Q73 / 150 · MHT CET 2026 (15 April Shift 2) · Mathematics

If $\sec 4\theta - \sec 2\theta = 2$, then $\theta =$

  • A
    $n\pi + \dfrac{\pi}{8}, \dfrac{n\pi}{5} + \dfrac{\pi}{6}, n \in Z$
  • B
    $n\pi + \dfrac{\pi}{6}, \dfrac{n\pi}{5} + \dfrac{\pi}{8}, n \in Z$
  • Correct answer
    C
    $n\pi + \dfrac{\pi}{2}, \dfrac{n\pi}{5} + \dfrac{\pi}{10}, n \in Z$
  • D
    $n\pi + \dfrac{\pi}{3}, n\pi + \dfrac{\pi}{10}, n \in Z$

Solution

Given equation: $\sec 4\theta - \sec 2\theta = 2$

$\Rightarrow \dfrac{1}{\cos 4\theta} - \dfrac{1}{\cos 2\theta} = 2$

$\Rightarrow \cos 2\theta - \cos 4\theta = 2 \cos 4\theta \cos 2\theta$

Using the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:

$\Rightarrow \cos 2\theta - \cos 4\theta = \cos 6\theta + \cos 2\theta$

$\Rightarrow -\cos 4\theta = \cos 6\theta$

$\Rightarrow \cos 6\theta + \cos 4\theta = 0$

Using the formula $\cos C + \cos D = 2 \cos\left(\dfrac{C+D}{2}\right) \cos\left(\dfrac{C-D}{2}\right)$:

$\Rightarrow 2 \cos 5\theta \cos \theta = 0$

$\Rightarrow \cos 5\theta = 0$ or $\cos \theta = 0$

For $\cos \theta = 0$, we get $\theta = n\pi + \dfrac{\pi}{2}, n \in Z$

For $\cos 5\theta = 0$, we get $5\theta = n\pi + \dfrac{\pi}{2} \Rightarrow \theta = \dfrac{n\pi}{5} + \dfrac{\pi}{10}, n \in Z$

Both sets of solutions are valid as they do not make $\cos 2\theta = 0$ or $\cos 4\theta = 0$.

Answer: $n\pi + \dfrac{\pi}{2}, \dfrac{n\pi}{5} + \dfrac{\pi}{10}, n \in Z$

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

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