theta}+\frac{1}{\tan\theta}$
$\displaystyle =\tan\theta+1+\cot\theta$
$\displaystyle =\frac{\sec^2\theta}{\tan\theta}+1$
$\displaystyle =\frac{1}{\sin\theta\cos\theta}+1$
$\displaystyle =\sec\theta\mathrm{cosec}\theta+1$
$\displaystyle =\text{RHS}$
$\\$

$\displaystyle \textbf{Question 9. }\text{If }\cos(\alpha+\beta)=0,\text{ then value of }\cos\left(\frac{\alpha+\beta}{2}\right)\text{ is equal to:} \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{(a) }\frac{1}{\sqrt{2}} \qquad \text{(b) }\frac{1}{2} \qquad \text{(c) }0 \qquad \text{(d) }\sqrt{2}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(a)}$
$\displaystyle \text{Given }\cos(\alpha+\beta)=0$
$\displaystyle \Rightarrow \cos(\alpha+\beta)=\cos90^\circ$
$\displaystyle \Rightarrow \alpha+\beta=90^\circ$
$\displaystyle \Rightarrow \frac{\alpha+\beta}{2}=\frac{90^\circ}{2}=45^\circ$
$\displaystyle \text{Now, }\cos\left(\frac{\alpha+\beta}{2}\right)=\cos45^\circ=\frac{1}{\sqrt{2}}$
$\\$

$\displaystyle \textbf{Question 10. }\text{Evaluate: }2\sqrt{2}\cos45^{\circ}\sin30^{\circ}+2\sqrt{3}\cos30^{\circ}. \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle 2\sqrt{2}\cos45^\circ\sin30^\circ+2\sqrt{3}\cos30^\circ$
$\displaystyle =2\sqrt{2}\times\frac{1}{\sqrt{2}}\times\frac{1}{2}+2\sqrt{3}\times\frac{\sqrt{3}}{2}$
$\displaystyle =1+3=4$
$\\$

$\displaystyle \textbf{Question 11. }\text{If }A=60^{\circ}\text{ and }B=30^{\circ},\text{ verify that: }\sin(A+B)=\sin A\cos B+\cos A\sin B. \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }A=60^\circ\text{ and }B=30^\circ$
$\displaystyle \text{LHS}=\sin(A+B)=\sin(60^\circ+30^\circ)=\sin90^\circ=1$
$\displaystyle \text{RHS}=\sin A\cos B+\cos A\sin B$
$\displaystyle =\sin60^\circ\cos30^\circ+\cos60^\circ\sin30^\circ$
$\displaystyle =\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}+\frac{1}{2}\times\frac{1}{2}$
$\displaystyle =\frac{3}{4}+\frac{1}{4}=1$
$\displaystyle \text{So, LHS = RHS, hence verified.}$
$\\$

$\displaystyle \textbf{Question 12. }\sec\theta\text{ when expressed in terms of }\cot\theta,\text{ is equal to:} \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }\frac{1+\cot^{2}\theta}{\cot\theta} \qquad \text{(b) }\sqrt{1+\cot^{2}\theta}$
$\displaystyle \text{(c) }\frac{\sqrt{1+\cot^{2}\theta}}{\cot\theta} \qquad \text{(d) }\frac{\sqrt{1-\cot^{2}\theta}}{\cot\theta}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(c)}$
$\displaystyle \sec^2\theta=1+\tan^2\theta=\frac{\cot^2\theta+1}{\cot^2\theta}$
$\displaystyle \Rightarrow \sec\theta=\frac{\sqrt{\cot^2\theta+1}}{\cot\theta}$
$\\$

$\displaystyle \textbf{Question 13. }\left(\sec^{2}\theta-1\right)\left(1-\mathrm{cosec}^{2}\theta\right)\text{ is equal to:} \hspace{0.2cm}\text{[CBSE 2023(C)]}$
$\displaystyle \text{(a) }1 \qquad \text{(b) }-1 \qquad \text{(c) }2 \qquad \text{(d) }-2$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(b)}$
$\displaystyle \text{Given, }(\sec^2\theta-1)(1-\mathrm{cosec}^2\theta)$
$\displaystyle =\tan^2\theta\times(-\cot^2\theta)$
$\displaystyle =\tan^2\theta\times\left(-\frac{1}{\tan^2\theta}\right)$
$\displaystyle =-1$
$\\$

$\displaystyle \textbf{Question 14. }\text{Prove that: }\frac{\sin A-2\sin^{3}A}{2\cos^{3}A-\cos A}=\tan A. \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\frac{\sin A-2\sin^3A}{2\cos^3A-\cos A}$
$\displaystyle =\frac{\sin A(1-2\sin^2A)}{\cos A(2\cos^2A-1)}$
$\displaystyle =\frac{\sin A(1-2\sin^2A)}{\cos A[2(1-\sin^2A)-1]}$
$\displaystyle =\frac{\sin A(1-2\sin^2A)}{\cos A(1-2\sin^2A)}$
$\displaystyle =\frac{\sin A}{\cos A}$
$\displaystyle =\tan A=\text{RHS}$
$\\$

