Question

Prove that sec^2a-tan^2a=1

Answer 1

Answer:-

We have to prove:

Sec² A - tan² A = 1

Using sec² A = 1/Cos² A and tan² A = sin² A/Cos² A in LHS we get,

→ (1/Cos² A) - (Sin² A/Cos² A) = 1

→ (1 - sin² A)/Cos² A = 1

We know that,

sin² A + cos² A = 1

→ cos² A = 1 - sin²A

Hence,

cos² A/Cos² A = 1

→ 1 = 1

→ LHS = RHS

Hence, Proved.

How can we say that Sin² A + Cos² A = 1 ?

Let us take a right angled ∆PQR , right angled at Q .

Let PQ = x , QR = y .

Using Pythagoras Theorem,

→ (PR)² = (PQ)² + (QR)²

→ (PR)² = x² + y²

→ PR = √x² + y²

Let A be an angle at P.

We know that,

sin ∅ = Opposite side/Hypotenuse

→ Sin A = (PQ)/(PR) = y/√x² + y²

cos A = Adjacent side/Hypotenuse

→ Cos A = (QR)/(PR) = x/√x² + y²

→ sin² A + Cos² A = 1

→ (y/√x² + y²)² + (x/√x² + y²)² = 1

→ y²/(x² + y² )+ x²/(x² + y²) = 1

→ (x² + y²)/(x² + y²) = 1

→ 1 = 1

Hence, we can say that sin² A + Cos² A = 1.