In a right triangle $A B C$, right-angled at $B$. if $\tan A =1,$ then verify that $2 \sin A \cos A=1$
In $\triangle ABC , \tan A =\frac{ BC }{ AB }=1$ (see $Fig.$)
i.e. $BC = AB$
Let $AB = BC =k,$ where $k$ is a positive number.
Now,$AC=\sqrt{ AB ^{2}+ BC ^{2}}$
$=\sqrt{(k)^{2}+(k)^{2}}=k \sqrt{2}$
Therfore, $\sin A=\frac{ BC }{ AC }=\frac{1}{\sqrt{2}} \quad$ and $\cos A =\frac{ AB }{ AC }=\frac{1}{\sqrt{2}}$
So, $\quad 2 \sin A \cos A =2\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)=1,$ which is the required value.
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 $B$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A$
If $\tan A =\cot B ,$ prove that $A + B =90^{\circ}$
$\sin 2 A=2 \sin A$ is true when $A=$
Evaluate the following:
$\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$