$\displaystyle \textbf{Question 15. }\text{Prove that }\sec A(1-\sin A)(\sec A+\tan A)=1. \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\sec A(1-\sin A)(\sec A+\tan A)$
$\displaystyle =\frac{1}{\cos A}(1-\sin A)\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)$
$\displaystyle =\frac{1}{\cos A}(1-\sin A)\left(\frac{1+\sin A}{\cos A}\right)$
$\displaystyle =\frac{1}{\cos^2A}(1-\sin^2A)$
$\displaystyle =\frac{1}{\cos^2A}\cdot\cos^2A$
$\displaystyle =1=\text{RHS}$
$\\$

$\displaystyle \textbf{Question 16. }\text{If }\tan\theta+\sin\theta=m\text{ and }\tan\theta-\sin\theta=n,\text{ then show that } \\ (m^{2}-n^{2})=4\sqrt{mn}. \hspace{0.2cm}\text{[CBSE 2023(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given,}$
$\displaystyle \tan\theta+\sin\theta=m \qquad \cdots (i)$
$\displaystyle \tan\theta-\sin\theta=n \qquad \cdots (ii)$
$\displaystyle \text{LHS}=(m^2-n^2)$
$\displaystyle =(\tan\theta+\sin\theta)^2-(\tan\theta-\sin\theta)^2$
$\displaystyle =\tan^2\theta+\sin^2\theta+2\tan\theta\sin\theta$
$\displaystyle \qquad -\tan^2\theta-\sin^2\theta+2\tan\theta\sin\theta$
$\displaystyle =4\tan\theta\sin\theta$
$\displaystyle \text{RHS}=4\sqrt{mn}$
$\displaystyle =4\sqrt{(\tan\theta+\sin\theta)(\tan\theta-\sin\theta)}$
$\displaystyle =4\sqrt{\tan^2\theta-\sin^2\theta}$
$\displaystyle =4\sqrt{\frac{\sin^2\theta}{\cos^2\theta}-\sin^2\theta}$
$\displaystyle =4\sqrt{\frac{\sin^2\theta(1-\cos^2\theta)}{\cos^2\theta}}$
$\displaystyle =4\sqrt{\frac{\sin^4\theta}{\cos^2\theta}}$
$\displaystyle =4\tan\theta\sin\theta$
$\displaystyle \therefore \text{LHS}=\text{RHS}$
$\displaystyle \therefore (m^2-n^2)=4\sqrt{mn}\qquad \text{Hence proved}$
$\\$

$\displaystyle \textbf{Question 17. }\text{Evaluate }\frac{5\cos^{2}60^{\circ}+4\sec^{2}30^{\circ}-\tan^{2}45^{\circ}}{\sin^{2}30^{\circ}+\cos^{2}30^{\circ}}. \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{5\cos^{2}60^\circ+4\sec^{2}30^\circ-\tan^{2}45^\circ}{\sin^{2}30^\circ+\cos^{2}30^\circ}$
$\displaystyle =\frac{5\left(\frac{1}{2}\right)^{2}+4\left(\frac{2}{\sqrt{3}}\right)^{2}-(1)^{2}}{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}$
$\displaystyle =\frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{1}{4}+\frac{3}{4}}$
$\displaystyle =\frac{\frac{67}{12}}{1}=\frac{67}{12}$
$\\$

$\displaystyle \textbf{Question 18. }\text{If }A\text{ and }B\text{ are acute angles such that }\sin(A-B)=0\text{ and } \\ 2\cos(A+B)-1=0,\text{ then find angles }A\text{ and }B. \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \sin(A-B)=0$
$\displaystyle \text{Also, }\sin0^\circ=0$
$\displaystyle \Rightarrow A-B=0^\circ \qquad (i)$
$\displaystyle 2\cos(A+B)-1=0$
$\displaystyle \Rightarrow \cos(A+B)=\frac{1}{2}$
$\displaystyle \text{Also, }\cos60^\circ=\frac{1}{2}$
$\displaystyle \Rightarrow A+B=60^\circ \qquad (ii)$
$\displaystyle \text{Adding }(i)\text{ and }(ii),$
$\displaystyle A-B=0^\circ$
$\displaystyle A+B=60^\circ$
$\displaystyle \Rightarrow 2A=60^\circ$
$\displaystyle \Rightarrow A=30^\circ$
$\displaystyle \text{When }A=30^\circ,\text{ equation }(i)\text{ becomes}$
$\displaystyle 30^\circ-B=0^\circ$
$\displaystyle \Rightarrow B=30^\circ$
$\\$

