# Finding the value of $\frac{\cos^4\beta}{\cos^2\alpha} + \frac{\sin^4\beta}{\sin^2\alpha}$.

Trigonometry

$\dfrac{\cos^4 \alpha}{\cos^2 \beta}+ \dfrac{\sin^4\alpha}{\sin^2\beta} = 1$

then the value of

$\dfrac{\cos^4\beta}{\cos^2\alpha}+ \dfrac{\sin^4\beta}{\sin^2\alpha}$ is?

NOTE: can somebody help me $\cos^2\alpha \left(\frac{\cos^2 \alpha}{\cos^2 \beta}\right)+ \sin^2\alpha \left(\frac{\sin^2 \alpha}{\sin^2\beta}\right)$

• Use $\cos$ (instead of $cos$) for $\cos$. Similarly for $\sin$. EX: $\sin^2(\alpha)+\cos^2(\alpha)=1$ $\to$ $\sin^2(\alpha)+\cos^2(\alpha)=1$ Aug 31 '14 at 3:58
• Aug 31 '14 at 4:08

Let $t=\sin^2(\alpha)$ and $s=\sin^2(\beta)$. Then, multiplying both sides of the given identity by $s(1-s)$ gives: $$(1-t)^2s+t^2(1-s)=s(1-s).$$ Bringing the RHS over to the LHS simplifies to $(s-t)^2=0$ so $s=t$. Now, the expression you want to evaluate is just $$\frac{(1-s)^2}{1-s}+\frac{s^2}{s}=1-s+s=1.$$
• $t$ and $s$ are just notational convenience. The rest is just simple manipulations. Keep in mind $\cos^2x+\sin^2x=1$ for any angle $x$. Aug 31 '14 at 4:46
$\cos^2 \alpha \left(\frac{\cos^2\alpha}{\cos^2 \beta}\right) + \sin^2 \alpha \left(\frac{\sin^2 \alpha}{\sin^2 \beta}\right)=1$.
It can be noticed that $\frac{\cos^2\alpha}{\cos^2 \beta}$ and $\frac{\sin^2 \alpha}{\sin^2 \beta}$ must be $1$ (Why?).
This tells us that $\cos^2 \alpha = \cos^2 \beta$ and $\sin^2 \alpha = \sin^2\beta$.