Question

If $\text{cosec x}-cotx=$ $\frac{1}{3}$ , where x $\ne$ 0, then the value of $co{s}^{2}x-si{n}^{2}x$ is

• A. $\frac{16}{25}$
• B. $\frac{9}{25}$
• C. $\frac{8}{25}$
• D. $\frac{7}{25}$

Answer 1

The correct option is D $\frac{7}{25}$

Given : cosec x - cot x $=\frac{1}{3}$ .......... (1)

We know, $cose{c}^{2}x-co{t}^{2}x=1$

$⟹$ (cosec x - cot x)(cosec x + cot x) $=1$

$⟹\frac{1}{3}$ (cosec x + cot x) $=1$

$⟹$ cosec x + cot x $=3$ ............ (2)

Solving (1) and (2) we get,

cosec x $=\frac{10}{6}=\frac{5}{3}$ and cot x $=\frac{4}{3}$

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

$\frac{cotx}{cosecx}=\frac{4}{5}$

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

$\therefore {\cos }^{2}x-{\sin }^{2}x=\left(\frac{16}{25}\right)-\left(\frac{9}{25}\right)=\frac{7}{25}$

Hence, option D is correct.