If (^{-1}) = \{ ^{-1} ≤ft( {a}{{b² - a²} ) \} : b>a, {then } x=

Correct Answer is : b2b2−a2{b}{{2b² - a²}}2b2−a2b

Correct Answer is : b2b2−a2{b}{{2b² - a²}}2b2−a2b

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AP EAMCET Engineering 26 Apr 2017 Shift 1 Paper

Section: Mathematics

Question : 24 of 160

Marks: +1, -0

\cot(\cos^{-1}) = \sec \{ \tan^{-1} \left( \frac{a}{\sqrt{b^2 - a^2}} \right) \} : b>a, \text{then } x=

Solution:

It is given that,

\cot (\cos^{-1} x) = \sec \{ \tan^{-1} \left( \frac{a}{\sqrt{b^{2} - a^{2}}} \right) \}

The above expression is simplified as,

\cot \left( \cot^{-1} \frac{x}{\sqrt{1 - x^{2}}} \right)

= \sec \{ \sec^{-1} \sqrt{1 + \left( \frac{a}{\sqrt{b^{2} - a^{2}}} \right)^{2}} \}

\frac{x}{\sqrt{1 - x^{2}}} = \sqrt{1 + \left( \frac{a}{\sqrt{b^{2} - a^{2}}} \right)^{2}}

\frac{x}{\sqrt{1 - x^{2}}} = \sqrt{ \frac{b^{2} - a^{2} + a^{2}}{b^{2} - a^{2}} }

\frac{x}{\sqrt{1 - x^{2}}} = \frac{b}{\sqrt{b^{2} - a^{2}}}

Square both sides,

\frac{x^{2}}{1 - x^{2}} = \frac{b^{2}}{b^{2} - x^{2}}

x^{2} b^{2} - x^{2} a^{2} = b^{2} - b^{2} x^{2}

x^{2} (2 b^{2} - a^{2}) = b^{2}

x = \frac{b}{\sqrt{2 b^{2} - a^{2}}}

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