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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 $\cos^2\alpha \left(\frac{\cos^2 \alpha}{\cos^2 \beta}\right)+ \sin^2\alpha \left(\frac{\sin^2 \alpha}{\sin^2\beta}\right)$

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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 $$ \frac{(1-s)^2}{1-s}+\frac{s^2}{s}=1-s+s=1. $$

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  • $\begingroup$ cant able to follow @kim jong un.It seems hard to me $\endgroup$
    – jyothika1
    Aug 31 '14 at 4:29
  • $\begingroup$ $t$ and $s$ are just notational convenience. The rest is just simple manipulations. Keep in mind $\cos^2x+\sin^2x=1$ for any angle $x$. $\endgroup$ Aug 31 '14 at 4:46
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$\cos^2 \alpha \left(\frac{\cos^2\alpha}{\cos^2 \beta}\right) + \sin^2 \alpha \left(\frac{\sin^2 \alpha}{\sin^2 \beta}\right)=1$.

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 $\sin^2 \alpha = \sin^2\beta$.

Plug those two results in the required expression. It will then become a well-known expression with a well-known value.

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