Question

How do you simplify `tanA - cscA secA (1 - 2cos^2 A)`?

Answer 1

Start by putting all of the terms into sine and cosine. This will make it easier to spot identities and see what might cancel out.

`sin A / cos A - 1/sinA 1/cos A ( 1-2cos^2A)`

Multiply through the parenthesis..

`sin A / cos A - 1/(sin A cos A)+(2cos^2A) /(sin Acos A)`

Now we should find a common denominator. If we multiply the top and bottom of the tangent term by `sinA`, we get a denominator of `sin A cos A`.

`sin^2 A / (sin A cos A) - 1/(sin A cos A)+(2cos^2A) /(sin A cos A)`

Combine the numerators.

`(sin^2A -1 + 2cos^2A)/(sinAcosA)`

Now we split the `2cosA` term into `cosA + cosA`. I rearranged a little to make the next step more aparent.

`(sin^2A + cos^2A +cos^2A - 1)/(sinAcosA)`

The Pythagorean theorem states that;

`sin^2A + cos^2A = 1`

Making this substitution into the numerator we get.

`(1 +cos^2A - 1)/(sinAcosA)`

The `1`s cancel out leaving;

`cos^2A/(sinAcosA)`

The bottom `cosA` cancels one from the top, leaving;

`cosA/sinA`

This is the identity for;

`cotA`