CBSE

MATHEMATICS · Multi correct · CBSE · Mathematics · Inverse Trigonometric Functions · Inverse Trigonometric Functions

Problem

cos⁡−1x=tan⁡−1x\cos^{-1}x=\tan^{-1}x, then (1) x2=5−12x^2=\frac{\sqrt{5}-1}{2} (2) x2=5+12x^2=\frac{\sqrt{5}+1}{2} (3) sin⁡(cos⁡−1x)=5−12\sin(\cos^{-1}x)=\frac{\sqrt{5}-1}{2} (4) tan⁡(cos⁡−1x)=5−12\tan(\cos^{-1}x)=\frac{\sqrt{5}-1}{2}

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