Question

1+tan theta the whole square + 1+cot theta the whole square = sec theta+ cosec theta the whole square​

Answer 1

Step-by-step explanation:

Given---->

( 1 + tanθ )² + ( 1 + Cotθ )² = ( Secθ + Cosecθ )²

Proof ----> LHS

= ( 1 + tanθ )² + ( 1 + Cotθ )²

we know that ,

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

= 1 + tan²θ + 2tanθ + 1 + Cot²θ + 2Cotθ

= ( 1 + tan²θ ) + ( 1 + Cot²θ ) + 2tanθ + 2Cotθ

We know that,

1 + tan²θ = Sec²θ

1 + Cot²θ = Cosec²θ

applying these formulee , we get,

= Sec²θ + Cosec²θ + 2 ( tanθ + Cotθ )

We know that,

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

= Sec²θ + Cosec²θ + 2(Sinθ / Cosθ + Cosθ / Sinθ)

= Sec²θ + Cosec²θ + 2 (Sin²θ + Cos²θ/Sinθ Cosθ)

We know that,

Sin²θ + Cos²θ = 1 , applying it we get,

= Sec²θ + Cosec²θ + 2 ( 1 / Sinθ Cosθ )

= Sec²θ + Cosec²θ + 2 ( 1 / Sinθ ) ( 1 / Cosθ )

We know that,

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

= Sec²θ + Cosec²θ + 2 Cosecθ Secθ

= ( Secθ )² + ( Cosecθ )² + 2 ( Cosecθ ) ( Secθ )

= ( Secθ + Cosecθ )²

Answer 2

Answer:

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