Evaluate: tan ^2(sec ^{-1} 3) + operatorname{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the expression be $E = 	an ^2(\sec ^{-1} 3) + \operatorname{cosec}^2(\cot ^{-1} 2) + \cos ^2(\cos ^{-1} \frac{2}{3} + \sin ^{-1} \frac{2}{3})

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(D) Let the expression be $E = 	an ^2(\sec ^{-1} 3) + \operatorname{cosec}^2(\cot ^{-1} 2) + \cos ^2(\cos ^{-1} \frac{2}{3} + \sin ^{-1} \frac{2}{3})$.Step $1$: Simplify $	an ^2(\sec ^{-1} 3)$.Let $\sec ^{-1} 3 = 	heta_1$,so $\sec 	heta_1 = 3$. Then $	an ^2 	heta_1 = \sec ^2 	heta_1 - 1 = 3^2 - 1 = 9 - 1 = 8$.Step $2$: Simplify $\operatorname{cosec}^2(\cot ^{-1} 2)$.Let $\cot ^{-1} 2 = 	heta_2$,so $\cot 	heta_2 = 2$. Then $\operatorname{cosec}^2 	heta_2 = 1 + \cot ^2 	heta_2 = 1 + 2^2 = 1 + 4 = 5$.Step $3$: Simplify $\cos ^2(\cos ^{-1} \frac{2}{3} + \sin ^{-1} \frac{2}{3})$.We know that $\cos ^{-1} x + \sin ^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$.Thus,$\cos ^2(\frac{\pi}{2}) = 0^2 = 0$.Step $4$: Calculate the total sum.$E = 8 + 5 + 0 = 13$.Therefore,the correct option is $D$.

Evaluate: $\tan ^2(\sec ^{-1} 3) + \operatorname{cosec}^2(\cot ^{-1} 2) + \cos ^2(\cos ^{-1} \frac{2}{3} + \sin ^{-1} \frac{2}{3}) = $ . . . . . . .

Similar Questions

$\tan ^{-1} \sqrt{3} - \cot ^{-1}(-\sqrt{3}) = $ . . . . . . .

The value of $\cos \left(\tan ^{-1}\left(\sin \left(\cot ^{-1} x\right)\right)\right)$ is

If $\theta = \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{13}\right) + \tan^{-1}\left(\frac{1}{21}\right) + \tan^{-1}\left(\frac{1}{31}\right)$,then $\tan \theta =$

If $\sin^{-1} x + \sin^{-1} y = \frac{2\pi}{3}$,then $\cos^{-1} x + \cos^{-1} y = $

If $\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{Bx+C}{x^{2}+1}$,then $\sin^{-1} A + \tan^{-1} B + \sec^{-1} C$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module