Evaluate the following:
$\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$
$\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$
$=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{3}{2}+\frac{2}{\sqrt{3}}}$
$=\frac{\frac{3 \sqrt{3}-4}{2 \sqrt{3}}}{\frac{3 \sqrt{3}+4}{2 \sqrt{3}}}=\frac{(3 \sqrt{3}-4)}{(3 \sqrt{3}+4)}$
$=\frac{(3 \sqrt{3}-4)(3 \sqrt{3}-4)}{(3 \sqrt{3}+4)(3 \sqrt{3}-4)}=\frac{(3 \sqrt{3}-4)^{2}}{(3 \sqrt{3})^{2}-(4)^{2}}$
$=\frac{27+16-24 \sqrt{3}}{27-16}=\frac{43-24 \sqrt{3}}{11}$
Prove that
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta},$ using the identity
$\sec ^{2} \theta=1+\tan ^{2} \theta$
Given $\sec \theta=\frac{13}{12},$ calculate all other trigonometric ratios.
If $\angle B$ and $\angle Q$ are acute angles such that $\sin B =\sin Q$, then prove that $\angle B =\angle Q$.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}=\tan \theta$
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 $B$