Question

$\frac{cosec\theta }{(cosec\theta -1)}+\frac{cosec\theta }{(cosec\theta +1)}=2se{c}^2\theta$

• A. True
• B. False

Answer 1

$=\frac{cosec\theta }{(cosec\theta -1)}+\frac{cosec\theta }{(cosec\theta +1)}$

$=\frac{cosec\theta (cosec\theta +1)+cosec\theta (cosec\theta -1)}{(cose{c}^2\theta -1)}$

$=\frac{2cose{c}^2\theta }{(1+co{t}^2\theta -1)}$ $[\because cose{c}^2\theta =1+co{t}^2\theta ]$

$=\frac{2cose{c}^2\theta }{co{t}^2\theta }$

$=2cose{c}^2\theta ta{n}^2\theta$

$=2\times \frac{1}{si{n}^2\theta }\times \frac{si{n}^2\theta }{co{s}^2\theta }$

$=\frac{2}{co{s}^2\theta }$

$=2se{c}^2\theta$