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

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(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 = $

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