prove that 1+SecA/SecA = Sin^2A/1 - CosA
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Answered by
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Step-by-step explanation:
How can I prove that 1+sec A/sec A=sin^2A/1-cos A?
LHS=1+secAsecA=1secA+secAsecA
=cosA+1
[∵1secA=cosA]
=1+cosA
Multiply and divide by the conjugate of the term, here, by 1−cosA
=(1+cosA)×(1−cosA)(1−cosA)
=(1+cosA)(1−cosA)1−cosA
=1−cos2A1−cosA
[∵(a+b)(a−b)=a2−b2]
=sin2A1−cosA=RHS
[∵1−cos2A=sin2A].
LHS from RHS is even easier.
RHS=sin2A1−cosA
=1−cos2A1−cosA
[∵sin2A=1−cos2A].
=(1−cosA)(1+cosA)1−cosA
[∵(a−b)(a+b)=a2−b2]
=1+cosA
[∵By cancelling 1−cosA]
=1+1secA
[∵cosA=1secA]
=secAsecA+1secA
=secA+1secA=LHS
Answered by
6
Answer:
Step-by-step explanation:
To Prove :-
Solution :-
Taking L.H.S :-
= cosA + 1
= R.H.S
Hence Proved
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