8. Introduction to Trigonometry
medium

If $\tan ( A + B )=\sqrt{3}$ and $\tan ( A - B )=\frac{1}{\sqrt{3}} ; 0^{\circ}< A + B \leq 90^{\circ} ; A > B ,$ find $A$ and $B$

Option A
Option B
Option C
Option D

Solution

$\tan (A+B)=\sqrt{3}$

$\Rightarrow \tan (A+B)=\tan 60$

$\Rightarrow A+B=60 \ldots(1)$

$\tan ( A – B )=\frac{1}{\sqrt{3}}$

$\Rightarrow \tan (A-B)=\tan 30$

$\Rightarrow A-B=30 \ldots(2)$

On adding both equations, we obtain

$2 A =90$

$\Rightarrow A=45$

From equation $(1),$ we obtain

$45+B=60$

$B=15$

Therefore, $\angle A =45^{\circ}$ and $\angle B =15^{\circ}$

Standard 10
Mathematics

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