If sec thita- tan thita=p. Obtain the values of tan thita ,sec thita and sin thita in term of p

Question

I want solution stepwise like first to find value of sec thita and then tan thita and then cos thita and then sin thita

Answer 1

Let t=tanθ, then sec²θ=1+t² and:

√(1+t²)-t=p, √(1+t²)=p+t. Squaring:

1+t²=p²+2pt+t², p²+2pt=1, t=tanθ=(1-p²)/2p.

In a right triangle tanθ=opposite/adjacent=(1-p²)/2p.

Therefore the hypotenuse²=(1-p²)²+4p²=(1+p²)².

So the hypotenuse = 1+p². sinθ = opposite/hypotenuse = (1-p²)/(1+p²);

cosθ = adjacent/hypotenuse=2p/(1+p²); secθ=(1+p²)/2p.

CHECK

secθ-tanθ=(1+p²)/2p-(1-p²)/2p=(1+p²-(1-p²))/2p=2p²/2p=p OK.