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: Simplify the left side:
Let \(\theta=\cos^{-1}x\), so \(\cos\theta=x\), \(\sin\theta=\sqrt{1-x^2}\). Then \(\cot\theta=\dfrac{x}{\sqrt{1-x^2}}\).

Step 2: Simplify the right side:
Let \(\phi=\tan^{-1}\dfrac{a}{\sqrt{b^2-a^2}}\). Then \(\sec\phi=\sqrt{1+\dfrac{a^2}{b^2-a^2}}=\dfrac{b}{\sqrt{b^2-a^2}}\).

Step 3: Equate and solve:
\[ \dfrac{x}{\sqrt{1-x^2}}=\dfrac{b}{\sqrt{b^2-a^2}} \]
Square both sides: \(x^2(b^2-a^2)=b^2(1-x^2)\), so \(x^2(2b^2-a^2)=b^2\).
\[ x=\dfrac{b}{\sqrt{2b^2-a^2}} \]

Step 4: Why the other options are wrong.
Option B and C give values with \(\sqrt{2b^2-a^2}\) in the numerator, which makes \(x>1\) for typical values. Option A uses a plus sign inside the root, which comes from a sign error when squaring.

Final Answer:
x equals b / sqrt(2b^2 - a^2). \[ \boxed{\text{(D) }\dfrac{b}{\sqrt{2b^2-a^2}}} \]