### If $cosec \space \theta + cot \space \theta = m,$ show that $\dfrac { m^2 - 1 } { m^2 + 1 } = cos \space \theta.$

$cos \space \theta$
1. It is given that $cosec \space \theta + cot \space \theta = m.$ \begin{aligned} \therefore \space (m^2 - 1) &= (cosec \space \theta + cot \space \theta )^2 - 1 \\ &= cosec^2 \space \theta + cot^2 \space \theta + 2cosec \space \theta \space cot \space \theta - 1 \\ &= (cosec^2 \space \theta -1) + cot^2 \space \theta + 2cosec \space \theta \space cot \space \theta \\ &= 2cot^2 \space \theta + 2cosec \space \theta \space cot \space \theta && [\because cosec^2 \space \theta - 1 = cot^2 \space \theta ] \\ &= 2cot \space \theta \space (cot \space \theta + cosec \space \theta) \end{aligned}
2. Similarly, \begin{aligned}(m^2 + 1) &= (cosec \space \theta + cot \space \theta )^2 + 1 \\ &= cosec^2 \space \theta + cot^2 \space \theta + 2cosec \space \theta \space cot \space \theta + 1 \\ &= (cot^2 \space \theta +1) + cosec^2 \space \theta + 2cosec \space \theta \space cot \space \theta \\ &= 2cosec^2 \space \theta + 2cosec \space \theta \space cot \space \theta && [\because 1 + cot^2 \space \theta = cosec^2 \space \theta ] \\ &= 2cosec \space \theta \space (cosec \space \theta + cot \space \theta) \end{aligned}
3. From step 1 and step 2, we get $$\dfrac { m^2 - 1 } { m^2 + 1 } = \dfrac { cot \space \theta } { cosec \space \theta } = \dfrac { \dfrac { cos \space \theta } { sin \space \theta } } { \dfrac { 1 } { sin \space \theta } } = cos \space \theta$$
4. Thus, the value of $\dfrac { m^2 - 1 } { m^2 + 1 }$ is $cos \bf {\space \theta}.$