$\displaystyle \textbf{Question 19. }\text{If in an acute angle }\triangle ABC,\ \sec(B+C-A)=2\text{ and } \\ \tan(C+A-B)=\frac{1}{\sqrt{3}},\text{ find the three angles of }\triangle ABC. \hspace{0.2cm}\text{[CBSE 2023(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\sec(B+C-A)=2$
$\displaystyle \Rightarrow \sec(B+C-A)=\sec60^\circ$
$\displaystyle \Rightarrow B+C-A=60^\circ$
$\displaystyle \Rightarrow B+C=60^\circ+A \qquad (i)$
$\displaystyle \text{and }\tan(C+A-B)=\frac{1}{\sqrt{3}}$
$\displaystyle \Rightarrow \tan(C+A-B)=\tan30^\circ$
$\displaystyle \Rightarrow C+A-B=30^\circ$
$\displaystyle \Rightarrow B-C=A-30^\circ \qquad (ii)$
$\displaystyle \text{On solving }(i)\text{ and }(ii),\ 2B=2A+30^\circ$
$\displaystyle \Rightarrow B=A+15^\circ \qquad (iii)$
$\displaystyle \text{Put this value in equation }(i),\text{ we get}$
$\displaystyle C=60^\circ-15^\circ=45^\circ$
$\displaystyle \text{We know}$
$\displaystyle A+B+C=180^\circ$
$\displaystyle \therefore A+B=135^\circ \qquad (iv)\quad [\because C=45^\circ]$
$\displaystyle \text{On solving }(iii)\text{ and }(iv),$
$\displaystyle B=75^\circ\text{ and }A=60^\circ$
$\\$

$\displaystyle \textbf{Question 20. }\text{In }\triangle ABC\text{ right angled at }B,\ \sin A=\frac{7}{25},\text{ then the value of }\cos C\text{ is:} \\ \hspace{0.2cm}\text{[CBSE 2021]}$
$\displaystyle \text{(a) }\frac{7}{25} \qquad \text{(b) }\frac{24}{25} \qquad \text{(c) }\frac{7}{24} \qquad \text{(d) }\frac{24}{7}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(a)}$
$\displaystyle \text{Given}$
$\displaystyle \sin A=\frac{BC}{AC}=\frac{7}{25}$
$\displaystyle \therefore \cos C=\frac{\text{Base }(BC)}{\text{Hypotenuse }(AC)}=\frac{7}{25}$
$\\$

$\displaystyle \textbf{Question 21. }\text{If }a\cot\theta+b\,\mathrm{cosec}\,\theta=p\text{ and }b\cot\theta+a\,\mathrm{cosec}\,\theta=q,\text{ then } \\ p^{2}-q^{2}= \hspace{0.2cm}\text{[CBSE 2021]}$
$\displaystyle \text{(a) }a^{2}-b^{2} \qquad \text{(b) }b^{2}-a^{2}$
$\displaystyle \text{(c) }a^{2}+b^{2} \qquad \text{(d) }b-a$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(b)}$
$\displaystyle \text{ATQ,}$
$\displaystyle p^2-q^2=(a\cot\theta+b\mathrm{cosec}\theta)^2-(b\cot\theta+a\mathrm{cosec}\theta)^2$
$\displaystyle =[(a\cot\theta+b\mathrm{cosec}\theta)+(b\cot\theta+a\mathrm{cosec}\theta)]$
$\displaystyle \qquad \times[(a\cot\theta+b\mathrm{cosec}\theta)-(b\cot\theta+a\mathrm{cosec}\theta)]$
$\displaystyle =(a+b)(\cot\theta+\mathrm{cosec}\theta)(a-b)(\cot\theta-\mathrm{cosec}\theta)$
$\displaystyle =(a^2-b^2)(\cot^2\theta-\mathrm{cosec}^2\theta)$
$\displaystyle =(a^2-b^2)(-1)$
$\displaystyle =b^2-a^2$
$\\$

$\displaystyle \textbf{Question 22. }\text{If }\sec\theta+\tan\theta=p,\text{ then }\tan\theta\text{ is:} \hspace{0.2cm}\text{[CBSE 2021]}$
$\displaystyle \text{(a) }\frac{p^{2}+1}{2p} \qquad \text{(b) }\frac{p^{2}-1}{2p}$
$\displaystyle \text{(c) }\frac{p^{2}-1}{p^{2}+1} \qquad \text{(d) }\frac{p^{2}+1}{p^{2}-1}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(b)}$
$\displaystyle \text{Given}$
$\displaystyle \sec\theta+\tan\theta=p \qquad \cdots (i)$
$\displaystyle \text{We know}$
$\displaystyle (\sec^2\theta-\tan^2\theta)=1$
$\displaystyle \Rightarrow (\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1$
$\displaystyle \Rightarrow \sec\theta-\tan\theta=\frac{1}{p}\qquad \cdots (ii)$
$\displaystyle \text{On solving (i) and (ii),}$
$\displaystyle 2\tan\theta=p-\frac{1}{p}$
$\displaystyle \therefore \tan\theta=\frac{p^2-1}{2p}$
$\\$

