How do you verify tanx / (1-cosx) = cscx (1+secx)?

2 Answers

Use trick algebra while working with the RHS.

Explanation:

There probably is a more efficient method. Nonetheless, here is my work to make the right hand side look like the left:

RHS =csc x(1 + sec x)

= csc x + csc x sec x --distribution
= 1/sinx + 1/(cosx sinx) -- definition

= (cosx+1)/(cosx sinx) -- common denom

Multiply both numerator and denominator by (cos x -1)
rArr (-(1-cos^2x))/(cos^2x sinx - cosx sinx)

--also took out a negative to fit with Pythagorean identity #1

Then multiply both the numerator and the denominator by (1/sin x )
rArr(-sinx)/(cos^2x-cosx)

=(-sinx)/(cosx(cosx-1))

Now do same with (1/cos x )
rArr(tanx)/-(cosx -1) --moved negative to bottom

=(tanx)/(1-cosx)= LHS, that we needed to verify!

Hope this helps :-)

May 20, 2017

LHS=tanx/(1-cosx)

=(tanx(1+cosx))/((1-cosx)(1+cosx)

=(sinx(1+cosx))/(cosx(1-cos^2x))

=(sinx(1+cosx))/(cosx(sin^2x))

=(sinx/sin^2x) xx(1+cosx)/cosx

=(1/sinx) xx(1/cosx+cosx/cosx)

=cscx xx(secx+1)=RHS

Proved