1.

Prove that (1+tanA-secA)×(1+tanA+secA)​

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tanA−secA+1

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.



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