How do you verify the identity $tan2theta=2/(cottheta-tantheta)$ ?

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Answer 1 · 2016-12-22T00:22:49

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 *

![ mathsgenius.co.za )

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!

How do you verify the identity tan2theta=2/(cottheta-tantheta)tan2theta=2/(cottheta-tantheta)?