$\displaystyle \textbf{Question 23. }\text{If }\cot\theta=\frac{1}{\sqrt{3}},\text{ the value of }\sec^{2}\theta+\mathrm{cosec}^{2}\theta\text{ is} \hspace{0.2cm}\text{[CBSE 2021]}$
$\displaystyle \text{(a) }1 \qquad \text{(b) }\frac{40}{9} \qquad \text{(c) }\frac{38}{9} \qquad \text{(d) }5\frac{1}{3}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(d)}$
$\displaystyle \text{Given }\cot\theta=\frac{1}{\sqrt{3}}=\cot60^\circ$
$\displaystyle \therefore \theta=60^\circ$
$\displaystyle \therefore \sec^{2}\theta+\mathrm{cosec}^{2}\theta=(\sec60^\circ)^{2}+(\mathrm{cosec}60^\circ)^{2}$
$\displaystyle =(2)^{2}+\left(\frac{2}{\sqrt{3}}\right)^{2}$
$\displaystyle =4+\frac{4}{3}=5\frac{1}{3}$
$\\$

$\displaystyle \textbf{Question 24. }\text{Given that }\sec\theta=\sqrt{2},\text{ the value of }\frac{1+\tan\theta}{\sin\theta}\text{ is} \hspace{0.2cm}\text{[CBSE 2021]}$
$\displaystyle \text{(a) }2\sqrt{2} \qquad \text{(b) }\sqrt{2}$
$\displaystyle \text{(c) }3\sqrt{2} \qquad \text{(d) }2$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(a)}$
$\displaystyle \text{Given }\sec\theta=\sqrt{2}=\sec45^\circ$
$\displaystyle \therefore \theta=45^\circ$
$\displaystyle \therefore \text{The value of }\frac{1+\tan\theta}{\sin\theta}$
$\displaystyle =\frac{1+\tan45^\circ}{\sin45^\circ}$
$\displaystyle =\frac{1+1}{\frac{1}{\sqrt{2}}}=2\sqrt{2}$
$\\$

$\displaystyle \textbf{Question 25. }\text{If }\theta\text{ is an acute angle and }\tan\theta+\cot\theta=2,\text{ then the value of } \\ \sin^{3}\theta+\cos^{3}\theta\text{ is} \hspace{0.2cm}\text{[CBSE 2021]}$
$\displaystyle \text{(a) }1 \qquad \text{(b) }\frac{1}{2}$
$\displaystyle \text{(c) }\frac{\sqrt{2}}{2} \qquad \text{(d) }\sqrt{2}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(c)}$
$\displaystyle \tan\theta+\cot\theta=2$
$\displaystyle \therefore \theta=45^\circ$
$\displaystyle \therefore \text{The value of }\sin^{3}\theta+\cos^{3}\theta$
$\displaystyle =(\sin45^\circ)^{3}+(\cos45^\circ)^{3}$
$\displaystyle =\left(\frac{1}{\sqrt{2}}\right)^{3}+\left(\frac{1}{\sqrt{2}}\right)^{3}$
$\displaystyle =\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$
$\\$

$\displaystyle \textbf{Question 26. }\text{The value of }\left(\sin^{2}\theta+\frac{1}{1+\tan^{2}\theta}\right)=\underline{\hspace{2cm}} \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \sin^2\theta+\frac{1}{1+\tan^2\theta}$
$\displaystyle =\sin^2\theta+\frac{1}{\sec^2\theta}$
$\displaystyle =\sin^2\theta+\cos^2\theta=1$
$\\$

$\displaystyle \textbf{Question 27. }\text{The value of }(1+\tan^{2}\theta)(1-\sin\theta)(1+\sin\theta)=\underline{\hspace{2cm}} \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle (1+\tan^2\theta)(1-\sin\theta)(1+\sin\theta)$
$\displaystyle =\sec^2\theta(1-\sin^2\theta)$
$\displaystyle =\sec^2\theta\cos^2\theta=1$
$\\$

$\displaystyle \textbf{Question 28. }\text{If }\sin\theta+\cos\theta=\sqrt{3},\text{ then prove that }\tan\theta+\cot\theta=1. \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Consider }\sin\theta+\cos\theta=\sqrt{3}$
$\displaystyle \text{Squaring, we get}$
$\displaystyle (\sin\theta+\cos\theta)^2=(\sqrt{3})^2$
$\displaystyle \Rightarrow \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=3$
$\displaystyle \Rightarrow 1+2\sin\theta\cos\theta=3$
$\displaystyle \Rightarrow 2\sin\theta\cos\theta=2$
$\displaystyle \Rightarrow \sin\theta\cos\theta=1 \qquad \cdots (i)$
$\displaystyle \text{Consider }\tan\theta+\cot\theta=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}$
$\displaystyle =\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}$
$\displaystyle =\frac{1}{\sin\theta\cos\theta}$
$\displaystyle =1 \qquad [\text{from (i)}]$
$\\$

