Question

Can anyone help me solve this

Answer 1

Answer:

²

Step-by-step explanation:

Given,

f ( θ ) = tan θ + sec θ

g ( θ ) = cosθ/1-sinθ

To prove : -

f ( θ ) = g ( θ )

Proof : -

LHS = f ( θ )

RHS = g ( θ )

LHS

= f ( θ )

= tan θ + sec θ

{ Identities : -

tan θ = sin θ/cos θ

sec θ = 1/cos θ }

= sin θ/cos θ + 1/cos θ

= ( sin θ + 1 )/cos θ

= ( 1 + sin θ )/cos θ

Multiply ( 1 - sin θ ) to both numerator and denominator,

= ( 1 + sin θ ) ( 1 - sin θ ) / ( cos θ ) ( 1 - sin θ )

{ ( 1 + sin θ ) ( 1 - sin θ ) is in the form a² - b² Identity,

so,

( 1 + sin θ ) ( 1 - sin θ ) = 1² - sin²θ = 1 - sin²θ

Formula : -

sin²θ + cos²θ = 1

cos²θ = 1 - sin²θ

So,

1 - sin²θ = cos²θ }

= cos²θ / ( cos θ ) ( 1 - sin θ )

{ cos²θ = cos θ . cos θ }

= cos θ . cos θ / ( cos θ ) ( 1 - sin θ )

= cos θ / 1 - sin θ

= RHS

= g ( θ )

Hence proved.