Rewrite #tantheta# and #cottheta# as sines and cosines using #color(magenta)(tan theta = sintheta/costheta and cot theta = costheta/sintheta#.

#(sin2theta)/(cos2theta) = 2/(costheta/sintheta - sintheta/costheta)#

I would recommend you simplify the right hand side prior to expanding the left.

#(sin2theta)/(cos2theta) = 2/((cos^2theta - sin^2theta)/(costhetasintheta)#

#(sin2theta)/(cos2theta) = (2costhetasintheta)/(cos^2theta - sin^2theta)#

We know this is true because #sin2theta= 2sinthetacostheta# and #cos2theta# can be written as #cos^2theta - sin^2theta#.

**Practice exercises:**

*Use the following table of trig identities to help you answer the next questions *

- Prove the following trig identities:

a) #(sin^2theta + cos^2theta + cot^2theta)/(1 + tan^2theta) = cot^2theta#

b) #cos(x + y) + cos(x - y) = 2cosxcosy#

c) #csc(2alpha) - cot(2alpha) = tan alpha#

Solve the following equation for x in the interval #0 ≤ x ≤ 2pi#:

#cos(2x) = 2sin^2x#

Hopefully this helps, and good luck!