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All previous year questions from Chapter: Inverse Trigonometric Functions
2 questions found
If 2cos−1x=y2\cos^{-1}x = y2cos−1x=y, then
(a) Simplify: tan−1[cos2x−sin2xcos2x+sin2x]\tan^{-1}\left[\dfrac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x}\right]tan−1[cos2x+sin2xcos2x−sin2x], 0<x<π/40 < x < \pi/40<x<π/4. OR (b) Evaluate: tan[sin−11−cos−1(−1/2)]\tan\left[\sin^{-1} 1 - \cos^{-1}(-1/2)\right]tan[sin−11−cos−1(−1/2)].