prove that. sec(theeta)/sec(theeta)-1 + sec(theeta)/sec(theeta)+1 = 2 cosec*2( theeta)
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Answer:
We have
secθ+tan
3
θcosecθ
=secθ[
secθ
secθ+tan
3
θcosecθ
]
=secθ[1+tan
3
θ⋅
sinθ
cosθ
]
=secθ[1+tan
3
θ×cotθ]
=
1+tan
2
θ
[1+tan
2
θ]
=[1+tan
2
θ]
3/2
=[1−(1−a
2
)]
3/2
=(2−a
2
)
3/2
[∵tan
2
θ=1−a
2
]
Step-by-step explanation:
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