How do you show that #(cscx - cotx)^2 = (1 - cosx)/(1 + cosx)#?
2 Answers
Convert to sine and cosine using
#(1/sinx - cosx/sinx)^2 = (1 - cosx)/(1 + cosx)#
#((1 - cosx)/sinx)^2 = (1 - cosx)/(1 + cosx)#
#(1 - 2cosx + cos^2x)/sin^2x = (1 - cosx)/(1 + cosx)#
Now use
#(1 - cosx)^2/(1 - cos^2x) = (1 - cosx)/(1 + cosx)#
Note the difference of squares in
#(1- cosx)^2/((1 + cosx)(1 - cosx)) = (1 - cosx)/(1 + cosx)#
#(1 - cosx)/(1 + cosx) = (1 - cosx)/(1 + cosx)#
This identity has been proven to be true.
Hopefully this helps!
Multiply the RHS by
Explanation:
Whenever you see a trig identity proof that has something like
In our case,
Note:
You can also convert everything to