If 2 tan-1 x -sin-12x-1-x2- then-

Putting θ =tan^ 1x, π/2 lt; θ lt; π/2, we have R.H.S. sin^ 12 tan θ/1+tan^2 θ=sin^ 1sin 2 θ =2 θ =2 tan^ 1x When π/2≤ 2 θ≤π/2 i.e., π/4≤θ≤π/4 i.e., 1 ...

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Standard XI

Mathematics

Basic Inverse Trigonometric Functions

If 2 tan-1 x ...

Question

If $2 t a n^{- 1} x = s i n^{- 1} \frac{2 x}{1 + x^{2}}$ , then:

x \in R

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x \geq 1

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x \leq - 1

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- 1 \leq x \leq 1

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Solution

The correct option is D $- 1 \leq x \leq 1$
Putting $θ = t a n^{- 1} x, - \frac{π}{2} < θ < \frac{π}{2}$ , we have
R.H.S. $s i n^{- 1} \frac{2 t a n θ}{1 + t a n^{2} θ} = s i n^{- 1} s i n 2 θ = 2 θ = 2 t a n^{- 1} x$
When $- \frac{π}{2} \leq 2 θ \leq \frac{π}{2}$ i.e., $- \frac{π}{4} \leq θ \leq \frac{π}{4}$
i.e., $- 1 \leq x = t a n θ \leq 1$ .

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If 2tan−1x=sin−12x1+x2, then:

The correct option is D −1≤x≤1Putting θ=tan−1x,−π2<θ<π2, we have R.H.S. sin−12tanθ1+tan2θ=sin−1sin2θ=2θ=2tan−1x When −π2≤2θ≤π2 i.e., −π4≤θ≤π4 i.e., −1≤x=tanθ≤1.

If $2 t a n^{- 1} x = s i n^{- 1} \frac{2 x}{1 + x^{2}}$ , then: