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 $\sec A=\frac{1}{\cos A}=\frac{5}{3}$
State whether the following are true or false. Justify your answer.
The value of $\cos \theta$ increases as $\theta$ increases
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 $\operatorname{cosec}^{2} A=1+\cot ^{2} A$
Show that:
$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=1$
$(ii)$ $\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}=0$
Prove that $\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}$
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 $B$