Question

5. Prove : tan²0 + cot²0 + 2 = sec²0.cosec²0.​

Answer 1

Step-by-step explanation:

To prove ----->

tan²θ + Cot²θ + 2 = Sec²θ Cosec²θ

Proof----->

LHS= tan²θ + Cot²θ + 2

Multiplying and dividing by tanθ in third term , we get,

= ( tanθ )² + ( Cotθ )² + 2 tanθ ( 1 / tanθ )

We know that , Cotθ = 1 / tanθ , applying it we get,

= ( tanθ )² + ( Cotθ )² + 2 tanθ Cotθ

We know that,

a² + b² + 2ab = ( a + b )² , applying it , we get,

= ( tanθ + Cotθ )²

We know that,

tanθ = Sinθ / Cosθ , Cotθ = Cosθ / Sinθ , applying it we get,

= { ( Sinθ / Cosθ ) + ( Cosθ / Sinθ ) }²

= { ( Sin²θ + Cos²θ ) / Sinθ Cosθ }²

We know that, Sin²θ + Cos²θ = 1 , applying it we get,

= ( 1 / Sinθ Cosθ )²

= 1 / Sin²θ Cos²θ

= ( 1 / Sin²θ ) ( 1 / Cos²θ )

We know that ,

1 / Sinθ = Cosecθ , 1 / Cosθ = Secθ , applying it , we get,

= Cosec²θ Sec²θ = RHS