Question

If 2tan⁻¹ (cos x) = tan⁻¹ (2 cosec x) then sin x + cos x =

• A. 2√2
• B. √2
• C. 1/√2
• D. 1/2

Answer 1

Answer: (B)

√2

Answer 2

The given equation is $2\tan^{-1} (\cos x) = \tan^{-1} (2 \operatorname{cosec} x)$.

Using the identity $2\tan^{-1} y = \tan^{-1}\left(\frac{2y}{1-y^2}\right)$, the left side becomes: $2\tan^{-1}(\cos x) = \tan^{-1}\left(\frac{2\cos x}{1-\cos^2 x}\right) = \tan^{-1}\left(\frac{2\cos x}{\sin^2 x}\right)$

So, the equation becomes: $\tan^{-1}\left(\frac{2\cos x}{\sin^2 x}\right) = \tan^{-1}(2\operatorname{cosec} x)$

For the principal values, if $\tan^{-1} A = \tan^{-1} B$, then $A=B$. $\frac{2\cos x}{\sin^2 x} = 2\operatorname{cosec} x$ $\frac{2\cos x}{\sin^2 x} = \frac{2}{\sin x}$

Assuming $\sin x \neq 0$ (which must be true for $\operatorname{cosec} x$ to be defined), we can multiply both sides by $\sin x$: $\frac{2\cos x}{\sin x} = 2$ $\cot x = 1$

The general solution is $x = n\pi + \frac{\pi}{4}$, where $n$ is an integer.

For the identity $2\tan^{-1} y = \tan^{-1}\left(\frac{2y}{1-y^2}\right)$ to be valid, we must have $y^2 < 1$, which means $\cos^2 x < 1$. This implies $\cos x \neq \pm 1$, so $x \neq n\pi$.

The solution $x = \frac{\pi}{4}$ satisfies this condition.

Now, we need to find $\sin x + \cos x$: For $x = \frac{\pi}{4}$: $\sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$