Prove the given identity
$\displaystyle \frac{Sin\;x}{Csc\;x}\;+\;\frac{Cos\;x}{Sec\;x}\;= \;1$
For some strange reason I get $\displaystyle 2\;=\;1$
$\displaystyle
\frac{{\sin (x)}}
{{\csc (x)}} + \frac{{\cos (x)}}
{{\sec (x)}} = 1
$
$\displaystyle
\frac{{\sin (x)}}
{{(\frac{1}
{{\sin (x)}})}} + \frac{{\cos (x)}}
{{(\frac{1}
{{\cos (x)}})}} = 1
$
$\displaystyle
\sin (x) \times \sin (x) + \cos (x) \times \cos (x) = 1
$
$\displaystyle
\sin ^2 (x) + \cos ^2 (x) = 1
$
$\displaystyle 1 = 1$
If you need to show the last step, that $\displaystyle sin^2(x)+cos^2(x)=1$, then draw a circle, we'll choose it to have a radius of 1, so it is a unit circle. (we can do this since trig functions are proportions, we can choose one side to be a certain length and the other sides will change accordingly.) Then the sine is the opposite over the hypotenuse. The hypotenuse is 1 since it is the unit circle, so the sine is just the opposite side, or the height. And the cosine is the adjacent over the hypotenuse which is 1, so it is just the adjacent, or the length. Then since sine is height, and cosine is length, $\displaystyle sin^2(x) +cos^2(x)$ can be a and b in the pythagorean theorem, which says $\displaystyle a^2+b^2=c^2$. Since c is the hypotenuse, c=1 and $\displaystyle c^2=1$. So $\displaystyle sin^2+cos^2=1$