Question

If \(cot(cos^{-1}x) = sec(tan^{-1}\frac{a}{\sqrt{b^2-a^2}})\), then the value of \(x\) is

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

Hint:

Express both sides algebraically using right triangles, then square and solve for x.

Answer 1

The Correct Option is D

Step 1: Use right triangles:
For $\phi$: opposite $a$, adjacent $\sqrt{b^2-a^2}$, hypotenuse $b$. So $\sec\phi=\dfrac{b}{\sqrt{b^2-a^2}}$.

Step 2: Left side from triangle:
For $\cos^{-1}x$: adjacent $x$, hypotenuse 1, opposite $\sqrt{1-x^2}$. So $\cot=\dfrac{x}{\sqrt{1-x^2}}$.

Step 3: Solve by taking reciprocals:
$\dfrac{1-x^2}{x^2}=\dfrac{b^2-a^2}{b^2}$ gives $\dfrac{1}{x^2}=1+\dfrac{b^2-a^2}{b^2}=\dfrac{2b^2-a^2}{b^2}$, so $x=\dfrac{b}{\sqrt{2b^2-a^2}}$. Option D.

Final Answer:
Equating the right-triangle ratios gives x = b / sqrt(2b^2 - a^2). \[ \boxed{\text{(D) }\dfrac{b}{\sqrt{2b^2-a^2}}} \]