How do you prove #(cotA-1)/(cotA+1) = (cosA-sinA)/(cosA+sinA)#?

3 Answers
Apr 7, 2018

Please refer to the Explanation.

Explanation:

#(cotA-1)/(cotA+1)#,

#={cosA/sinA-1}-:{cosA/sinA+1}#,

#={(cosA-sinA)/cancelsinA}-:{(cosA+sinA)/cancelsinA}#,

#=(cosA-sinA)/(cosA+sinA)#, as desired!

Apr 7, 2018

See below.

Explanation:

#LHS=(cotA-1)/(cotA+1)#

#=(cosA/sinA-1)/(cosA/sinA+1)#

#=(cosA/sinA-1)/(cosA/sinA+1)timessinA/sinA#

#=(cosA-sinA)/(cosA+sinA)#

#=RHS#

Apr 7, 2018

Refer to explanation.

Explanation:

Starting from #color(red)(LHS)#

#(cot A - 1)/(cotA+1)#

#rArr (cosA/sin A -1)/(cosA/sinA + 1)#

Since, #cotA= cosA/sin A#

#rArr ((cosA-sinA)/sinA)/((cosA+sinA)/sinA)#

#rArr((cosA-sinA)/cancelsinA)/((cosA+sinA)/cancelsinA)#

#rArr (cosA-sinA)/(cosA+sinA)#

We get #color(red)(RHS).#

#color(white)(aaaaaaaaaaaaaaaaaaaaĆ aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaasasaaaaaaaasaaaaaaaaaaaaaaaaa)#
Hope this helps :)