In $\triangle$ $OPQ$, right-angled at $P$, $OP =7\, cm$ and $OQ - PQ =1\, cm$ (see $Fig.$). Determine the values of $\sin Q$ and $\cos Q$.
In $\triangle$ OPQ, we have
$OQ ^{2}= OP ^{2}+ PQ ^{2}$
i.e., $\quad(1+ PQ )^{2}= OP ^{2}+ PQ ^{2}$
i.e., $\quad 1+ PQ ^{2}+2 PQ = OP ^{2}+ PQ ^{2}$
i.e., $\quad 1+2 PQ =7^{2}$
i.e., $\quad PQ =24\,cm$ and $OQ =1+ PQ =25 \,cm$
So, $\sin Q =\frac{7}{25}$ and $\cos Q =\frac{24}{25}$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}$
Evaluate:
$\frac{\sin 18^{\circ}}{\cos 72^{\circ}}$
Prove that $\sec A(1-\sin A)(\sec A+\tan A)=1$
Express $\sin 67^{\circ}+\cos 75^{\circ}$ in terms of trigonometric ratios of angles between $0^{\circ}$ and $45^{\circ}$
If $\tan 2 A=\cot \left(A-18^{\circ}\right),$ where $2 A$ is an acute angle, find the value of $A .$ (in $^{\circ}$)