08.
Prove that
(1 + tan A - sec A) #(1 + tan A + sec A) = 2 tan A
Answers
Answered by
0
Answer:
L.H.S
(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.
Similar questions