Given $\tan A=\frac{4}{3},$ find the other trigonometric ratios of the $\angle A$
Let us first draw a right $\Delta ABC$ (see $Fig.$).
Now, we know that $\tan A =\frac{ BC }{ AB }=\frac{4}{3}$
Therefore, if $BC =4 k,$ then $AB =3 k,$ where $k$ is a positive number.
Now, by using the Pythagoras Theorem, we have
$AC ^{2}= AB ^{2}+ BC ^{2}=(4 k)^{2}+(3 k)^{2}=25 k ^{2}$
$AC =5 k$
Now, we can write all the trigonometric ratios using their definitions.
$\sin A=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$
$\cos A=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$
Therefore, $\cot A=\frac{1}{\tan A}=\frac{3}{4}, \operatorname{cosec} A=\frac{1}{\sin A}=\frac{5}{4}$ and $\sec A=\frac{1}{\cos A}=\frac{5}{3}$
If $\tan 2 A=\cot \left(A-18^{\circ}\right),$ where $2 A$ is an acute angle, find the value of $A .$ (in $^{\circ}$)
$(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)=..........$
$\sin 2 A=2 \sin A$ is true when $A=$
$9 \sec ^{2} A-9 \tan ^{2} A=..........$
If $A , B$ and $C$ are interior angles of a triangle $ABC ,$ then show that
$\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$