Question

Prove that sec2theta-cos2theta=sin2theta(sec2theta+1)

Answer 1

To prove---> Sec²θ - Cos²θ = Sin²θ ( Sec²θ + 1 )

Proof---> LHS = Sec²θ - Cos²θ

We know that Cosθ = 1 / Secθ , applying it here we get,

= Sec²θ - ( 1 / Sec²θ )

Taking Sec²θ as LCM , we get,

= ( Sec⁴θ - 1 ) / Sec²θ

= { ( Sec²θ )² - ( 1 )² } / Sec²θ

We have an identity , a² - b² = ( a + b ) ( a - b ) , applying it here , we get ,

= ( Sec²θ + 1 ) ( Sec²θ - 1 ) / Sec²θ

Putting Sec²θ - 1 = tan²θ , we get,

= ( Sec²θ + 1 ) tan²θ ( 1 / Sec²θ )

We know that , tanθ = Sinθ / Cosθ and

Cosθ = 1 /Secθ , applying it here , we get.

= ( Sec²θ + 1 ) ( Sin²θ / Cos²θ ) Cos²θ

= ( Sec²θ + 1 ) Sin²θ

= Sin²θ ( Sec²θ + 1 ) = RHS

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Answer 2

Answer:

Step-by-step explanation:

To prove---> Sec²θ - Cos²θ = Sin²θ ( Sec²θ + 1 )

Proof---> LHS = Sec²θ - Cos²θ

We know that Cosθ = 1 / Secθ , applying it here we get,

= Sec²θ - ( 1 / Sec²θ )

Taking Sec²θ as LCM , we get,

= ( Sec⁴θ - 1 ) / Sec²θ

= { ( Sec²θ )² - ( 1 )² } / Sec²θ

We have an identity , a² - b² = ( a + b ) ( a - b ) , applying it here , we get ,

= ( Sec²θ + 1 ) ( Sec²θ - 1 ) / Sec²θ

Putting Sec²θ - 1 = tan²θ , we get,

= ( Sec²θ + 1 ) tan²θ ( 1 / Sec²θ )

We know that , tanθ = Sinθ / Cosθ and

Cosθ = 1 /Secθ , applying it here , we get.

= ( Sec²θ + 1 ) ( Sin²θ / Cos²θ ) Cos²θ

= ( Sec²θ + 1 ) Sin²θ

= Sin²θ ( Sec²θ + 1 ) = RHS