Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
Let $\sin (\mathrm{A}+\mathrm{B})=1$ and $\sin (\mathrm{A}-\mathrm{B})=\frac{1}{2}$ where $\mathrm{A}, \mathrm{B} \in\left[0, \frac{\pi}{2}\right]$.
What is the value of A?
MathematicsTrigonometric Ratios & IdentitiesNDANDA 2012 (Phase 1)
Options:
  • A $\frac{\pi}{6}$
  • B $\frac{\pi}{3}$
  • C $\frac{\pi}{4}$
  • D $\frac{\pi}{8}$
Solution:
1234 Upvotes Verified Answer
The correct answer is: $\frac{\pi}{3}$
$\begin{array}{l}\sin (\mathrm{A}+\mathrm{B})=1 \\ \Rightarrow \sin (\mathrm{A}+\mathrm{B})=\sin 90^{\circ} \\ \Rightarrow \mathrm{A}+\mathrm{B}=90^{\circ} \\ & \text { Given } \sin (\mathrm{A}-\mathrm{B})=\frac{1}{2}=\sin 30^{\circ} \\ \Rightarrow \mathrm{A}-\mathrm{B}=30^{\circ}\end{array}$
On solving $(1)$ and $(2)$, we get $\mathrm{A}=60$
$\mathrm{B}=30$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.