Question

show that (tan tetha +sec tetha)^2+(tan tetha -sec tetha)^2=2(1+sin^2 tetha/1-sin^2 tetha)

Answer 1

To prove--->

(tanθ + Secθ)² + (tanθ - Secθ)²

= 2 ( 1 + Sin²θ ) / ( 1 - Sin²θ )

Proof--->

LHS= (tanθ + Secθ )² + ( tanθ - Secθ )²

We have two identity as follows

( a + b )² = a² + b² + 2ab

( a - b )² = a² + b² - 2ab , applying these identities here , we get

= tan²θ + Sec²θ + 2tanθ Secθ + tan²θ + Sec²θ

- 2 tanθ Secθ

- 2tanθSecθ and 2tanθ Secθ cancel out each other

= 2 tan²θ + 2 Sec²θ

Taking 2 common , we get,

= 2 ( tan²θ + Sec²θ )

We know that,

tanθ = Sinθ / Cosθ , Secθ = 1 / Cosθ , applying it here , we get,

= 2 { ( Sin²θ / Cos²θ ) + ( 1 / Cos²θ ) }

Taking Cos²θ as LCM, we get,

= 2 { ( Sin²θ + 1 ) / Cos²θ }

We know that ,

Cos²θ =1 - Sin²θ , applying it we get,

= 2 ( 1 + Sin²θ ) / ( 1 - Sin²θ ) = RHS

Answer 2

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