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 \( x \): \[ x = \frac{b}{\sqrt{2b^2 - a^2}}. \]

Answer 2

Step 1: Simplify \( \cot(\cos^{-1} x) \)
Let \( \theta = \cos^{-1} x \), so that \( \cos \theta = x \). Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can find \( \sin \theta \): \[ \sin^2 \theta = 1 - x^2 \quad \Rightarrow \quad \sin \theta = \sqrt{1 - x^2}. \] Now, we can compute \( \cot(\cos^{-1} x) \) as follows: \[ \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x}{\sqrt{1 - x^2}}. \] Thus: \[ \cot(\cos^{-1} x) = \frac{x}{\sqrt{1 - x^2}}. \] Step 2: Simplify \( \sec \left( \tan^{-1} \left( \frac{a}{\sqrt{b^2 - a^2}} \right) \right) \)
Let \( \phi = \tan^{-1} \left( \frac{a}{\sqrt{b^2 - a^2}} \right) \). Then \( \tan \phi = \frac{a}{\sqrt{b^2 - a^2}} \). Using the identity \( \sec^2 \phi = 1 + \tan^2 \phi \), we can find \( \sec \phi \): \[ \sec^2 \phi = 1 + \left( \frac{a}{\sqrt{b^2 - a^2}} \right)^2 = 1 + \frac{a^2}{b^2 - a^2}. \] Simplifying: \[ \sec^2 \phi = \frac{b^2 - a^2 + a^2}{b^2 - a^2} = \frac{b^2}{b^2 - a^2}. \] Thus: \[ \sec \phi = \frac{b}{\sqrt{b^2 - a^2}}. \] Step 3: Equate the two sides
We are given the equation: \[ \cot(\cos^{-1} x) = \sec \left( \tan^{-1} \left( \frac{a}{\sqrt{b^2 - a^2}} \right) \right). \] From Step 1, we have: \[ \frac{x}{\sqrt{1 - x^2}} = \frac{b}{\sqrt{b^2 - a^2}}. \] Now, solving for \( x \), we cross-multiply: \[ x \sqrt{b^2 - a^2} = b \sqrt{1 - x^2}. \] Squaring both sides: \[ x^2 (b^2 - a^2) = b^2 (1 - x^2). \] Expanding both sides: \[ x^2 b^2 - x^2 a^2 = b^2 - b^2 x^2. \] Rearranging terms: \[ x^2 (b^2 - a^2 + b^2) = b^2. \] Thus: \[ x^2 (2b^2 - a^2) = b^2. \] Solving for \( x^2 \): \[ x^2 = \frac{b^2}{2b^2 - a^2}. \] Finally, taking the square root: \[ x = \frac{b}{\sqrt{2b^2 - a^2}}. \] Final Answer:
The value of \( x \) is:

\( \frac{b}{\sqrt{2b^2 - a^2}} \)