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. ∴ (m2−1)=(cosec θ+cot θ)2−1=cosec2 θ+cot2 θ+2cosec θ cot θ−1=(cosec2 θ−1)+cot2 θ+2cosec θ cot θ=2cot2 θ+2cosec θ cot θ[∵cosec2 θ−1=cot2 θ]=2cot θ (cot θ+cosec θ)
- Similarly, (m2+1)=(cosec θ+cot θ)2+1=cosec2 θ+cot2 θ+2cosec θ cot θ+1=(cot2 θ+1)+cosec2 θ+2cosec θ cot θ=2cosec2 θ+2cosec θ cot θ[∵1+cot2 θ=cosec2 θ]=2cosec θ (cosec θ+cot θ)
- From step 1 and step 2, we get m2−1m2+1=cot θcosec θ=cos θsin θ1sin θ=cos θ
- Thus, the value of m2−1m2+1 is cos θ.