Question

If $cosec\theta -cot\theta =\frac{1}{2},0<\theta <{90}^{\circ }$ then $cos\theta$ is equal to _______.

• A. $\frac{5}{3}$
• B. $\frac{3}{5}$
• C. $-\frac{3}{5}$
• D. $-\frac{5}{3}$

Answer 1

The correct option is B $\frac{3}{5}$

Given:
$cosec\theta -cot\theta =\frac{1}{2}$ --------- (i)
Using identity
$cose{c}^2\theta -cot{}^2\theta =1$
$(cosec\theta +\cot \theta )(cosec\theta -cot\theta )=1$ -------(ii)

Substituting, (i) in (ii) we get,
$(cosec\theta +\cot \theta )\left(\frac{1}{2}\right)=1$

$(cosec\theta +\cot \theta )=2$ ----------- (iii)

On adding (i) and (iii),
$2cosec\theta =\frac{1}{2}+2=\frac{5}{2}\Rightarrow cosec\theta =\frac{5}{4}$

Since, $cosec\theta =\frac{1}{sin\theta },$
$\Rightarrow sin\theta =\frac{4}{5}\Rightarrow si{n}^2\theta =\frac{4^2}{5^2}=\frac{16}{25}$

By the identity, $si{n}^2\theta =1-co{s}^2\theta ,$
$\Rightarrow 1-co{s}^2\theta =\frac{16}{25}\Rightarrow co{s}^2\theta =1-\frac{16}{25}=\frac{9}{25}$
$\Rightarrow cos\theta =\pm \frac{3}{5}$

$\because 0<\theta <{90}^{\circ },cos\theta =\frac{3}{5}$