If
$$a\sin^2\theta-b\cos^2\theta=a-b$$
then prove
$$a\cos^4 \theta +b\sin^4 \theta=\frac{ab}{a+b}$$
I have tried some ways to solve the answer but didn't succeed. Such as:
$$\begin{align}
a\cos^4 \theta +b\sin^4 \theta &= b(\cos^4\theta+\sin^4\theta)+(a-b)\cos^4\theta \\
&=b(1-2\cos^2\theta\sin^2\theta)+(a\sin^2\theta-b\cos^2\theta)\cos^4\theta
\end{align}$$
and here I'm stuck because I'll have $b\cos^6\theta$ and somethings like these.
Or this way:
$$\begin{align}
a\sin^2 \theta -b\cos^2 \theta &= a-b \\
\Rightarrow\qquad a\sin^2\theta-b(1-\sin^2\theta)&=a-b \\
\Rightarrow\qquad a\sin^2\theta + \sin^2\theta &=a-b \\
\Rightarrow\qquad \sin\theta(a+1)&=a
\end{align}$$
But when I want to use $\sin^2\theta=1-\cos^2\theta$ in my main proof, I can't because there is $\sin^4$ and $\cos^4$, and it'll be a long polynomial. I don't find a way to the answer and the proof.
Best Answer
We are given $$a\sin^2\theta-b\cos^2\theta=a-b$$
Note that the given equation implies $$\begin{align} a(1-\cos^2\theta)-b(1-\sin^2\theta)&=a-b \tag{1}\\[4pt] b\sin^2\theta-a\cos^2\theta&=0 \tag{2} \\[4pt] \tan^2\theta&=\frac{a}{b} \tag{3} \\[4pt] \tan\theta&=\sqrt{\frac{a}{b}} \tag{4} \end{align}$$
We can construct a triangle and deduce: $$\cos\theta=\frac{\sqrt{b}}{\sqrt{a+b}} \qquad \sin\theta=\frac{\sqrt{a}}{\sqrt{a+b}} \tag{5}$$
Then, we can sub the result above into $a\cos^4\theta+b\sin^4\theta$ and obtain the desired result
$$\begin{align} a\cos^4\theta+b\sin^4\theta &= a\left(\frac{\sqrt{b}}{\sqrt{a+b}}\right)^4+b\left(\frac{\sqrt{a}}{\sqrt{a+b}}\right)^4 \tag{6}\\[4pt] &=ab\left(\frac{b}{(a+b)^2}+\frac{a}{(a+b)^2}\right) \tag{7}\\[4pt] &=ab\cdot \frac{1}{a+b} \tag{8}\\[4pt] &=\frac{ab}{a+b} \tag{9} \end{align}$$