$\displaystyle \textbf{Question 29. }\text{Prove that }2(\sin^{6}\theta+\cos^{6}\theta)-3(\sin^{4}\theta+\cos^{4}\theta)+1=0. \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)+1$
$\displaystyle =2[(\sin^2\theta)^3+(\cos^2\theta)^3]-3[(\sin^2\theta)^2+(\cos^2\theta)^2]+1$
$\displaystyle \text{Using }a^3+b^3=(a+b)^3-3ab(a+b)\text{ and }a^2+b^2=(a+b)^2-2ab$
$\displaystyle =2[(1)^3-3\sin^2\theta\cos^2\theta(1)]$
$\displaystyle \qquad -3[(1)^2-2\sin^2\theta\cos^2\theta]+1$
$\displaystyle =2-6\sin^2\theta\cos^2\theta-3+6\sin^2\theta\cos^2\theta+1$
$\displaystyle =0=\text{RHS}$
$\displaystyle \therefore \text{Hence proved}$
$\\$

$\displaystyle \textbf{Question 30. }\text{Prove that: }\frac{\cot\theta+\mathrm{cosec}\,\theta-1}{\cot\theta-\mathrm{cosec}\,\theta+1}=\frac{1+\cos\theta}{\sin\theta}. \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\frac{\cot\theta+\mathrm{cosec}\theta-1}{\cot\theta-\mathrm{cosec}\theta+1}$
$\displaystyle =\frac{(\cot\theta+\mathrm{cosec} \theta)-(\mathrm{cosec}^2 \theta-\cot^2\theta)}{\cot\theta-\mathrm{cosec}\theta+1}$
$\displaystyle =\frac{(\cot\theta+\mathrm{cosec} \theta)-(\mathrm{cosec} \theta+\cot\theta)(\mathrm{cosec}\theta-\cot\theta)}{\cot\theta-\mathrm{cosec}\theta+1}$
$\displaystyle =\frac{(\cot\theta+\mathrm{cosec}\theta)(1-\mathrm{cosec}\theta+\cot\theta)}{\cot\theta-\mathrm{cosec}\theta+1}$
$\displaystyle =\cot\theta+\mathrm{cosec} \theta$
$\displaystyle =\frac{\cos\theta}{\sin\theta}+\frac{1}{\sin\theta}$
$\displaystyle =\frac{\cos\theta+1}{\sin\theta}$
$\displaystyle =\text{RHS}$
$\\$

$\displaystyle \textbf{Question 31. }\text{The rod }AC\text{ of a TV disc antenna is fixed at right angles to the} \\ \text{wall }AB\text{ and a rod }CD\text{ is supporting the disc as shown in Figure.} \\ \text{If }AC=1.5\text{ m long and } CD=3\text{ m, find (i) }\tan\theta\text{ (ii) }\sec\theta+\mathrm{cosec}\,\theta. \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle AC=1.5\text{ m},\quad CD=3\text{ m}$
$\displaystyle \therefore \sin\theta=\frac{AC}{CD}=\frac{1.5}{3}=\frac{1}{2}$
$\displaystyle \Rightarrow \theta=30^\circ$
$\displaystyle \text{(i)}\quad \tan\theta=\tan30^\circ=\frac{1}{\sqrt{3}}$
$\displaystyle \text{(ii)}\quad \sec\theta+\mathrm{cosec}\theta=\sec30^\circ+\mathrm{cosec}30^\circ$
$\displaystyle =\frac{2}{\sqrt{3}}+2$
$\displaystyle =\frac{2}{\sqrt{3}}(1+\sqrt{3})$
$\\$

$\displaystyle \textbf{Question 32. }\text{Prove that }(\sin\theta+\mathrm{cosec}\,\theta)^{2}+(\cos\theta+\sec\theta)^{2}=7+\tan^{2}\theta+\cot^{2}\theta. \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Taking LHS }(\sin\theta+\mathrm{cosec}\theta)^2+(\cos\theta+\sec\theta)^2$
$\displaystyle =\sin^2\theta+\mathrm{cosec}^2\theta+2+\cos^2\theta+\sec^2\theta+2$
$\displaystyle =\sin^2\theta+\cos^2\theta+\mathrm{cosec}^2\theta+\sec^2\theta+4$
$\displaystyle =1+(1+\cot^2\theta)+(1+\tan^2\theta)+4$
$\displaystyle =7+\cot^2\theta+\tan^2\theta$
$\displaystyle =\text{RHS}$
$\\$

$\displaystyle \textbf{Question 33. }\text{Prove that }(1+\cot A-\mathrm{cosec}\,A)(1+\tan A+\sec A)=2. \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle (1+\cot A-\mathrm{cosec} A)(1+\tan A+\sec A)=2$
$\displaystyle \text{Taking LHS}$
$\displaystyle =\left(1+\frac{\cos A}{\sin A}-\frac{1}{\sin A}\right)\left(1+\frac{\sin A}{\cos A}+\frac{1}{\cos A}\right)$
$\displaystyle =\frac{\sin A+\cos A-1}{\sin A}\cdot\frac{\cos A+\sin A+1}{\cos A}$
$\displaystyle =\frac{(\sin A+\cos A)^2-1}{\sin A\cos A}$
$\displaystyle =\frac{\sin^2A+\cos^2A+2\sin A\cos A-1}{\sin A\cos A}$
$\displaystyle =\frac{1+2\sin A\cos A-1}{\sin A\cos A}$
$\displaystyle =\frac{2\sin A\cos A}{\sin A\cos A}$
$\displaystyle =2$
$\displaystyle =\text{RHS}$
$\\$

