Question

Prove that : 1/secA+tanA=secA-tanA


help me please​

Answer 1

Step-by-step explanation:

Given,

secA−tanA

1

−

cosA

1

=

cosA

1

−

secA+tanA

1

or

secA−tanA

1

+

secA+tanA

1

=

cosA

1

+

cosA

1

Here, R.H.S.=

cosA

2

Now,

L.H.S.=

secA−tanA

1

+

secA+tanA

1

=

(secA−tanA)(secA+tanA)

secA+tanA+secA−tanA

=

cosA

2

Thus, L.H.S.=R.H.S.

Answer 2

$\frac{1}{seca + tana} \\ \\ = \frac{1}{seca + tana} \times \frac{seca - tana}{seca - tana} \\ \\ = \frac{seca - tana}{ {sec}^{2}a - {tan}^{2} a } \\ \\ = seca - tana$