Been 20 years since I've identified trig identities. We are stuck on a particular type and would appreciate some explanation. 1/sin(-x) * (1-cos^2x) that's cos(squared x) and 1+1/cos(-x) / -sinx-tanx
Because $\displaystyle \cos^2(x)+\sin^2(x)=1$
This can be derived multiple ways...the most intuitive being
On the unit circle which has a radius of one any of the points on the circle have coordinates
$\displaystyle (\cos(x),\sin(x))$
So the distance from the origin to any point on the circle which I stated earlier is 1
So setting up the distance equation we have
$\displaystyle \sqrt{(0-\cos(x))^2+(0-\sin(x))^2}=1^2$
Simplifying we get
$\displaystyle \sqrt{\cos^2(x)+\sin^2(x)}=1^2\Rightarrow{\cos^2(x )+\sin^2(x)=1}$
Usually in trig classes this indentity is just taken to be true
First, convert everything to sines and cosines using the identities
$\displaystyle \sec x = \frac1{\cos x}$
$\displaystyle \tan x = \frac{\sin x}{\cos x}$
Then, combine the fractions in the numerator and denominator, reduce, factor and cancel, and you should be left with $\displaystyle -\csc x$ if you do it right. When working this one, it may be helpful to observe that, since sine is an odd function and cosine is even,
$\displaystyle \sin\left(-x\right) = -\sin x$
$\displaystyle \cos\left(-x\right) = \cos x$
If you have difficulty with the simplification, let us know.