Question

Prove that sintheta-costheta+1/sintheta+costheta-1=1/sec theta-tan theta

Answer 1

HELLO DEAR,

Prove that :- (sinθ - cosθ+1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ).

NOW,

"Solving LHS"

(sinθ - cosθ + 1) / (sinθ + cosθ - 1)

[ divide both Numerator and denominator by cosθ ]

(sinθ/cosθ - cosθ/cosθ + 1/cosθ) / (sinθ/cosθ + cosθ/cosθ - 1/cosθ)

(tanθ + secθ - 1) / (tanθ - secθ + 1)

Multiply by (tanθ - secθ) in Numerator and denominator

we get,

(tanθ + secθ - 1)(tanθ - secθ)/(tanθ - secθ + 1)(tanθ - secθ)

[(tan²θ - sec²θ) - (tanθ - secθ)] / (tanθ - secθ + 1)(tanθ - secθ)

(-1 - tanθ + secθ) / (tanθ - secθ + 1)(tanθ - secθ)

$\bold{\to\to\to\to\to\to\to\boxed{ [ sec^2x - tan^2x = 1 ]}}$

-1/(tanθ - secθ)

1/(secθ - tanθ)

HENCE, LHS = RHS


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

Your answer attached in the photo