Question

prove that 1+SecA/SecA = Sin^2A/1 - CosA​

Answer 1

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

Answer 2

Answer:

Step-by-step explanation:

To Prove :-

$\dfrac{1+secA}{secA}=\dfrac{sin^2A}{1-cosA}$

Solution :-

Taking L.H.S :-

$=\dfrac{1+secA}{secA}$

$=\dfrac{1+\dfrac{1}{cosA}}{\dfrac{1}{cosA}}$

$=\dfrac{\dfrac{cosA+1}{cosA}}{\dfrac{1}{cosA}}$

$=\dfrac{cosA+1}{1}$

= cosA + 1

$= 1+cosA\times \dfrac{1-cosA}{1-cosA}$

$=\dfrac{1-cos^2A}{1-cosA}$

$=\dfrac{sin^2A}{1-cosA}$

= R.H.S

Hence Proved