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 $B$
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 $B$
$\tan (A+B)=\sqrt{3}$
$\Rightarrow \tan (A+B)=\tan 60$
$\Rightarrow A+B=60 \ldots(1)$
$\tan ( A - B )=\frac{1}{\sqrt{3}}$
$\Rightarrow \tan (A-B)=\tan 30$
$\Rightarrow A-B=30 \ldots(2)$
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 $\angle B =15^{\circ}$
Similar Questions
If $\cot \theta=\frac{7}{8},$ evaluate:
$(i)$ $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$(ii)$ $\cot ^{2} \theta$
If $\cot \theta=\frac{7}{8},$ evaluate:
$(i)$ $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$(ii)$ $\cot ^{2} \theta$
If $\sin ( A - B )=\frac{1}{2}, \cos ( A + B )=\frac{1}{2}, 0^{\circ} < A + B \leq 90^{\circ}, A > B ,$ find $A$ and $B$
If $\sin ( A - B )=\frac{1}{2}, \cos ( A + B )=\frac{1}{2}, 0^{\circ} < A + B \leq 90^{\circ}, A > B ,$ find $A$ and $B$
$\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=$
$\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=$
If $\sin 3 A =\cos \left( A -26^{\circ}\right),$ where $3 A$ is an acute angle, find the value of $A= . . . . ^{\circ}$.
If $\sin 3 A =\cos \left( A -26^{\circ}\right),$ where $3 A$ is an acute angle, find the value of $A= . . . . ^{\circ}$.
If $3 \cot A=4,$ check whether $\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A$ or not.
If $3 \cot A=4,$ check whether $\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A$ or not.