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

Question

4842 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-07T13:26:10

Please refer to the Explanation.

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

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

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

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

Answer 2 · 2018-04-07T13:27:25

See below.

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

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

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

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

#=RHS#

Answer 3 · 2018-04-07T13:38:47

Refer to 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)(aaaaaaaaaaaaaaaaaaaaàaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaasasaaaaaaaasaaaaaaaaaaaaaaaaa)#
Hope this helps :)