Question : If $\sec x+\cos x=\frac{5}{2}$, where $x$ lies between $0^{\circ}$ and $90^{\circ}$, then what is the value of $\sin ^2 x$ ?
Option 1: $\frac{3}{4}$
Option 2: $\frac{1}{2}$
Option 3: $1$
Option 4: $\frac{1}{4}$
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: $\frac{3}{4}$
Solution : Given: $\sec x + \cos x = \frac{5}{2}$ Now, $\cos\theta = \frac{1}{\sec\theta}$ ⇒ $\frac{1}{\cos x} + \cos x = \frac{5}{2}$ The above equation satisfies for $x = 60^\circ$ Thus, $\sin^2x = \sin^2 60^\circ= (\frac{\sqrt3}{2})^2$ $\therefore \sin^2x = \frac{3}{4}$ Hence, the correct answer is $\frac{3}{4}$.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : If $\sin (\theta +18^{\circ})=\cos 60^{\circ}(0< \theta < 90^{\circ})$, then the value of $\cos 5\theta$ is:
Question : If $\sin (x - y) = \frac{1}2$ and $\cos (x + y) = \frac{1}2$, then what is the value of $\sin x \cos x + 2\sin^2x + cos^3x \sec x$?
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:
Question : If $\frac{x-x\tan^{2}30^{\circ}}{1+\tan^{2}30^{\circ}}=\sin^{2}30^{\circ}+4\cot^{2}45^{\circ}-\sec^{2}60^{\circ}$, then value of $x$ is:
Question : If $\sin(60^{\circ}-x)=\cos(y+60^{\circ})$, then the value of $\sin(x-y)$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile