Question

Find the values of other five trigonometric functions if $\sin x=\frac{3}{5},x$ lies in second quadrant.

Answer 1

Step 1: Solve for value of $\cos x$
$\sin x=\frac{3}{5}$ and $x$ lies in second quadrant
We know that
${\sin }^{2}x+{\cos }^{2}x=1\Rightarrow {\left(\frac{3}{5}\right)}^2+{\cos }^{2}x=1\Rightarrow \frac{9}{25}+{\cos }^{2}x=1\Rightarrow {\cos }^{2}x=1-\frac{9}{25}=\frac{16}{25}\Rightarrow \cos x=\pm \frac{4}{5}$
As $x$ is in second quadrant, $\cos x$ is negative.
$\therefore \cos x=-\frac{4}{5}$

Step 2: Solve for value of $\sec x$
We know that $\sec x=\frac{1}{\cos x}$
$\therefore \sec x=\frac{1}{\frac{-4}{5}}=-\frac{5}{4}$

Step 3: Solve for value of $\tan x$
We know that $\tan x=\frac{\sin x}{\cos x}$
$\therefore \tan x=\frac{\frac{3}{5}}{-\frac{4}{5}}=-\frac{3}{4}$

Step 4: Solve for value of $\cot x$
We know that $\cot x=\frac{1}{\tan x}$
$\therefore \cot x=\frac{1}{-\frac{3}{4}}$ =-\dfrac{4}{3}\)

Step 5: Solve for value of $\text{cosec}x$
We know that $\text{cosec}x=\frac{1}{\sin x}$
$\therefore \text{cosec}x=\frac{1}{\frac{3}{5}}=\frac{5}{3}$