$\displaystyle \textbf{Question 34. }\text{Prove that }\frac{\sin A-\cos A+1}{\sin A+\cos A-1}=\frac{1}{\sec A-\tan A}. \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\frac{\sin A-\cos A+1}{\sin A+\cos A-1}$
$\displaystyle \text{Dividing numerator and denominator by }\cos A,\text{ we have}$
$\displaystyle =\frac{\tan A-1+\sec A}{\tan A+1-\sec A}$
$\displaystyle =\frac{\tan A+\sec A-(\sec^2A-\tan^2A)}{\tan A-\sec A+1}$
$\displaystyle =\frac{(\tan A+\sec A)(1-\sec A+\tan A)}{\tan A-\sec A+1}$
$\displaystyle =\tan A+\sec A$
$\displaystyle =(\tan A+\sec A)(\sec A-\tan A)$
$\displaystyle \qquad [\because \sec^2A-\tan^2A=1]$
$\displaystyle =\frac{1}{\sec A-\tan A}=\text{RHS}$
$\\$

$\displaystyle \textbf{Question 35. }\text{Prove that }\frac{\tan^{2}A}{\tan^{2}A-1}+\frac{\mathrm{cosec}^{2}A}{\sec^{2}A-\mathrm{cosec}^{2}A}=\frac{1}{1-2\cos^{2}A}. \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\frac{\tan^2A}{\tan^2A-1}+\frac{\mathrm{cosec}^2A}{\sec^2A-\mathrm{cosec}^2A}$
$\displaystyle =\frac{\sin^2A}{\sin^2A-\cos^2A}+\frac{\cos^2A}{1-\cos^2A-\cos^2A}$
$\displaystyle =\frac{\sin^2A}{\sin^2A-\cos^2A}+\frac{\cos^2A}{\sin^2A-\cos^2A}$
$\displaystyle =\frac{\sin^2A+\cos^2A}{\sin^2A-\cos^2A}$
$\displaystyle =\frac{1}{\sin^2A-\cos^2A}$
$\displaystyle =\frac{1}{1-2\cos^2A}=\text{RHS}$
$\\$

$\displaystyle \textbf{Question 36. }\text{If }\sec\theta=x+\frac{1}{4x},\ x\neq0,\text{ find }(\sec\theta+\tan\theta). \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \sec\theta=x+\frac{1}{4x}\qquad \cdots (i)$
$\displaystyle \text{We know that}$
$\displaystyle \sec^2\theta=1+\tan^2\theta$
$\displaystyle \Rightarrow 1+\tan^2\theta=\left(x+\frac{1}{4x}\right)^2$
$\displaystyle \Rightarrow 1+\tan^2\theta=x^2+\frac{1}{16x^2}+\frac{1}{2}$
$\displaystyle \Rightarrow \tan^2\theta=\left(x-\frac{1}{4x}\right)^2$
$\displaystyle \Rightarrow \tan\theta=\pm\left(x-\frac{1}{4x}\right)\qquad \cdots (ii)$
$\displaystyle \text{On adding (i) and (ii),}$
$\displaystyle \sec\theta+\tan\theta=x+\frac{1}{4x}+x-\frac{1}{4x}=2x$
$\displaystyle \text{or}$
$\displaystyle \sec\theta+\tan\theta=x+\frac{1}{4x}-x+\frac{1}{4x}=\frac{1}{2x}$
$\\$

$\displaystyle \textbf{Question 37. }\text{If }4\tan\theta=3,\text{ evaluate }\left(\frac{4\sin\theta-\cos\theta+1}{4\sin\theta+\cos\theta-1}\right). \hspace{0.2cm}\text{[CBSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle 4\tan\theta=3\Rightarrow \tan\theta=\frac{3}{4}$
$\displaystyle \text{In }\triangle ABC,\ \angle B=90^\circ$
$\displaystyle \tan\theta=\frac{AB}{BC}=\frac{3k}{4k}$
$\displaystyle \therefore AB=3k,\ BC=4k$
$\displaystyle \text{By Pythagoras theorem,}$
$\displaystyle AC^2=AB^2+BC^2$
$\displaystyle =(3k)^2+(4k)^2$
$\displaystyle =9k^2+16k^2$
$\displaystyle =25k^2$
$\displaystyle \therefore AC=5k$
$\displaystyle \sin\theta=\frac{AB}{AC}=\frac{3k}{5k}=\frac{3}{5}$
$\displaystyle \cos\theta=\frac{BC}{AC}=\frac{4k}{5k}=\frac{4}{5}$
$\displaystyle \text{Now,}$
$\displaystyle \frac{4\sin\theta-\cos\theta+1}{4\sin\theta+\cos\theta-1}$
$\displaystyle =\frac{4\times\frac{3}{5}-\frac{4}{5}+1}{4\times\frac{3}{5}+\frac{4}{5}-1}$
$\displaystyle =\frac{\frac{12}{5}-\frac{4}{5}+1}{\frac{12}{5}+\frac{4}{5}-1}$
$\displaystyle =\frac{\frac{13}{5}}{\frac{11}{5}}$
$\displaystyle =\frac{13}{11}$
$\\$

