SinA-cosA+1÷sinA+cosA-1 = 1÷secA-tanA
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Answer:
LHS =
sinθ+cosθ−1
sinθ−cosθ+1
=
tanθ+1−secθ
tanθ−1+secθ
=
(tanθ−secθ)+1
(tanθ+secθ)−1
=
{(tanθ−secθ)+1}(tanθ−secθ)
{(tanθ+secθ)−1}(tanθ−secθ)
=
{tanθ−secθ+1}(tanθ−secθ)
(tan
2
θ−sec
2
θ)−(tanθ−secθ)
=
tanθ−secθ
−1
[∵tan
2
θ−sec
2
θ=−1]
=
secθ−tanθ
1
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