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Find the value of x for which { ^{ - 1}}x + { ^{ - 1}}x = { }{

Question

Find the value of $x$ for which $sec^{- 1} x + sin^{- 1} x = \frac{π}{2} .$

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Harper Discussed

Find the value of

x

for which

sec^{- 1} x + sin^{- 1} x = \frac{π}{2} .

9 mins ago

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9 mins ago

Text solutionVerified

Solution: given,

\Rightarrow sec^{- 1} x + sin^{- 1} x = \frac{π}{2}

We know

sin^{- 1} x + cos^{- 1} x = \frac{π}{2}

and

sec^{- 1} x = cos^{- 1} \frac{1}{x}

\Rightarrow cos^{- 1} \frac{1}{x} + sin^{- 1} x = \frac{π}{2}

\Rightarrow \frac{π}{2} - sin^{- 1} \frac{1}{x} + sin^{- 1} x = \frac{π}{2}

\Rightarrow sin^{- 1} x - sin^{- 1} \frac{1}{x} = 0

\Rightarrow sin^{- 1} (x 1 - \frac{1}{x ^{2}} - \frac{1}{x} 1 - x^{2}) = 0

\Rightarrow x \times \frac{x ^{2} - 1}{x} - \frac{1}{x} 1 - x^{2} = sin 0

\Rightarrow x^{2} - 1 - \frac{1}{x} 1 - x^{2} = 0

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

squaring both side

\Rightarrow (x^{2} - 1) = \frac{1 - x ^{2}}{x ^{2}}

\Rightarrow (x^{2} - 1) = \frac{- ( x ^{2} - 1 )}{x ^{2}}

\Rightarrow x^{2} = - 1

\Rightarrow x = \pm i

Ans (here

i^{2} = - 1

)

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Question Text	Find the value of $x$ for which $sec^{- 1} x + sin^{- 1} x = \frac{π}{2} .$
Answer Type	Text solution:1
Upvotes	150