Question

PROVE THAT:
secA+tanA-1/tanA-secA+1=cosA/1-sinA​

Answer 1

Answer:

tanA+secA−1

=

cosA

1+sinA

Taking L.H.S.-

tanA−secA+1

tanA+secA−1

=

tanA−secA+1

(tanA+secA)−(sec

2

A−tan

2

A)

[∵1+tan

2

A=sec

2

A]

=

tanA−secA+1

(tanA+secA)−(secA+tanA)(secA−tanA)

=

tanA−secA+1

(tanA+secA)(1−(secA−tanA))

=

tanA−secA+1

(tanA+secA)(1−secA+tanA)

=tanA+secA

=

cosA

sinA

+

cosA

1

=

cosA

1+sinA

= R.H.S.

Hence proved.

Answer 2

$\frac{tanA + secA - 1}{1 + tanA - secA} \\ \\ = \frac{secA + tanA - 1}{( {sec}^{2}A - {tan}^{2} A) - (secA - tanA) } \\ \\ = \frac{secA + tanA - 1}{(secA - tanA)(secA + tanA - 1)} \\ \\ = \frac{1}{secA - tanA} \\ \\ = \frac{1}{ \frac{1 - sinA}{cosA} } \\ \\ = \frac{cosA}{1 - sinA}$