Question

if cosec theta + cot theta = p , prove that (p2+1)) cos theta = p2-1

Answer 1

Step-by-step explanation:

given, cosecθ+cotθ = p

To prove that (p²+1)cosθ = p²-1

take, cosecθ+cotθ = p

=> 1/p = $\frac{1}{cosec\theta+cot\theta}$ * $\frac{cosec\theta-cot\theta}{cosec\theta-cot\theta}$  

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

          = $\frac{cosec\theta-cot\theta}{cosec^2\theta-cot^2\theta}$

          = cosecθ-cotθ

consider, p+$\frac{1}{p}$

=> p+$\frac{1}{p}$  = cosecθ+cotθ+cosecθ-cotθ = 2cosecθ

=> $\frac{p^2+1}{p}$ = 2cosecθ

=> sinθ = $\frac{2p}{p^2+1}$

=> sin²θ = $(\frac{2p}{p^2+1})^{2}$

=> 1-sin²θ = 1-$\frac{(2p)^2}{(p^2+1)^2}$

=> cos²θ = $\frac{(p^2+1)^2 - (2p)^2}{(p^2+1)^2}$

=> cos²θ = $\frac{(p^2+1+2p)(p^2+1-2p)}{(p^2+1)^2}$

=> cos²θ = $\frac{(p+1)^2(p-1)^2}{(p^2+1)^2}$

=> cosθ = $\frac{(p+1)(p-1)}{(p^2+1)}$

=> cosθ = $\frac{(p^2-1)}{(p^2+1)}$

=> (p²+1)cosθ = p²-1    (required equation)

Hence proved