Question

answer this............

Answer 1

Answer:

(p² + 1)/(p² - 1)

Step-by-step explanation:

Given: sec θ + tan θ = p   ----- (i)

We know that sec²θ - tan²θ = 1

⇒ (secθ + tanθ)(secθ - tanθ) = 1

⇒ (p)(secθ - tanθ) = 1

⇒ secθ - tanθ = (1/p)   ----- (ii)

On solving (i) & (ii), we get

⇒ secθ + tanθ + secθ - tanθ = p + 1/p

⇒ 2secθ = p² + 1/p

⇒ secθ = (p² + 1)/2p

⇒ cosθ = (1/secθ)

            = 2p/p² + 1


Sin²θ = 1 - cos²θ

         = 1 - (2p/p² + 1)²

         = 1 - (4p²)/p⁴ + 1 + 2p²

         = (p⁴ + 1 + 2p² - 4p²)/p⁴ + 1 + 2p

         = (p⁴ + 1 - 2p²)/p⁴ + 1 + 2p

         = (p² - 1)²/(p² + 1)²

sin θ= p² - 1/p² + 1.


Now,

We know that cosecθ = (1/sinθ)

⇒ (p² + 1)/p² - 1.


Hope it helps!

Answer 2

Given : Secθ - Tanθ = p ------------- [1]

Multiplying both sides with (Secθ + Tanθ)

⇒ (Secθ - Tanθ)(Secθ + Tanθ) = (p)(Secθ + Tanθ)

⇒ (Sec²θ - Tan²θ) = (p)(Secθ + Tanθ)

We know that : (Sec²θ - Tan²θ) = 1

⇒ (p)(Secθ + Tanθ) = 1

⇒ $(Sec\theta + Tan\theta) = \frac{1}{p}$ -------- [2]

Adding Both Equations [1] and [2], We get :

⇒ $(Sec\theta - Tan\theta) + (Sec\theta + Tan\theta) = p + \frac{1}{p}$

⇒ $Sec\theta = \frac{p^2 + 1}{2p}$

Subtracting Equation [1] from Equation [2], We get :

⇒ $(Sec\theta + Tan\theta) - (Sec\theta - Tan\theta) = \frac{1}{p} - p$

⇒ $Tan\theta = \frac{1 - p^2}{2p}$

We know that : $Tan\theta = \frac{Sec\theta}{Cosec\theta} \implies Cosec\theta = \frac{Sec\theta}{Tan\theta}$

Substituting the Values of Secθ and Tanθ in the Above Equation, We get :

⇒ $Cosec\theta = \frac{\frac{1 + p^2}{2p}}{\frac{1 - p^2}{2p}}$

⇒ $Cosec\theta = \frac{1 + p^2}{1 - p^2}$