Question : The value of $x$ in the expression $\tan^{2}\frac{\pi }{4}-\cos^{2}\frac{\pi }{3}=x\sin\frac{\pi }{4}\cos\frac{\pi }{4}\tan\frac{\pi }{3}$ is:
Option 1: $\frac{2}{\sqrt{3}}$
Option 2: $\frac{3\sqrt{3}}{4}$
Option 3: $\frac{1}{\sqrt{3}}$
Option 4: $\frac{\sqrt{3}}{2}$
New: SSC CHSL Tier 2 answer key released | 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
Correct Answer: $\frac{\sqrt{3}}{2}$
Solution :
$\tan^{2}\frac{\pi }{4}-\cos^{2}\frac{\pi }{3}=x\sin\frac{\pi }{4}\cos\frac{\pi }{4}\tan\frac{\pi }{3}$
Taking L.H.S.
$⇒\tan^{2}\frac{\pi}{4} - \cos^{2}\frac{\pi}{3} = 1 - \left(\frac{1}{2}\right)^{2} = 1 - \frac{1}{4} = \frac{3}{4}$
Taking R.H.S.
$⇒x\sin\frac{\pi}{4}\cos\frac{\pi}{4}\tan\frac{\pi}{3} = x \times \frac{1}{2} \times\sqrt{3} = \frac{x\sqrt{3}}{2}$
$⇒\frac{3}{4} = \frac{x\sqrt{3}}{2}$
$⇒x = \frac{3}{4} \times \frac{2}{\sqrt{3}} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2}$
Hence, the correct answer is $\frac{\sqrt{3}}{2}$.
Related Questions
Know More about
Staff Selection Commission Combined High ...
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Get Updates Brochure
Your Staff Selection Commission Combined Higher Secondary Level Exam brochure has been successfully mailed to your registered email id “”.
