Question

1 + cot A upon sec A = 2 sin square A upon 1 - Cos A

Answer 1

hii!! mate here is ur ans!

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

hope it helps!