Question : If $\sin(A+B)=\sin A\cos B+\cos A \sin B$, then the value of $\sin75°$ is:
Option 1: $\frac{\sqrt{3}+1}{\sqrt{2}}$
Option 2:
$\frac{\sqrt{2}+1}{2\sqrt{2}}$
Option 3:
$\frac{\sqrt{3}+1}{2\sqrt{2}}$
Option 4:
$\frac{\sqrt{3}+1}{2}$
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
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer:
Solution : Given: $\sin(A+B)=\sin A\cos B+\cos A\sin B$ So, $\sin75°=\sin(45°+30°)$ = $\sin 45°\cos30°+\cos45°\sin30°$ = $\frac{1}{\sqrt{2}}×\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}×\frac{1}{2}$ = $\frac{\sqrt{3}+1}{2\sqrt{2}}$ Hence, the correct answer is $\frac{\sqrt{3}+1}{2\sqrt{2}}$.
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 $(r\cos \theta -\sqrt{3})^{2}+(r\sin \theta -1)^{2}=0$, then the value of $\frac{r\tan \theta +\sec \theta}{r\sec \theta +\tan\theta}$ is equal to:
Question : If $\alpha$ and $\beta$ are positive acute angles, $\sin (4\alpha -\beta )=1$ and $\cos (2\alpha +\beta)=\frac{1}{2}$, then the value of $\sin (\alpha +2\beta)$ is:
Question : If $\frac{\cos\alpha}{\cos\beta}=a$, $\frac{\sin\alpha}{\sin\beta}=b$, then $\sin^{2}\beta$ is equal to:
Question : If $\sin A-\cos A=\frac{\sqrt{3}-1}{2}$, then the value of $\sin A\cdot \cos A$ is:
Question : If $\theta$ is an acute angle and $\sin \theta \cos \theta=2 \cos ^3 \theta-\frac{1}{4} \cos \theta$, then the value of $\sin \theta$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile