Question #c2684

2 Answers
Narad T.
Oct 2, 2017

See the proof below

Explanation:

We need

#tanx=sinx/cosx#

#sin2x=2sinxcosx#

Therefore

#LHS=(tan(2x)cos2x)/sinx#

#=sin(2x)/cancelcos(2x)*cancelcos(2x)/sinx#

#=(2cancelsinxcosx)/cancelsinx#

#=2cosx#

#=RHS#

#QED#

Pankaj Solanki
Oct 2, 2017

L.H.S.
#=(tan2xcos2x)/sinx#
Using Formula #tan2x=(sin2x)/(cos2x)#
#=(sin2x)/cancel(cos2x)xxcancel(cos2x)/sinx#

Using Formula #sin2x=2sinxcosx#
#=(2cancelsinxcosx)/cancelsinx## #=2cosx#

Related questions
Impact of this question
1037 views around the world
You can reuse this answer
Creative Commons License