How do you express #cos theta - cos^2 theta + cot theta # in terms of #sin theta #?

1 Answer
sankarankalyanam
Apr 18, 2018

#color(blue)(=>(sqrt(1-sin^2 theta) / sin theta) * ( sin theta - sin(2theta) / 2 - 1)#

Explanation:

#cos theta - cos^@ theta - cot theta#

![https://www.onlinemathlearning.com/http://trigonometric-identities.html](https://useruploads.socratic.org/n2blswKSWms5UU0QoTY5_trigonometric%20identities.png)

#=>cos theta - cos^2 theta - (cos theta / sin theta)#

#=>cos theta * ( 1 - cos theta - (1/sin theta))#

#=>cos theta (sin theta - sin theta cos theta - 1 ) / sin theta#

We will use the following identities to convert into sine form.

#color(crimson)(sin 2 theta = 2 sin theta cos theta, cos^2 theta = 1 - sin^2 theta#

#color(blue)(=>(sqrt(1-sin^2 theta) / sin theta) * ( sin theta - sin(2theta) / 2 - 1)#

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