Question

sec²θ+cosec²θ =sec²θ*cosec²θ,Prove it

Answer 1

$\bf{LHS}$ = sec²θ + cosec²θ
We know, secα = 1/cosα and cosecα = 1/sinα , use here
= 1/cos²θ + 1/sin²θ
= (sin²θ + cos²θ)/cos²θ.sin²θ

As you know, sin²α + cos²α = 1 from Trigonometric identity.
so, sin2θ + cos²θ = 1

= 1/cos²θ.sin²θ = (1/cosθ)²(1/sinθ)²
= (secθ)²(cosecθ)²
= sec²θcosec²θ = $\bf{RHS}$

Answer 2

Answer:

= sec²θ + cosec²θ

We know, secα = 1/cosα and cosecα = 1/sinα , use here

= 1/cos²θ + 1/sin²θ

= (sin²θ + cos²θ)/cos²θ.sin²θ

As you know, sin²α + cos²α = 1 from Trigonometric identity.

so, sin2θ + cos²θ = 1

= 1/cos²θ.sin²θ = (1/cosθ)²(1/sinθ)²

= (secθ)²(cosecθ)²

= sec²θcosec²θ =

Step-by-step explanation: