Question : If $\sin (x - y) = \frac{1}2$ and $\cos (x + y) = \frac{1}2$, then what is the value of $\sin x \cos x + 2\sin^2x + cos^3x \sec x$?
Option 1: $2$
Option 2: $\sqrt{2}+1$
Option 3: $1$
Option 4: $\frac{3}{4}$
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: $2$
Solution : $\sin (x - y) = \frac{1}2=\sin30^\circ$ $⇒(x-y)=30^\circ$---(1) $\cos (x + y) = \frac{1}2=\cos60^\circ$ $⇒(x+y)=60^\circ$---(2) Solving equation 1 and 2, we get, ⇒ $x=45^\circ$ Now, Putting the value of $x$, we get: $\sin x\cos x + 2\sin^2x + \cos^3 x \sec x$ $=\sin 45^\circ\cos45^\circ + 2\sin^2 45^\circ + \cos^3 45^\circ \sec 45^\circ$ $=\frac{1}{\sqrt2}\times \frac{1}{\sqrt2}+2\times\frac{1}{2}+\frac{1}{2\sqrt2}\times\sqrt2$ $=\frac{1}{2}+1+\frac{1}{2}$ $= 1 + 1$ $= 2$ Hence, the correct answer is $2$.
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 $x=\sqrt{3}-\frac{1}{\sqrt{3}}, y=\sqrt{3}+\frac{1}{\sqrt{3}}$, then the value of $\frac{x^2}{y}+\frac{y^2}{x}$ is:
Question : If $\sin A-\cos A=\frac{\sqrt{3}-1}{2}$, then the value of $\sin A\cdot \cos A$ is:
Question : If $\sin(60^{\circ}-x)=\cos(y+60^{\circ})$, then the value of $\sin(x-y)$ is:
Question : If $x\cos \theta -y\sin \theta =\sqrt{x^{2}+y^{2}}$ and $\frac{\cos ^2{\theta }}{a^{2}}+\frac{\sin ^{2}\theta}{b^{2}}=\frac{1}{x^{2}+y^{2}},$ then the correct relation is:
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:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile