tan ∝ = P tan β, sin ∝ = Q sin β, then cos² ∝  = ?

\(\frac{Q^2 - 1}{P^2 - 1}\)

tan ∝ = P tan β,

\(\frac{1}{cot ∝}\) = \(\frac{P}{cot β}\)

cot β = P cot ∝ = P \(\frac{cos ∝}{sin ∝}\)

&

sin ∝ = Q sin β

\(\frac{1}{cosec ∝}\) = \(\frac{Q}{cosec β}\)

cosec β = Q cosec ∝ = Q \(\frac{1}{sin ∝}\)

Now,

{cosec² θ - cot² θ = 1}

cosec² β - cot² β = 1

\(\frac{Q^2}{sin^2∝}\) - \(\frac{P^2 cos^2 ∝}{sin^2∝}\) = 1

Q² - P² cos² ∝ = sin² ∝ = (1 - cos² ∝)

Q² - 1 = (P² -1) cos² ∝

cos² ∝ = \(\frac{Q^2- 1}{P^2- 1}\)