Q176 / 270 · NDA 2026 (Phase 1) · Mathematics

If $2\sec 4\beta=\tan 2\alpha+\cot 2\alpha$, then which one of the following is a possible value of $(\alpha+\beta)$ ?

  • A
    $\dfrac{\pi}{2}$
  • B
    $\dfrac{\pi}{4}$
  • C
    $\dfrac{\pi}{6}$
  • Correct answer
    D
    $\dfrac{\pi}{8}$

Solution

The given equation is $2\sec 4\beta = \tan 2\alpha + \cot 2\alpha$

We know that $\tan \theta + \cot \theta = \dfrac{\sin \theta}{\cos \theta} + \dfrac{\cos \theta}{\sin \theta} = \dfrac{1}{\sin \theta \cos \theta} = \dfrac{2}{\sin 2\theta}$

Substituting $\theta = 2\alpha$, we get $\tan 2\alpha + \cot 2\alpha = \dfrac{2}{\sin 4\alpha}$

The equation becomes $2\sec 4\beta = \dfrac{2}{\sin 4\alpha}$

$\Rightarrow \dfrac{1}{\cos 4\beta} = \dfrac{1}{\sin 4\alpha}$

$\Rightarrow \cos 4\beta = \sin 4\alpha$

$\Rightarrow \cos 4\beta = \cos\left(\dfrac{\pi}{2} - 4\alpha\right)$

The general solution is $4\beta = 2n\pi \pm \left(\dfrac{\pi}{2} - 4\alpha\right)$ where $n$ is an integer.

Taking the positive sign, we get:

$4\beta = 2n\pi + \dfrac{\pi}{2} - 4\alpha$

$\Rightarrow 4(\alpha + \beta) = 2n\pi + \dfrac{\pi}{2}$

$\Rightarrow \alpha + \beta = \dfrac{n\pi}{2} + \dfrac{\pi}{8}$

For $n = 0$, a possible value is $\alpha + \beta = \dfrac{\pi}{8}$

Answer: $\dfrac{\pi}{8}$

Hence, the correct answer is option: 'D'.

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