${\sin^2(x)\over \tan^2(x)}$
I did this and then got stuck. Could someone give me some hints please?
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Sign up to join this community${\sin^2(x)\over \tan^2(x)}$
I did this and then got stuck. Could someone give me some hints please?
Good job at arriving at $${1-\cos^2(x)\over \sec^2(x)-1}= {\sin^2(x)\over \tan^2(x)}$$ We know that $\tan x = \dfrac {\sin x}{\cos x}.$ So.... $${\sin^2(x)\over \tan^2(x)} = \frac{\sin^2 x}{\frac{\sin^2 x}{\cos^2 x}}= \cos^2 x = 1-\sin^2 x$$
For the equation: $$\frac{1-\cos^2x}{\sec^2x-1}$$
Multiply the numerator and denominator by $\cos^2x$
We now get: $$\frac{\cos^2x-\cos^4x}{1-\cos^2x}$$
Separate this into two fractions: $$\frac{\cos^2x}{\sin^2x} - \frac{\cos^4x}{\sin^2x}$$
This can then be converted to: $$\cot^2x - \cot^2x\cos^2x$$
We take $\cot^2x$ common, and get: $\cot^2x\sin^2x$
When this is multiplied, this gives us: $\cos^2x$, or rather, $1-\sin^2x$
Hence proved.
\cos, \sin, \cot, \tan...
to format $\cos, \sin, \cot, \tan...$ For example, not the difference between $tan x$
=$ tan x$, versus, $\tan x$ = $\tan x$
$\endgroup$
– amWhy
Nov 29 '16 at 19:35
\ln, \det, \sin, \cos, \gcd
etc, yields $\ln, \det,\sin, \cos, \gcd$, etc.$
$\endgroup$
– amWhy
Nov 29 '16 at 19:50
Notice that $\sec^2(x)=\frac{1}{\cos^2(x)}$ Putting it in our equation on R.H.S. it becomes: $${1-\cos^2(x)\over \sec^2(x)-1}={1-\cos^2(x)\over\frac{1}{\cos^2(x)}-1}$$ $$=\frac{(1-\cos^2)(\cos^2(x))}{1-\cos^2(x)}$$ $$=\cos^2(x)=1-\sin^2(x)$$
we have $$\frac{1}{\sec(x)^2-1}=\frac{\cos(x)^2}{1-\cos(x)^2}$$ and from both we get $$\cos(x)^2=1-\sin(x)^2$$