Question

evaluate: cot(sec inverse x+sin inverse(1/X))

Answer 1

\[\cot [\sec^{-1}(x) + \sin^{-1}(\frac{1}{x})]\]

Is this what the question is?

Answer 2

yes

Answer 3

prove it @Siddy

Answer 4

can we use calculus?

Answer 5

yeh @mukushla

Answer 6

thts not true @Siddy

Answer 7

cot(pi/2) = 0

Answer 8

\[\cot(\sec^{-1}x + cosec^{-1}x ) \implies \cot(\frac{\pi}{2}) = 0\]

Answer 9

let
\[f(x)=arcsec \ x+ \arcsin \ \frac{1}{x} \\ f'(x)=\frac{1}{\left| x \right| \sqrt{x^2-1}}-\frac{1}{\left| x \right| \sqrt{x^2-1}}=0 \]
so f(x) is constant

Answer 10

\[\sin^{-1} \frac{1}{x} = cosec^{-1}x\]

Answer 11

\[\sin^{-1}x + \cos^{-1} x = \frac{\pi}{2}\]

\[\sec^{-1}x + cosec^{-1} x =\frac{\pi}{2}\]

\[\tan^{-1}x + \cot^{-1} x = \frac{\pi}{2}\]

Answer 12

Using these formula the term in the bracket can be changed to \(\sec^{-1}x + \csc^{-1} x = \frac{\pi}{2}\) and you will get your answer..

Answer 13

will the answer be 0 ?

Answer 14

Yes...

Answer 15

fine.thanks.

Answer 16

Welcome..