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Trigonometric Ratios of Some Particular Angles — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTrigonometric Ratios of Some Particular Angles4 Marks
Question
Evaluate the following:
If $\sin(\text{A}+\text{B})=1$ and $\cos(\text{A}-\text{B})=1,0^\circ\leq(\text{A}+\text{B})\leq90^\circ$ and $\text{A}>\text{B}$ then find A and B.
✓

Answer

Here, $\sin(\text{A}+\text{B})=1$
$\Rightarrow\sin(\text{A}+\text{B})=\sin90^\circ$ $[\because\ \sin90^\circ=1]$
$\Rightarrow\text{A}+\text{B}=90^\circ\dots(\text{i})$
Also, $\cos(\text{A}-\text{B})=1$
$\Rightarrow\cos(\text{A}-\text{B})=\cos0^\circ$ $[\because\ \cos0^\circ=1]$
$\Rightarrow\text{A}-\text{B}=0^\circ\dots(\text{ii})$
Solving (i) and (ii), we get:
A = 45° and B = 45°.

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