If 2 tan -1 x -sin -12 x -1-x2- then-A- x - RB- x - 1C- x -1D- -1 - x - 1

The correct option is D 1 ≤ x ≤ 1Putting 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 ...

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Byju's Answer

Standard XII

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., $- 1 \leq t a n θ \leq 1$
i.e., $- 1 \leq x \leq 1$

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Similar questions

Q. Prove the following

(1) {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 {tan}^{- 1} x, | x | \leq 1

(2) {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = 2 {tan}^{- 1} x, x \geq 0

(3) {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) = 2 {tan}^{- 1} x, - 1 < x < 1

c o s^{- 1} x = t a n^{- 1} \frac{2 x}{1 + x^{2}}

, then:

Q. If

x \geq 1, t h e n 2 T a n^{- 1} x + S i n^{- 1} (\frac{2 x}{1 + x^{2}})

=___

Q. If

x > 1

, then

2 {tan}^{- 1} x + {sin}^{- 1} (\frac{2 x}{1 + x^{2}})

is equal to

Q. If

2 {tan}^{- 1} x + {sin}^{- 1} \frac{2 x}{1 + x^{2}}

is independent of

x

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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−12tan θ1+tan2 θ=sin−1sin 2θ=2 θ=2 tan−1x When −π2≤2θ≤π2 i.e., −π4≤θ≤π4 i.e., −1≤tan θ≤1 i.e.,−1≤x≤1

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