Solving this trigo Question ?

How do you solve this Trigo problem?

Can (1-cosx)/sinx be = tan(x/2)

1 - cos x = 2 ( sin (x/2) )^2 ----- (1)
sinx = 2 sin (x/2) cos (x/2) ------ (2)

dividing (1) / (2) gives you tan (x/2)

Can (tanx/2 sinx/2 + cosx/2-1 ) = (secx/2 - 1) ?

Let x/2 be A.

LHS
= tan A sin A + cos A - 1
=/= sec A - 1
=/= RHS

Neutral

tan A sin A + cos A -1
= (sin A )^2 / cos A + cos A -1
= { (sin A )^2 + (cos A )^2 } / cos A - 1
= 1 / cos A - 1
= sec A - 1

*dies* Laughing Confused

Thanks !!

don't worry

It's eagle that's why Laughing