<s>[INST] 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.
 [/INST] 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
$\implies {\cos ^2 \alpha-\sin^2 \alpha}=\large \frac{1-r^2}{1+r^2}$
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