Question

(1 + tan A -sec A)(1+tan A+ sec A) =

(A)2 tan A (B) 2 sin A (C) 2 sec A (d) 2 cot A​

Answer 1

Answer:

1+tanA−secA)×(1+tanA+secA)

=(1+tanA)

2

−(secA)

2

[∵(a+b)(a−b)=a

2

−b

2

)]

=1+tan

2

A+2tanA−sec

2

A

=sec

2

A+2tanA−sec

2

A [∵sec

2

θ=1+tan

2

θ]

=2tanA

=R.H.S.

∴L.H.S=R.H.S

(1+tanA−secA)×(1+tanA+secA)=2tanA

Hence proved

therefore A)2tanA is correct

Answer 2

Answer:

a) 2 tan A is the correct answer.

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