Value of $\cos^2\alpha-\sin^2\alpha$ My problem is from Israel Gelfand's Trigonometry textbook.
Page 48. Exercise 8: b) If $\tan\alpha=r$, write an expression in terms of $r$ that represents the value of $\cos^2\alpha-\sin^2\alpha$.
The attempt at a solution:
Well I solved a) which was similar, question was: a) If $\tan\alpha=2/5$, find the numerical value of $\cos^2\alpha-\sin^2\alpha$. Solution was this, since $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{2}{5}$, then $\cos^2\alpha-\sin^2\alpha=5^2-2^2=21$. b) part is more general version of a) I guess, but I can't seem to find identity that would allow me to solve it. I would appreciate some hints, thank you in advance.
 A: Two Hints:
$$\cos^2 x = \frac{\cos^2x}{1}=\frac{\cos^2x}{\sin^2 x+\cos^2 x}=\frac{1}{\tan^2 x+1}$$
And:
$$\cos^2 x - \sin^2 x=\cos^2 x-\sin^2 x +(\cos^2x +\sin^2x)-1=2\cos^2x-1$$
A: $$\cos^2⁡x-\sin^2x=\frac{\cos^2x-\sin^2x}{\cos^2x+\sin^2x}=\frac{1-\tan^2x}{1+\tan^2x}.$$
A: We have
$$\cos^2\alpha-\sin^2\alpha=2\cos^2\alpha-1$$
and by 
$$\cos^2\alpha=\frac1{1+\tan^2\alpha}$$
we find
$$\cos^2\alpha-\sin^2\alpha=\frac2{1+r^2}-1=\frac{1-r^2}{1+r^2}$$
A: We have:
$$\sin \alpha=\pm\frac{\tan \alpha}{\sqrt{1 + \tan^2 \alpha}}\!$$
and
$$\cos \alpha=\pm\frac{1}{\sqrt{1 + \tan^2 \alpha}}\!$$
then
$$\sin^2 \alpha=\frac{\tan^2 \alpha}{1 + \tan^2 \alpha}=\frac{r^2}{1 + r^2}\!$$
and
$$\cos^2 \alpha=\frac{1}{1 + \tan^2 \alpha}\!=\frac{1}{1 + r^2}$$
A: My Approach:
$\tan \alpha=r$
$\implies \tan^2 \alpha=r^2$
$\implies \large \frac{\sin^2 \alpha}{\cos^2 \alpha}=r^2$
Apply Componendo and Dividendo
$\implies \large \frac{\sin^2 \alpha+\cos^2 \alpha}{\sin^2 \alpha-\cos^2 \alpha}=\large \frac{r^2+1}{r^2-1}$
$\implies \large \frac{1}{\sin^2 \alpha - \cos ^2 \alpha}=\large \frac{r^2+1}{r^2-1}$
$\implies {\sin^2 \alpha - \cos ^2 \alpha}=\large \frac{r^2-1}{r^2+1}$
Multiply both sides by -1
$\implies {\cos ^2 \alpha-\sin^2 \alpha}=\large \frac{1-r^2}{1+r^2}$
A: You start from $r=\frac{\sin\alpha}{\cos\alpha}$ and must evaluate $\cos^2\alpha-{\sin^2\alpha}$.
This should push you to set $c=\cos^2\alpha, s=\sin^2\alpha$, use $$c+s=1$$
and rewrite the first relation as
$$r^2c-s=0.$$
Solve this linear system for $c$ and $s$:
$$(1+r^2)\ c=1\\(1+r^2)\ s=r^2.$$
Then$$c-s=\frac{1-r^2}{1+r^2}.$$
