Question : The given expression is equal to: $1-\frac{\tan ^2 \phi}{\sec ^2 \phi}$
Option 1: $\operatorname{sin}^2 \phi\cos ^2 \phi$
Option 2: $\operatorname{sin}^2 \phi\cot ^2 \phi$
Option 3: $\cot^2 \phi \cos^2 \phi$
Option 4: $\tan^2 \phi\cos^2 \phi$
New: SSC CHSL tier 1 answer key 2024 out | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $\operatorname{sin}^2 \phi\cot ^2 \phi$
Solution : We know, $\sec^2 \phi-\tan^2 \phi$ = 1 and $\frac{\cos^2 \phi}{\sin^2 \phi}$ = $\cot^2 \phi$ So, $1-\frac{\tan^2 \phi}{\sec^2 \phi} $ = $\frac{\sec^2 \phi-\tan^2 \phi}{\sec^2 \phi}$ = $\frac{1}{\sec^2 \emptyset}$ = $\cos^2 \phi$ = $\frac{\sin^2 \phi\times{\cos^2 \phi}}{\sin^2 \phi}$ = $\sin^2 \phi\cot^2 \phi$ Hence, the correct answer is $\sin^2 \phi\cot^2 \phi$.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : Find the value of the following expression $\frac{(1+\sec\phi)}{\sec\phi}(1-\cos\phi)$.
Question : Simplify the given expression. $\sqrt{\frac{1+\cos P}{1-\cos P}}$
Question : The given expression is equal to: $(\cot B-\tan B)\sin B\cos B$:
Question : The given expression is equal to: $\frac{\left(1+\tan^2 A\right)}{\operatorname{cosec}^2 A \cdot \tan A}$
Question : The value of $\frac{\operatorname{sin} 58^{\circ}}{\cos 32^{\circ}}+\frac{\sin 55^{\circ} \sec 35^{\circ}}{\tan 5^{\circ} \tan 45^{\circ} \tan 85^{\circ}}$ is equal to:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile