Find the values of other five trigonometric functions if $\sin x=\frac{3}{5}, x$ lies in second quadrant.
$\sin x=\frac{3}{5}$
$\csc x=\frac{1}{\sin x}=\frac{1}{\left(\frac{3}{5}\right)}=\frac{5}{3}$
$\sin ^{2} x+\cos ^{2} x=1$
$\Rightarrow \cos ^{2} x=1-\sin ^{2} x$
$\Rightarrow \cos ^{2} x=1-\left(\frac{3}{5}\right)^{2}$
$\Rightarrow \cos ^{2} x=1-\frac{9}{25}$
$\Rightarrow \cos ^{2} x=\frac{16}{25}$
$\Rightarrow \cos x=\pm \frac{4}{5}$
since $x$ lies in the $2^{\text {nd }}$ quadrant, the value of $\cos x$ will be negative
$\therefore \cos x=-\frac{4}{5}$
$\sec x=\frac{1}{\cos x}=\frac{1}{\left(-\frac{4}{5}\right)}=-\frac{5}{4}$
$\tan x=\frac{\sin x}{\cos x}=\frac{\left(\frac{3}{5}\right)}{\left(-\frac{4}{5}\right)}=-\frac{3}{4}$
$\cot x=\frac{1}{\tan x}=-\frac{4}{3}$
Find the value of:
$\tan 15^{\circ}$
Which of the following relations is correct
If $\sin \theta + \cos \theta = 1$, then $\sin \theta \cos \theta = $
Find the values of other five trigonometric functions if $\cos x=-\frac{1}{2}, x$ lies in third quadrant.
Prove that :
$\cot ^{2} \frac{\pi}{6}+\cos ec \,\frac{5 \pi}{6}+3 \tan ^{2}\, \frac{\pi}{6}=6$