Q51 / 150 · MHT CET 2026 (15 April Shift 2) · Mathematics
The value of $\tan^{-1}\left(\dfrac{\cos\left(\frac{19\pi}{4}\right) - 1}{\sin\left(\frac{\pi}{4}\right)}\right)$ is equal to
- A$-\dfrac{\pi}{4}$
- B$-\dfrac{5\pi}{12}$
- Correct answerC$-\dfrac{3\pi}{8}$
- D$-\dfrac{4\pi}{9}$
Solution
$\cos\left(\dfrac{19\pi}{4}\right) = \cos\left(4\pi + \dfrac{3\pi}{4}\right) = \cos\left(\dfrac{3\pi}{4}\right) = -\dfrac{1}{\sqrt{2}}$
$\sin\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$
Substituting these values into the given expression:
$\tan^{-1}\left(\dfrac{-\dfrac{1}{\sqrt{2}} - 1}{\dfrac{1}{\sqrt{2}}}\right) = \tan^{-1}\left(\dfrac{-1 - \sqrt{2}}{1}\right) = \tan^{-1}(-(\sqrt{2} + 1))$
Using the property $\tan^{-1}(-x) = -\tan^{-1}(x)$:
$-\tan^{-1}(\sqrt{2} + 1)$
Since $\tan\left(\dfrac{3\pi}{8}\right) = \sqrt{2} + 1$, we get:
$-\tan^{-1}\left(\tan\left(\dfrac{3\pi}{8}\right)\right) = -\dfrac{3\pi}{8}$
Answer: $-\dfrac{3\pi}{8}$
$\sin\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$
Substituting these values into the given expression:
$\tan^{-1}\left(\dfrac{-\dfrac{1}{\sqrt{2}} - 1}{\dfrac{1}{\sqrt{2}}}\right) = \tan^{-1}\left(\dfrac{-1 - \sqrt{2}}{1}\right) = \tan^{-1}(-(\sqrt{2} + 1))$
Using the property $\tan^{-1}(-x) = -\tan^{-1}(x)$:
$-\tan^{-1}(\sqrt{2} + 1)$
Since $\tan\left(\dfrac{3\pi}{8}\right) = \sqrt{2} + 1$, we get:
$-\tan^{-1}\left(\tan\left(\dfrac{3\pi}{8}\right)\right) = -\dfrac{3\pi}{8}$
Answer: $-\dfrac{3\pi}{8}$
Hence, the correct answer is option: 'C'.