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Find the value of cos[tan-1{sin(cot-1x)}] = \sqrt {\frac{{{x^2} + 1}}{{{x^2} + 2}}}

Option: 1

x


Option: 2

\sqrt {\frac{{{x^2} + 1}}{{{x^2} + 2}}}


Option: 3

1/x


Option: 4

\sqrt {\frac{{{x^2} + 2}}{{{x^2} + 1}}}


Answers (1)

best_answer

As we learnt

 

Relation between all the Inverse Trigonometric Functions -

 

\sin ^{-1}x = \tan ^{-1}\frac{x}{\sqrt{1-x^{2}}} = cosec ^{-1}\frac{1}{x} = \cos ^{-1}\sqrt{1-x^{2}} = \cot ^{-1}\frac{\sqrt{1-x^{2}}}{x} = \sec ^{-1}\frac{1}{\sqrt{1-x^{2}}}

- wherein

This happens with domain of functions, x>0

 

 

Let cot–1 x =\theta \Rightarrow x = cot\theta, 0 < \theta < \pi.

            \therefore sin (cot^{-1}x) = sin\theta =\frac{1}{{\sqrt {1 + {x^2}} }} , as 0 < \theta < \pi

            Now let \tan^{-1}\frac{1}{{\sqrt {1 + {x^2}} }}= a \Rightarrow  \frac{1}{{\sqrt {1 + {x^2}} }} = \tan \alpha ,\,0 < \alpha < \pi /2$

                        =cos\left[ {{{\tan }^{ - 1}}\left( {\frac{1}{{\sqrt {1 + {x^2}} }}} \right)} \right] =  \cos \alpha = \sqrt {\frac{{1 + {x^2}}}{{2 + {x^2}}}} 

Posted by

manish painkra

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