Question

If $2 \tan^{-1}(\cos x) = \tan^{-1}(2 \csc x)$, then the value of $x$ is

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{12}$

Hint:

When answering multiple-choice questions, you can substitute the options directly back into the original equation. Plugging $x = \frac{\pi}{4}$ gives $2\tan^{-1}\left(\frac{1}{\sqrt{2}}\right)$ on the left and $\tan^{-1}(2\sqrt{2})$ on the right. Using the formula $\frac{2(1/\sqrt{2})}{1 - 1/2} = \frac{\sqrt{2}}{1/2} = 2\sqrt{2}$ confirms the match instantly!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given an equation containing inverse trigonometric functions. We need to solve for $x$ within the standard principal value branch.

Step 2: Key Formula or Approach:
We use the double-angle formula for the inverse tangent function: $$2 \tan^{-1}(\theta) = \tan^{-1}\left(\frac{2\theta}{1 - \theta^2}\right)$$ By substituting $\theta = \cos x$, we can drop the inverse tangent function from both sides of the equation and convert it into a standard trigonometric equation.

Step 3: Detailed Explanation:
Apply the double-angle identity to the left side of our equation: $$\tan^{-1}\left(\frac{2\cos x}{1 - \cos^2 x}\right) = \tan^{-1}(2\csc x)$$ Now apply the tangent function to both sides to remove the inverse tangent blocks: $$\frac{2\cos x}{1 - \cos^2 x} = 2\csc x$$ Using the fundamental identity $1 - \cos^2 x = \sin^2 x$ and rewriting $\csc x = \frac{1}{\sin x}$: $$\frac{2\cos x}{\sin^2 x} = \frac{2}{\sin x}$$ Divide both sides by 2 and multiply by $\sin^2 x$ (assuming $\sin x \neq 0$): $$\cos x = \frac{\sin^2 x}{\sin x}$$ $$\cos x = \sin x$$ Divide by $\cos x$ on both sides to convert the expression into a tangent form: $$\tan x = 1$$ The principal value satisfying this equation within the standard domain is: $$x = \frac{\pi}{4}$$

Step 4: Final Answer:
The value of $x$ is $\frac{\pi}{4}$, which corresponds to option (B).