Question 28
Answer fast

Question 28 Answer fast 25. tan 26. tan- I + cos a s — 3ax2 27. Solve for x, tan-I(x+ l) + I ) = tan i 8 Find the values of x which satisfy the equation sin-lx + sin-1(1 —x) = cos-lx. 29. Find the value of x, if sin [cot-I(x + l)] = cos(tan-l x) '201 INCE.kI A1 I Illiubanesl.w'il 2013 Inverse Trigonometric Functions 67

We have, \sin^{- 1} x + \sin^{- 1} (1 - x) = \cos^{- 1} x \Rightarrow \sin^{- 1} x - \cos^{- 1} x = - \sin^{- 1} (1 - x) \Rightarrow \sin^{- 1} x - \cos^{- 1} x = \sin^{- 1} (x - 1) . . . . . . . . (1) [as \sin^{- 1} (- x) = - \sin^{- 1} x] Put \sin^{- 1} x = θ a n d \cos^{- 1} x = ϕ \Rightarrow \sin θ = x a n d \cos ϕ = x Then, \cos θ = \sqrt{1 - \sin^{2} θ} and \sin ϕ = \sqrt{1 - \cos^{2} ϕ} \Rightarrow \cos θ = \sqrt{1 - x^{2}} and \sin ϕ = \sqrt{1 - x^{2}} Applying the formula : \sin (θ - ϕ) = \sin θ \cos ϕ - \cos θ \sin ϕ \Rightarrow \sin (θ - ϕ) = x . x - \sqrt{1 - x^{2}} . \sqrt{1 - x^{2}} \Rightarrow \sin (θ - ϕ) = x^{2} - (1 - x^{2}) \Rightarrow \sin (θ - ϕ) = x^{2} - 1 + x^{2} \Rightarrow \sin (θ - ϕ) = 2 x^{2} - 1 \Rightarrow (θ - ϕ) = \sin^{- 1} (2 x^{2} - 1) \Rightarrow \sin^{- 1} x - \cos^{- 1} x = \sin^{- 1} (2 x^{2} - 1) . . . . . . . . . . . . (2)

Hence, from (1) and (2) \sin^{- 1} (2 x^{2} - 1) = \sin^{- 1} (x - 1) \Rightarrow 2 x^{2} - 1 = x - 1 \Rightarrow 2 x^{2} - x - 1 + 1 = 0 \Rightarrow 2 x^{2} - x = 0 \Rightarrow x (2 x - 1) = 0 Either x = 0 or 2 x - 1 = 0 Either x = 0 or x = \frac{1}{2}

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