$\displaystyle \textbf{Question 38. }\text{Prove that } \left(\frac{1+\tan^{2}A}{1+\cot^{2}A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan^{2}A. \hspace{0.2cm}\text{[CBSE 2018(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{1+\tan^2A}{1+\cot^2A}$
$\displaystyle =\frac{\sec^2A}{\mathrm{cosec}^2A}$
$\displaystyle =\frac{\frac{1}{\cos^2A}}{\frac{1}{\sin^2A}}$
$\displaystyle =\frac{\sin^2A}{\cos^2A}$
$\displaystyle =\tan^2A$
$\displaystyle \text{Taking }\left(\frac{1-\tan A}{1-\cot A}\right)^2$
$\displaystyle =\left[\frac{\cos A-\sin A}{\cos A}\cdot\frac{\sin A}{\sin A-\cos A}\right]^2$
$\displaystyle =\left[\frac{\cos A-\sin A}{\cos A}\cdot\frac{-\sin A}{\cos A-\sin A}\right]^2$
$\displaystyle =\frac{\sin^2A}{\cos^2A}$
$\displaystyle =\tan^2A$
$\displaystyle \therefore \frac{1+\tan^2A}{1+\cot^2A}=\left(\frac{1-\tan A}{1-\cot A}\right)^2$
$\\$

$\displaystyle \textbf{Question 39. }\text{If }\sin(A+2B)=\frac{\sqrt{3}}{2}\text{ and }\cos(A+4B)=0,\ A>B,\text{ and } \\ A+4B\leq90^{\circ},\text{ then find }A\text{ and }B. \hspace{0.2cm}\text{[CBSE 2018(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \sin(A+2B)=\frac{\sqrt{3}}{2}$
$\displaystyle \text{So, }\sin(A+2B)=\sin60^\circ$
$\displaystyle \therefore A+2B=60^\circ \qquad (i)$
$\displaystyle \text{Also, we have }\cos(A+4B)=0$
$\displaystyle \cos(A+4B)=\cos90^\circ$
$\displaystyle \therefore A+4B=90^\circ \qquad (ii)$
$\displaystyle \text{Subtracting }(ii)\text{ from }(i),\text{ we have}$
$\displaystyle -2B=-30^\circ$
$\displaystyle \Rightarrow B=15^\circ$
$\displaystyle \text{Put }B=15^\circ\text{ in equation }(i),\text{ we have}$
$\displaystyle A+2(15^\circ)=60^\circ$
$\displaystyle A+30^\circ=60^\circ$
$\displaystyle \Rightarrow A=30^\circ$
$\\$

$\displaystyle \textbf{Question 40. }\text{If }(1+\cos A)(1-\cos A)=\frac{3}{4},\text{ find the value of }\sec A. \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle (1+\cos A)(1-\cos A)=\frac{3}{4}$
$\displaystyle \Rightarrow 1-\cos^2A=\frac{3}{4}$
$\displaystyle \Rightarrow \sin^2A=\frac{3}{4}$
$\displaystyle \Rightarrow \cos^2A=\frac{1}{4}$
$\displaystyle \Rightarrow \sec^2A=4$
$\displaystyle \Rightarrow \sec A=2$
$\\$

$\displaystyle \textbf{Question 41. }\text{If }\mathrm{cosec}\,\theta+\cot\theta=x,\text{ find the value of }\mathrm{cosec}\,\theta-\cot\theta. \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \mathrm{cosec}\theta+\cot\theta=x \qquad \text{(given)}$
$\displaystyle \text{As we know that}$
$\displaystyle \mathrm{cosec}^2\theta-\cot^2\theta=1$
$\displaystyle \Rightarrow (\mathrm{cosec}\theta-\cot\theta)(\mathrm{cosec}\theta+\cot\theta)=1$
$\displaystyle \Rightarrow (\mathrm{cosec}\theta-\cot\theta)x=1$
$\displaystyle \Rightarrow \mathrm{cosec}\theta-\cot\theta=\frac{1}{x}$
$\\$

$\displaystyle \textbf{Question 42. }\text{Prove that: }\frac{1}{\sec A-\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{\sec A+\tan A}. \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\frac{1}{\sec A-\tan A}-\frac{1}{\sec A+\tan A}$
$\displaystyle =\frac{\sec A+\tan A}{(\sec A-\tan A)(\sec A+\tan A)}-\frac{1}{\cos A}$
$\displaystyle =\sec A+\tan A-\sec A$
$\displaystyle =\tan A$
$\displaystyle \text{RHS}=\frac{1}{\cos A}-\frac{1}{\sec A+\tan A}$
$\displaystyle =\sec A-\frac{\sec A-\tan A}{(\sec A+\tan A)(\sec A-\tan A)}$
$\displaystyle =\sec A-(\sec A-\tan A)$
$\displaystyle =\tan A$
$\displaystyle \therefore \text{LHS}=\text{RHS}$
$\\$

