Question

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

Answer 1

Here sin x = $\frac{3}{5}$

Now ${\text{sin}}^{2}x+{\text{cos}}^2$ x=1

$\Rightarrow {\left(\frac{3}{5}\right)}^2+co{s}^{2}x=1$

$\Rightarrow {\text{cos}}^{2}x=1-\frac{9}{25}$

$\Rightarrow {\text{cos}}^{2}x=\frac{16}{25}$

$\Rightarrow \text{cos}x=\pm \frac{4}{5}$

But x lies in second quadrant.

$\therefore$ cos x =- $\frac{4}{5}$

Now cosec x = $\frac{1}{\text{sin x}}=\frac{5}{3}$

and sec x = $\frac{1}{\text{cos x}}=-\frac{5}{4}$

tan x = $\frac{\text{sin x}}{\text{cos x}}=\frac{\frac{3}{5}}{\frac{-4}{5}}=\frac{-3}{4}$

and cot x = $\frac{\text{cos x}}{\text{sin x}}=\frac{\frac{-4}{5}}{\frac{3}{5}}=\frac{-4}{3}$