Question

If \( \cot(\cos^{-1} x) = \sec \left( \tan^{-1} \left( \frac{a}{\sqrt{b^2 - a^2}} \right) \right) \), then:

• A. \( \frac{b}{\sqrt{2b^2 - a^2}} \)
• B. \( \frac{\sqrt{b^2 - a^2}}{ab} \)
• C. \( \frac{a}{\sqrt{2b^2 - a^2}} \)
• D. \( \frac{\sqrt{b^2 - a^2}}{a} \)

Hint:

For inverse trigonometric functions, rewrite in terms of known trigonometric identities to simplify expressions.

Answer 1

The Correct Option is A

Step 1: {Express cotangent and secant}
\[ \cot(\cos^{-1} x) = \frac{x}{\sqrt{1 - x^2}}. \] \[ \sec \left( \tan^{-1} \left( \frac{a}{\sqrt{b^2 - a^2}} \right) \right) = \sqrt{1 + \left( \frac{a}{\sqrt{b^2 - a^2}} \right)^2}. \] Step 2: {Equating expressions}
\[ \frac{x}{\sqrt{1 - x^2}} = \sqrt{1 + \frac{a^2}{b^2 - a^2}}. \] Simplify: \[ \frac{x}{\sqrt{1 - x^2}} = \sqrt{\frac{b^2 - a^2 + a^2}{b^2 - a^2}}. \] \[ \frac{x}{\sqrt{1 - x^2}} = \frac{b}{\sqrt{b^2 - a^2}}. \] Squaring both sides: \[ x^2(2b^2 - a^2) = b^2. \] Solving for \( x \): \[ x = \frac{b}{\sqrt{2b^2 - a^2}}. \]