Question

show that 1+seco /seco =sin²o/1-coso​

Answer 1

Step-by-step explanation:

To show--->

( 1 + Secθ ) / Secθ = Sin²θ / ( 1 - Cosθ )

Proof--->

LHS = ( 1 + Secθ ) / Secθ

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

= { 1 + ( 1 / Cosθ ) } / 1 / Cosθ

= Cosθ ( Cosθ + 1 ) / Cosθ

Multiplying in numerator and denominator by

( 1 - Cosθ ) , we get

= ( Cosθ + 1 ) ( 1 - Cosθ ) / ( 1 - Cosθ )

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

= { ( 1 )² - ( Cosθ )² } / ( 1 - Cosθ )

= ( 1 - Cos²θ ) / ( 1 - Cosθ )

We know that, Sin²θ = 1 - Cos²θ , we get,

= Sin²θ / ( 1 - Cosθ ) = RHS