Evaluate: {cot ^{ - 1}}3 + {csc ^{ - 1}}sqrt | vedclass

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

Question

(B) We know that ${\csc ^{ - 1}}x = {\cot ^{ - 1}}\sqrt {{x^2} - 1} $ for $x > 1$. Substituting $x = \sqrt 5$,we get ${\csc ^{ - 1}}\sqrt 5 = {\cot ^{

Vedclass · Accepted Answer

$\frac{\pi }{4}$

Vedclass · Answer

(B) We know that ${\csc ^{ - 1}}x = {\cot ^{ - 1}}\sqrt {{x^2} - 1} $ for $x > 1$. Substituting $x = \sqrt 5$,we get ${\csc ^{ - 1}}\sqrt 5 = {\cot ^{ - 1}}\sqrt {{(\sqrt 5 )^2} - 1} = {\cot ^{ - 1}}\sqrt {5 - 1} = {\cot ^{ - 1}}\sqrt 4 = {\cot ^{ - 1}}2$. Now,the expression becomes ${\cot ^{ - 1}}3 + {\cot ^{ - 1}}2$. Using the formula ${\cot ^{ - 1}}x + {\cot ^{ - 1}}y = {\cot ^{ - 1}}\left( {\frac{{xy - 1}}{{x + y}}} ight)$ for $xy > 1$: ${\cot ^{ - 1}}3 + {\cot ^{ - 1}}2 = {\cot ^{ - 1}}\left( {\frac{{3 	imes 2 - 1}}{{3 + 2}}} ight) = {\cot ^{ - 1}}\left( {\frac{{6 - 1}}{5}} ight) = {\cot ^{ - 1}}\left( {\frac{5}{5}} ight) = {\cot ^{ - 1}}(1)$. Since ${\cot ^{ - 1}}(1) = \frac{\pi }{4}$,the final answer is $\frac{\pi }{4}$.

Evaluate: ${\cot ^{ - 1}}3 + {\csc ^{ - 1}}\sqrt 5 = $

Similar Questions

$\cot ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) = $ . . . . . .

Find $\frac{dy}{dx}$,if $y = \sin^{-1}x + \sin^{-1}\sqrt{1-x^2}$,where $-1 \le x \le 1$.

Let $S$ be the set of all solutions of the equation $\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^2}\right)=\pi$,where $x \in\left[-\frac{1}{2}, \frac{1}{2}\right]$. Then $\sum_{x \in S} 2 \sin ^{-1}\left(x^2-1\right)$ is equal to

${\sin ^{ - 1}}\left[ {x\sqrt {1 - x} - \sqrt x \sqrt {1 - {x^2}} } \right] = $

Statement $I:$ The equation $(\sin^{-1} x)^3 + (\cos^{-1} x)^3 - a\pi^3 = 0$ has a solution for all $a \ge \frac{1}{32}.$
Statement $II:$ For any $x \in [-1, 1],$ $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ and $0 \le (\sin^{-1} x - \frac{\pi}{4})^2 \le \frac{9\pi^2}{16}.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module