The value of x for which s i n ( c o t − 1 ( 1 + x ) ) = c o s ( t a n − 1 x ) is: Options: A . &nbsp 1 2 B . &nbsp 1 C . &nbsp 0 D . &nbsp − 1 2

Answer: Option D : D s i n ( c o t − 1 ( 1 + x ) ) = c o s ( t a n − 1 x ) ⇒ c o t − 1 ( 1 + x ) = [ π 2 ± t a n − 1 x ] ⇒ π 2 − t a n − 1 ( 1 + x ) = π 2 ± t a n − 1 x ⇒ 1 + x = ± x x = − 1 2 Which, on verification, satisfies the equation.

The value of x for which sin(cot−1(1+x))=cos(tan−1x) is:

Discussion Forum : Inverse Trigonometric Functions

Question -

The value of x for which $s i n (c o t^{- 1} (1 + x)) = c o s (t a n^{- 1} x)$ is:

Options:

A .

\frac{1}{2}

B . 1

C . 0

D .

- \frac{1}{2}

Answer: Option D
:
D

s i n (c o t^{- 1} (1 + x)) = c o s (t a n^{- 1} x) \Rightarrow c o t^{- 1} (1 + x) = [\frac{π}{2} \pm t a n^{- 1} x] \Rightarrow \frac{π}{2} - t a n^{- 1} (1 + x) = \frac{π}{2} \pm t a n^{- 1} x \Rightarrow 1 + x = \pm x

x = - \frac{1}{2}

Which, on verification, satisfies the equation.

Was this answer helpful ?

Next Question

Discussion Forum : Inverse Trigonometric Functions

Submit Your Solution hear: