Question

sinA tanA /1-cosA =1+secA

Answer 1

LET angle A=x

LHS=
$\frac{ \sin(x) \times \frac{ \sin(x) }{ \cos(x) } }{1 - \cos(x) } = \frac{ { \sin(x) }^{2} }{ \cos(x)(1 - \cos(x)) } = \frac{1 - { \cos(x) }^{2} }{ \cos(x)(1 - \cos(x) }$
ie

$\frac{(1 + \cos(x) )(1 - \cos(x)) }{ \cos(x) (1 - \cos(x) )} = \frac{1 + \cos(x) }{ \cos(x) } = \frac{1}{ \cos(x) } + \frac{ \cos(x) }{ \cos(x) }$
ie
$1 + \sec(x) = \: rhs$

Answer 2

Answer:

L.H.S. = sinA tanA / (1 - cosA)

Multiplying and dividing by (1 + cosA)

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

= (sinA tanA)(1 + cosA) / (1 - cos^2 A)

= (sinA tanA)(1 + cosA) / sin^2 A

= [(sin^2A/cosA)(1 + cosA)] / sin^2 A

= sin^2A (1 + cosA) / cosA sin^2 A

= (1 + cosA) / cosA

= 1/cosA + cosA/cosA

= secA + 1

= 1 + secA

= R.H.S.

Thats it dear.

Step-by-step explanation: