If sin−1x+tan−1x=π2 , then the value of x2 is

sin−1x+tan−1x=π2 ⇒sin−1x+cos−111+x2=π2 x=11+x2⇒x2=11+x2⇒x2+x4=1 x4+x2−1=0⇒x2=−1±1−4(−1)2 x2=−1+52=2(5−14)=2sin18°

If sin−1x+tan−1x=π2 , then the value of x2 is

sin−1x+tan−1x=π2 ⇒sin−1x+cos−111+x2=π2 x=11+x2⇒x2=11+x2⇒x2+x4=1 x4+x2−1=0⇒x2=−1±1−4(−1)2 x2=−1+52=2(5−14)=2sin18°

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If $\sin^{- 1} x + \tan^{- 1} x = \frac{π}{2}$ , then the value of $x^{2}$ is

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2 \sin 18 °

\sin 18 °

\cos 36 °

2 \cos 36 °

answer is C.

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

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

$x = \frac{1}{\sqrt{1 + x^{2}}} \Rightarrow x^{2} = \frac{1}{1 + x^{2}} \Rightarrow x^{2} + x^{4} = 1$

$x^{4} + x^{2} - 1 = 0 \Rightarrow x^{2} = \frac{- 1 \pm \sqrt{1 - 4 (- 1)}}{2}$

$x^{2} = \frac{- 1 + \sqrt{5}}{2} = 2 (\frac{\sqrt{5} - 1}{4}) = 2 \sin 18 °$

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If $\sin^{- 1} x + \tan^{- 1} x = \frac{π}{2}$ , then the value of $x^{2}$ is