Question

Prove that

1/sec theta + tan theta = sec theta - tan theta

Answer 1

Answer:

We have,

LHS =

secθ−tanθ

1

⇒ LHS =

secθ−tanθ

1

×

secθ+tanθ

secθ+tanθ

⇒ LHS =

sec

2

θ−tan

2

θ

secθ+tanθ

=

1

secθ+tanθ

[∵sec

2

θ−tan

2

θ=1]

⇒ LHS = secθ+tanθ=RHS

Answer 2

$\frac{1}{sec \theta + tan \theta} \\ \\ = \frac{sec \theta - tan \theta}{(sec \theta + tan \theta)(sec \theta - tan \theta)} \\ \\ = \frac{sec \theta - tan \theta}{ {sec}^{2} \theta - {tan}^{2} \theta} \\ \\ = sec \theta - tan \theta$