SSC CGL Tier 1 2026 · 01 Oct 2026 10:00 AM (Shift-1)
Q6901 Oct 2026 10:00 AM (Shift-1) · Q69Trigonometric Identitiesmoderate
If $\tan\theta = \sqrt{3}$, find the value of $\dfrac{\operatorname{cosec}^{2}\theta+\sec^{2}\theta}{\operatorname{cosec}^{2}\theta-\sec^{2}\theta}$.
Answer: D
From $\tan\theta=\sqrt{3}$: $\tan^{2}\theta=3$, $\cot^{2}\theta=\frac13$. So $\operatorname{cosec}^{2}\theta=1+\cot^{2}\theta=\frac43$ and $\sec^{2}\theta=1+\tan^{2}\theta=4$. Value = $\frac{\frac43+4}{\frac43-4}=\frac{16/3}{-8/3}=-2$. Answer D. Option D.
यदि $\tan\theta = \sqrt{3}$, तो $\dfrac{\operatorname{cosec}^{2}\theta+\sec^{2}\theta}{\operatorname{cosec}^{2}\theta-\sec^{2}\theta}$ का मान ज्ञात कीजिए।
उत्तर: D
$\tan\theta=\sqrt{3}$ से $\tan^{2}\theta=3$ और $\cot^{2}\theta=\frac13$ है। अतः $\operatorname{cosec}^{2}\theta=1+\cot^{2}\theta=\frac43$ और $\sec^{2}\theta=1+\tan^{2}\theta=4$। मान = $\frac{\frac43+4}{\frac43-4}=-2$। सही उत्तर D है। विकल्प D।