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प्रश्न
If \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} \right)\] = α, then x2 =
पर्याय
sin 2 α
sin α
cos 2 α
cos α
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उत्तर
(a) sin 2α
\[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} \right) = \alpha\]
\[ \Rightarrow \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} = \tan\alpha\]
\[\]
\[ \Rightarrow \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} \times \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} = \tan\alpha\]
\[ \Rightarrow \frac{\left( \sqrt{1 + x^2} \right)^2 + \left( \sqrt{1 - x^2} \right)^2 - 2\sqrt{1 + x^2}\sqrt{1 - x^2}}{\left( \sqrt{1 + x^2} \right)^2 - \left( \sqrt{1 - x^2} \right)^2} = \tan\alpha\]
\[ \Rightarrow \frac{1 - \sqrt{1 - x^4}}{x^2} = \tan\alpha\]
\[ \Rightarrow x^2 \tan\alpha = 1 - \sqrt{1 - x^4}\]
\[ \Rightarrow \sqrt{1 - x^4} = 1 - x^2 \tan\alpha\]
\[ \Rightarrow 1 - x^4 = 1 + x^4 \tan^2 \alpha - 2 x^2 \tan\alpha\]
\[ \Rightarrow x^4 + x^4 \tan^2 \alpha - 2 x^2 \tan\alpha = 0\]
\[ \Rightarrow x^4 \sec^2 \alpha - 2 x^2 \tan\alpha = 0\]
\[ \Rightarrow x^2 \left( x^2 \sec^2 \alpha - 2\tan\alpha \right) = 0\]
\[ \Rightarrow x^2 \sec^2 \alpha - 2\tan\alpha = 0 \left[ \because x^2 \neq 0 \right]\]
\[ \Rightarrow x^2 \sec^2 \alpha = 2\tan\alpha\]
\[ \Rightarrow x^2 = \frac{2\tan\alpha}{\sec^2 \alpha} = 2\sin\alpha\cos\alpha = \sin2\alpha\]
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