Prove the identity. tanx-cotx/tanx+cotx=sin^2x-cos^2x Have to show the statements and the rules?

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Answer 1 · 2018-04-07T11:51:34

See below.

Identities:

#color(red)bb(tanx=sinx/cosx)#

#color(red)bb(cotx=cosx/sinx)#

Substituting these in the LHS:

#(sinx/cosx-cosx/sinx)/(sinx/cosx+cosx/sinx)#

Adding the terms in numerator and denominator:

#((sin^2x-cos^2x)/(sinxcosx))/((sin^2x+cos^2x)/(cosxsinx)#

Multiplying by #(cosxsinx)#

#(sin^2x-cos^2x)/(sin^2x+cos^2x)#

Identity:

#color(red)bb(sin^2x+cos^2x=1)#

Substituting this in the denominator:

#(sin^2x-cos^2x)/1#

#sin^2x-cos^2x#

As required:

#LHS-=RHS#

Answer 2 · 2018-04-07T15:10:43

#(tanx-cotx)/(tanx+cotx)=sin^2x-cos^2x#

#=>(tanx-1/tanx)/(tanx+1/tanx)#

#=>(tan^2x-1)/(tan^2x+1)#

#=>(tan^2x-1)/(sec^2x)# #color(white)(wwwwww# #["as "tan^2x-sec^2x = 1] #

#=>(tan^2x/sec^2x-1/sec^2x)#

#=>sin^2x-cos^2x#