If $\tan A =\cot B ,$ prove that $A + B =90^{\circ}$
Given that,
$\tan A =\cot B$
$\tan A=\tan \left(90^{\circ}-B\right)$
$A=90^{\circ}-B$
$A+B=90^{\circ}$
Express the ratios $\cos A ,$ tan $A$ and $\sec A$ in terms of $\sin A .$
Evaluate:
$\cos 48^{\circ}-\sin 42^{\circ}$
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.
$\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A$
In triangle $ABC ,$ right -angled at $B ,$ if $\tan A =\frac{1}{\sqrt{3}},$ find the value of:
$(i)$ $\sin A \cos C+\cos A \sin C$
$(ii)$ $\cos A \cos C-\sin A \sin C$
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