Question

If cosec 0 + cot 0 = p, then prove that cos 0 = P2-1/
p²+1​

Answer 1

Given-

cosec Ф + cot Ф = p

To prove-

cos Ф = p²-1/p²+1

Proof-

cosecФ + cotФ = p

⇒1/sinФ + cos Ф/sinФ = p

⇒(1+cosФ)/sinФ = p

⇒1+cosФ = psinФ

Squaring both sides we get,

⇒(1+cosФ)² = p²sin²Ф

⇒(1+cosФ)² = p²(1-cos²Ф)                             ∴(sin²Ф+cos²Ф=1)

⇒(1+cosФ)² = p²(1+cosФ)(1-cosФ)                ∴[a²-b²=(a+b)(a-b)]

⇒1+cosФ = p²(1-cosФ)

Last step⇒ cosФ = (p²-1)/p²+1)

Hope it helps , cheers !!

Answer 2

• $\sf{cos\:θ\:=\:{\dfrac{P^{2}\:-\:1}{P^{2}\:+\:1}}}$

Step-by-step explanation:

Given,

• $\sf{cosec\:θ\:+\:cot\:θ\:=\:P}$ -------- (1)

⟼ $\sf{cosec^{2}\:θ\:-\:cot^{2}\:θ\:=\:1}$ (Identity)

⟼ $\sf{cosec\:θ\:-\:cot\:θ\:=\:{\dfrac{1}{cosec\:θ\:+\:cot\:θ}}}$

⟼ $\sf{cosec\:θ\:-\:cot\:θ\:=\:{\dfrac{1}{P}}}$ --------- (2)

Adding (1) & (2),

⟼ $\sf{2\:cosec\:θ\:=\:P\:+{\dfrac{1}{P}}}$

⟼ $\sf{2\:cosec\:θ\:=\:{\dfrac{P^{2}\:+\:1}{P}}}$ ---------- (3)

Subtracting (2) from (1),

⟼ $\sf{cosec\:θ\:+\:cot\:θ\:=\:P}$ $\sf{-}$$\sf{cosec\:θ\:-\:cot\:θ\:=\:{\dfrac{1}{P}}}$

⟼ $\sf{2\:cot\:θ\:=\:P\:-\:{\dfrac{1}{P}}}$

⟼ $\sf{2\:cot\:θ\:=}$ $\sf{ \dfrac{P^2\:-\:1}{P}}$ ---------- (4)

Dividing (4) & (3),

⟼ $\sf\dfrac{2\:cot\:θ}{2\:cosec\:θ}\:=\:{\dfrac{P^{2}\:-\:1}{P}\:×\:{\dfrac{P}{P^{2}\:+\:1}}}$

⟼ $\sf{cos\:θ\:=\:{\dfrac{P^{2}\:-\:1}{P^{2}\:+\:1}}}$

• Hence Proved