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