If 3 s i n − 1 2 x 1 + x 2 − 4 c o s − 1 1 − x 2 1 + x 2 + 2 t a n − 1 2 x 1 + x 2 = π 3 then x= Options: A . &nbsp √ 3 B . &nbsp 1 √ 3 C . &nbsp 1 D . &nbsp None of these

Answer: Option B : B 3 s i n − 1 2 x 1 + x 2 − 4 c o s − 1 1 − x 2 1 + x 2 + 2 t a n − 1 2 x 1 + x 2 = π 3 Putting x = t a n θ 3 s i n − 1 ( 2 t a n θ 1 + t a n 2 θ ) − 4 c o s − 1 ( 1 − t a n 2 θ 1 + t a n 2 θ ) + 2 t a n − 1 ( 2 t a n θ 1 − t a n 2 θ ) = π 3 ⇒ 3 s i n − 1 ( s i n 2 θ ) − 4 c o s − 1 ( c o s 2 θ ) + 2 t a n − 1 ( t a n 2 θ ) = π 3 ⇒ 3 ( 2 θ ) − 4 ( 2 θ ) + 2 ( 2 θ ) = π 3 ⇒ 6 θ − 8 θ + 4 θ = π 3 ⇒ θ = π 6 ⇒ t a n − 1 x = π 6 ⇒ x = t a n π 6 = 1 √ 3

If 3sin−12x1+x2−4cos−11−x21+x2+2tan−12x1+x2

Discussion Forum : Inverse Trigonometric Functions

Question -

If $3 s i n^{- 1} \frac{2 x}{1 + x^{2}} - 4 c o s^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 t a n^{- 1} \frac{2 x}{1 + x^{2}} = \frac{π}{3}$ then x=

Options:

A .

\sqrt{3}

B .

\frac{1}{\sqrt{3}}

C . 1

D . None of these

Answer: Option B
:
B

3 s i n^{- 1} \frac{2 x}{1 + x^{2}} - 4 c o s^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 t a n^{- 1} \frac{2 x}{1 + x^{2}} = \frac{π}{3}

Putting

x = t a n θ

3 s i n^{- 1} (\frac{2 t a n θ}{1 + t a n^{2} θ}) - 4 c o s^{- 1} (\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ})

+ 2 t a n^{- 1} (\frac{2 t a n θ}{1 - t a n^{2} θ}) = \frac{π}{3}

\Rightarrow 3 s i n^{- 1} (s i n 2 θ) - 4 c o s^{- 1} (c o s 2 θ)

+ 2 t a n^{- 1} (t a n 2 θ) = \frac{π}{3}

\Rightarrow 3 (2 θ) - 4 (2 θ) + 2 (2 θ) = \frac{π}{3} \Rightarrow 6 θ - 8 θ + 4 θ = \frac{π}{3}

\Rightarrow θ = \frac{π}{6} \Rightarrow t a n^{- 1} x = \frac{π}{6} \Rightarrow x = t a n \frac{π}{6} = \frac{1}{\sqrt{3}}

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