How do you express #4 cos^2 theta - sec^2 theta + 2 cot theta # in terms of #sin theta #?

1 Answer
Feb 19, 2016

#4 - 4 sin^2 theta - 1 / (1 - sin^2 theta) + (2 sqrt(1 - sin^2 theta)) / sin theta#

Explanation:

We will use the following identities:

[1] #" " sin^2 theta + cos^2 theta = 1 " " <=> " " cos^2 theta = 1 - sin^2 theta#

[2] #" " sec theta = 1 / cos theta#

[3] #" " cot theta = cos theta / sin theta#

Thus, we can express the term as follows:

#4 cos^2 theta - sec^2 theta + 2 cot theta#

# = 4(1 - sin^2 theta) - 1 / cos^2 theta + (2 cos theta) / sin theta#

... apply [1] once again...

# = 4 - 4 sin^2 theta - 1 / (1 - sin^2 theta) + (2 cos theta) / sin theta#

... now, there is only one #cos theta# expression left which we can express as #sqrt(1 - sin^2 theta)#:

# = 4 - 4 sin^2 theta - 1 / (1 - sin^2 theta) + (2 sqrt(1 - sin^2 theta)) / sin theta#

Hope that this helped!