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 $\cos Q =\frac{24}{25}$
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 $B$
$(\sec A+\tan A)(1-\sin A)=..........$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec} A+\cot A,$ using the identity $\operatorname{cosec}^{2} A=1+\cot ^{2} A$
Prove that $\sec A(1-\sin A)(\sec A+\tan A)=1$
In $Fig.$ find $\tan P-\cot R .$