Question

If $\sin x=\frac{3}{5}$ and $x$ lies in second quadrant, then

• A. $\cos x=-\frac{4}{5}$
• B. $\tan x=\frac{4}{3}$ and $\cot x=\frac{3}{4}$
• C. $\sec x=\frac{5}{4}$ and $\mathrm{cosec}x=\frac{-5}{3}$
• D. All are correct

Answer 1

Answer: (A)

$\sin x=\frac{3}{5}$
Given that $x$ lies in second quadrant.
i.e., $\frac{\pi }{2}<x<\pi$
$\because {\sin }^{2}x+{\cos }^{2}x=1$
$\Rightarrow {\cos }^{2}x=1-{\sin }^{2}x=1-{\left(\frac{3}{5}\right)}^2$
$=1-\frac{9}{25}=\frac{25-9}{25}=\frac{16}{25}$
$\Rightarrow \cos x=\pm \frac{4}{5}$
$\because$ In second quadrant, $\cos x$ is negative, so we will leave its positive value.
i.e., $\cos x=-\frac{4}{5}$
$\Rightarrow \tan x=\frac{\sin x}{\cos x}$
$=\frac{\frac{3}{5}}{-\frac{4}{5}}=-\frac{3}{4}$
$\Rightarrow \cot x=\frac{1}{\tan x}=-\frac{4}{3}$
$\Rightarrow \cos x=\frac{1}{\cos x}=-\frac{5}{4}$
and $\mathrm{cosec}x=\frac{1}{\sin x}=\frac{5}{3}$