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

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Answer 1 · 2016-04-22T03:49:42

Use trick algebra while working with the RHS.

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 :-)

Answer 2 · 2017-05-20T04:38:15

#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