Question

If \(2 \tan^-1 (\cos x) = \tan^-1 (2 \csc x)\) then the value of \(x\) is:

• A. \(\frac3\pi4\)
• B. \(\frac\pi3\)
• C. \(\frac\pi4\)
• D. \(\frac\pi2\)

Hint:

When solving inverse trigonometric equations with multiples of \(\tan^-1\), use the double-angle identity to simplify before equating expressions.

Answer 1

The Correct Option is C

We start from the given equation:
\[ 2 \tan^-1(\cos x) = \tan^-1(2 \csc x) \] First, recall the double-angle identity for inverse tangent:
\[ 2 \tan^-1 u = \tan^-1 \left( \frac2u1 - u^2 \right), \textprovided u \neq \pm 1 \] Here, \(u = \cos x\).
Thus, the LHS becomes:
\[ \tan^-1 \left( \frac2 \cos x1 - \cos^2 x \right) \] Since \(1 - \cos^2 x = \sin^2 x\), we have:
\[ \frac2 \cos x1 - \cos^2 x = \frac2 \cos x\sin^2 x \] Also, \(\csc x = \frac1\sin x\), so \(2 \csc x = \frac2\sin x\).
Therefore, the equation becomes:
\[ \tan^-1 \left( \frac2 \cos x\sin^2 x \right) = \tan^-1 \left( \frac2\sin x \right) \] Since \(\tan^-1 A = \tan^-1 B\) implies \(A = B\) (within principal branch considerations), we equate:
\[ \frac2 \cos x\sin^2 x = \frac2\sin x \] Cancel \(2\) from both sides:
\[ \frac\cos x\sin^2 x = \frac1\sin x \] Multiply both sides by \(\sin^2 x\):
\[ \cos x = \sin x \] This gives \(\tan x = 1 \Rightarrow x = \frac\pi4 + n\pi\), where \(n\) is an integer.
Considering principal values and the domain of the inverse tangent expressions, we choose \(x = \frac\pi4\).