Question

prove that 1 - cosA + sinA / sinA + cosA - 1 = 1 + sinA / cosA'

Answer 1

(1+sinA)(1-sinA)/cosA(1-sinA)

=sin2A/cosA(1-sinA)

=cos2A/cosA(1-sinA)

=cosA/(1-sinA)

=1/1/cosA-sinA/cosA

=1/secA-tanA

Answer 2

$\bold{\frac{[\sin A+(1-\cos A)]}{[\sin A-(1-\cos A)]}=\frac{1+\sin A}{\cos A}}$ is proven.

Solution:

$\frac{[\sin A+(1-\cos A)]}{[\sin A-(1-\cos A)]}$

Multiplying  [sinA+(1-cosA)] with numerator and denominator, we get

$=\frac{[\sin A+(1-\cos A)]^{2}}{\left[\sin ^{2} A-(1-\cos A)^{2}\right]}$

Expanding the numerator in the form $(a-b)^{2}=a^{2}-2 a b+b^{2}$

$=\frac{\left[\left(\sin ^{2} A+2 \cdot \sin A \cdot(1-\cos A)+(1-\cos A)^{2}\right]\right.}{\left[\left(1-\cos ^{2} A\right)-(1-\cos A)^{2}\right]}$

Since $\sin ^{2} A=1-\cos ^{2} A$

$=\frac{\left[\left(1-\cos ^{2} A\right)+2 \cdot \sin A \cdot(1-\cos A)+(1-\cos A)^{2}\right]}{\left[(1-\cos A)(1+\cos A)-(1-\cos A)^{2}\right]} \because 1-\cos ^{2} A=(1-\cos A)(1+\cos A)$

Taking (1- cos A) in common

$\frac{=(1-\cos A)[1+\cos A+2 \sin A+1-\cos A]}{[1-\cos A][1+\cos A-1+\cos A]}$

Cancelling + cos A and – cos A in numerator and adding two cos A in denominator.

$=\frac{[2+2 \sin A]}{[2 \cdot \cos A]}$

Cancelling 2 in numerator and denominator.

$=\frac{2(1+\sin A)}{2 \cdot \cos A}$

$\frac{1+\sin A}{\cos A}$ = R.H.S

The derived L.H.S is equal to R.H.S. Hence Proved.