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 $\angle B =15^{\circ}$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}$
Consider $\triangle ACB$, right-angled at $C$, in which $AB =29$ units, $BC =21$ units and $\angle ABC =\theta$ (see $Fig.$). Determine the values of
$(i)$ $\cos ^{2} \theta+\sin ^{2} \theta$
$(ii)$ $\cos ^{2} \theta-\sin ^{2} \theta$
If $\sin 3 A =\cos \left( A -26^{\circ}\right),$ where $3 A$ is an acute angle, find the value of $A= . . . . ^{\circ}$.
Given $15 \cot A =8,$ find $\sin A$ and $\sec A .$
State whether the following are true or false. Justify your answer.
$\sin (A+B)=\sin A+\sin B$