If 2 t a n − 1 x = s i n − 1 2 x 1 + x 2 , then: Options: A . &nbsp x ∈ R B . &nbsp x ≥ 1 C . &nbsp x ≤ − 1 D . &nbsp − 1 ≤ x ≤ 1

Answer: Option D : D Putting θ = t a n − 1 x , − π 2 < θ < π 2 , we have R.H.S. s i n − 1 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 − π 2 ≤ 2 θ ≤ π 2 i.e., − π 4 ≤ θ ≤ π 4 i.e., − 1 ≤ x = t a n θ ≤ 1 .

If 2tan−1x=sin−12x1+x2, then:

Discussion Forum : Inverse Trigonometric Functions

Question -

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

Options:

A .

x \in R

B .

x \geq 1

C .

x \leq - 1

D .

- 1 \leq x \leq 1

Answer: Option D
:
D
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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Discussion Forum : Inverse Trigonometric Functions

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