Question

A value of x satisfying the equation $\sin [cot^{-1}(1 + x)] = \cos[tan^{-1}x]$, is

• A. $\frac{-1}{2}$
• B. 0
• C. -1
• D. -$\frac{1}{2}$

Answer 1

Answer: (A), (D)

x = -$\frac{1}{2}$

Answer 2

To solve the equation $\sin [\cot^{-1}(1 + x)] = \cos[\tan^{-1}x]$, we can use trigonometric identities.

First, we simplify both sides of the equation:

• Left-hand side (LHS): $\sin[\cot^{-1}(1 + x)] = \frac{1}{\sqrt{1 + (1+x)^2}}$

• Right-hand side (RHS): $\cos[\tan^{-1}x] = \frac{1}{\sqrt{1 + x^2}}$

Now, we set the LHS equal to the RHS:

$\frac{1}{\sqrt{1 + (1+x)^2}} = \frac{1}{\sqrt{1 + x^2}}$

Squaring both sides to eliminate the square roots gives:

$1 + (1+x)^2 = 1 + x^2$

Expanding and simplifying:

$1 + (1 + 2x + x^2) = 1 + x^2$ $2 + 2x + x^2 = 1 + x^2$ $2x = -1$ $x = -\frac{1}{2}$

To verify the solution, we substitute $x = -\frac{1}{2}$ back into the original equation. Both sides evaluate to $\frac{2}{\sqrt{5}}$, confirming that $x = -\frac{1}{2}$ is indeed the correct solution.