tan ^{-1} 2+tan ^{-1} 3= | vedclass

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

Question

(D) We use the formula $	an ^{-1} x + 	an ^{-1} y = \pi + 	an ^{-1} \left( \frac{x+y}{1-xy} ight)$ when $xy > 1$.Here,$x = 2$ and $y = 3$. Since

Vedclass · Accepted Answer

$\frac{3 \pi}{4}$

Vedclass · Answer

(D) We use the formula $	an ^{-1} x + 	an ^{-1} y = \pi + 	an ^{-1} \left( \frac{x+y}{1-xy} ight)$ when $xy > 1$.Here,$x = 2$ and $y = 3$. Since $xy = 2 	imes 3 = 6 > 1$,we apply the formula:$	an ^{-1} 2 + 	an ^{-1} 3 = \pi + 	an ^{-1} \left( \frac{2+3}{1-(2 	imes 3)} ight)$$= \pi + 	an ^{-1} \left( \frac{5}{1-6} ight)$$= \pi + 	an ^{-1} \left( \frac{5}{-5} ight)$$= \pi + 	an ^{-1} (-1)$Since $	an ^{-1} (-1) = -\frac{\pi}{4}$,we have:$= \pi - \frac{\pi}{4} = \frac{3 \pi}{4}$.

$\tan ^{-1} 2+\tan ^{-1} 3=$

Similar Questions

If $y=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)+\tan ^{-1}\left(\frac{4 x-4 x^3}{1-6 x^2+x^4}\right)$, then $\frac{d y}{d x}$ is equal to

The value of $\tan^{-1} \left( \frac{\sin 2 - 1}{\cos 2} \right)$ is equal to:

$\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3} = ?$

Prove that $\frac{9 \pi}{8} - \frac{9}{4} \sin^{-1} \frac{1}{3} = \frac{9}{4} \sin^{-1} \frac{2 \sqrt{2}}{3}$.

If $2 \sin^{-1} x = \sin^{-1}(2x \sqrt{1-x^2})$,then $x \in$ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module