Question

For \(x > 0\), if \(sin(cos^{-1}x+tan^{-1}x)-cos(sin^{-1}x+tan^{-1}x) = sin(cot^{-1}2)\) then \(x =\)

• A. \(\frac{1}{\sqrt{2}}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{\sqrt{3}}{2}\)
• D. \(1\)

Hint:

Put a = arcsin x and b = arctan x. The left side becomes cos(a-b) - cos(a+b) = 2 sin a sin b.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Rewrite \(\cos^{-1}x = \dfrac{\pi}{2} - \sin^{-1}x\) so that both terms use \(a = \sin^{-1}x\) and \(b = \tan^{-1}x\).

Step 2: Key Formula or Approach:
\(\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\), or directly \(\cos(a-b) - \cos(a+b) = 2\sin a\sin b\).

Step 3: Detailed Explanation:
\[ \sin\left(\cos^{-1}x + \tan^{-1}x\right) = \sin\left(\frac{\pi}{2} - a + b\right) = \cos(b - a) = \cos(a - b) \]
So the left side is
\[ \cos(a-b) - \cos(a+b) = 2\sin a\sin b \]
Now \(\sin a = x\), and \(\sin b = \sin(\tan^{-1}x) = \dfrac{x}{\sqrt{1+x^2}}\), so
\[ 2\sin a\sin b = \frac{2x^2}{\sqrt{1+x^2}} \]
The right side: \(\cot^{-1}2\) is an angle with \(\cot = 2\), so \(\sin = \dfrac{1}{\sqrt{5}}\).
Solve:
\[ \frac{2x^2}{\sqrt{1+x^2}} = \frac{1}{\sqrt5} \]
Test \(x = \tfrac12\): the left side is \(\dfrac{2\cdot\frac14}{\sqrt{5/4}} = \dfrac{1/2}{\sqrt5/2} = \dfrac{1}{\sqrt5}\), which matches. The left side is increasing for \(x > 0\), so this is the only positive root. Options (A), (C) and (D) give larger left-side values (for example \(x = 1\) gives \(2/\sqrt2 = \sqrt2\)).

Final Answer:
\(x = \dfrac12\), option (B). \[ \boxed{\frac{1}{2} \text{ (B)}} \]