$\displaystyle \textbf{Question 43. }\text{If }\sin\theta=\frac{12}{13},\ 0^{\circ}<\theta<90^{\circ},\text{ find the value of }$
$\displaystyle \frac{\sin^{2}\theta-\cos^{2}\theta}{2\sin\theta\cos\theta}\times\frac{1}{\tan^{2}\theta}. \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: }\sin\theta=\frac{12}{13}$
$\displaystyle \cos\theta=\sqrt{1-\sin^2\theta}$
$\displaystyle =\sqrt{1-\left(\frac{12}{13}\right)^2}$
$\displaystyle =\sqrt{\frac{169-144}{169}}$
$\displaystyle =\sqrt{\frac{25}{169}}$
$\displaystyle =\frac{5}{13}$
$\displaystyle \tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac{12}{13}}{\frac{5}{13}}=\frac{12}{5}$
$\displaystyle \text{Now, put values of }\sin\theta,\cos\theta\text{ and }\tan\theta\text{ in the given expression}$
$\displaystyle \frac{\sin^2\theta-\cos^2\theta}{2\sin\theta\cos\theta}\times\frac{1}{\tan^2\theta}$
$\displaystyle =\frac{\frac{144}{169}-\frac{25}{169}}{2\times\frac{12}{13}\times\frac{5}{13}}\times\frac{25}{144}$
$\displaystyle =\frac{\frac{119}{169}}{\frac{120}{169}}\times\frac{25}{144}$
$\displaystyle =\frac{119}{120}\times\frac{25}{144}$
$\displaystyle =\frac{595}{3456}$
$\\$

$\displaystyle \textbf{Question 44. }\text{Prove that: }\frac{\tan^{3}\theta}{1+\tan^{2}\theta}+\frac{\cot^{3}\theta}{1+\cot^{2}\theta}=\sec\theta\,\mathrm{cosec}\,\theta-2\sin\theta\cos\theta. \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\frac{\tan^3\theta}{1+\tan^2\theta}+\frac{\cot^3\theta}{1+\cot^2\theta}$
$\displaystyle =\frac{\tan^3\theta}{\sec^2\theta}+\frac{\cot^3\theta}{\mathrm{cosec}^2\theta}$
$\displaystyle =\frac{\sin^3\theta}{\cos\theta}+\frac{\cos^3\theta}{\sin\theta}$
$\displaystyle =\frac{\sin^4\theta+\cos^4\theta}{\sin\theta\cos\theta}$
$\displaystyle =\frac{(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta}{\sin\theta\cos\theta}$
$\displaystyle =\frac{1-2\sin^2\theta\cos^2\theta}{\sin\theta\cos\theta}$
$\displaystyle =\frac{1}{\sin\theta\cos\theta}-2\sin\theta\cos\theta$
$\displaystyle =\sec\theta\mathrm{cosec}\theta-2\sin\theta\cos\theta$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 45. }\text{If }\sec\theta-\tan\theta=x,\text{ show that: }\sec\theta=\frac{1}{2}\left(x+\frac{1}{x}\right)\text{ and } \\ \tan\theta=\frac{1}{2}\left(\frac{1}{x}-x\right). \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: }\sec\theta-\tan\theta=x \qquad (i)$
$\displaystyle \text{As we know, }\sec^{2}\theta-\tan^{2}\theta=1$
$\displaystyle \Rightarrow(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1$
$\displaystyle \Rightarrow x(\sec\theta+\tan\theta)=1$
$\displaystyle \Rightarrow \sec\theta+\tan\theta=\frac{1}{x} \qquad (ii)$
$\displaystyle \text{Adding }(i)\text{ and }(ii),\text{ we get}$
$\displaystyle 2\sec\theta=x+\frac{1}{x}$
$\displaystyle \Rightarrow \sec\theta=\frac{1}{2}\left(x+\frac{1}{x}\right)$
$\displaystyle \text{Subtracting }(i)\text{ from }(ii),\text{ we get}$
$\displaystyle 2\tan\theta=\frac{1}{x}-x$
$\displaystyle \Rightarrow \tan\theta=\frac{1}{2}\left(\frac{1}{x}-x\right)$
$\\$

$\displaystyle \textbf{Question 46. }\text{Write the values of }\sec0^{\circ},\sec30^{\circ},\sec45^{\circ},\sec60^{\circ}\text{ and }\sec90^{\circ}. \\ \text{What happens to }\sec x\text{ when }x\text{ increases from }0^{\circ}\text{ to }90^{\circ}? \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Angles }(\theta)&0^\circ&30^\circ&45^\circ&60^\circ&90^\circ\\ \hline \sec\theta&1&\frac{2}{\sqrt{3}}&\sqrt{2}&2&\text{not defined}\\ \hline \end{array}$
$\displaystyle \text{The value of }\sec x\text{ increases and finally reaches to not defined limit as }x\text{ increases from }0^\circ\text{ to }90^\circ.$
$\\$