If cosec θ+cot θ=m, show that m2−1m2+1=cos θ.
Answer:
cos θ
- It is given that cosec θ+cot θ=m. ∴
- 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}
- 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
- Thus, the value of \dfrac { m^2 - 1 } { m^2 + 1 } is cos \bf {\space \theta}.