If $a\sin^2\theta-b\cos^2\theta=a-b$, then prove $a\cos^4\theta+b\sin^4\theta=\frac{ab}{a+b}$

Question

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.

Answer 1

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

$$\begin{align} a\cos^4\theta+b\sin^4\theta &= a\left(\frac{\sqrt{b}}{\sqrt{a+b}}\right)^4+b\left(\frac{\sqrt{a}}{\sqrt{a+b}}\right)^4 \tag{6}\\[4pt] &=ab\left(\frac{b}{(a+b)^2}+\frac{a}{(a+b)^2}\right) \tag{7}\\[4pt] &=ab\cdot \frac{1}{a+b} \tag{8}\\[4pt] &=\frac{ab}{a+b} \tag{9} \end{align}$$

Answer 2

Hint:

$$b(1-\cos^2\theta)=a(1-\sin^2\theta)$$

$$\iff\dfrac a{\sin^2\theta}=\dfrac b{\cos^2\theta}=\dfrac{a+b}{\sin^2\theta+\cos^2\theta}$$

$\sin^2\theta=\dfrac a{a+b}\implies\sin^4\theta=\left(\dfrac a{a+b}\right)^2$

Similarly, $\cos^4\theta=\left(\dfrac b{a+b}\right)^2$

Answer 3

Variation of lab bhattacharjee's answer.

From the given: $$a\sin^2\theta-b\cos^2\theta=a-b \iff b\sin^2 \theta=a\cos^2 \theta \iff \frac{a}{b}=\tan^2\theta.$$ Use: $$\tan^2\theta +1=\frac{1}{\cos^2\theta} \Rightarrow \cos^2\theta =\frac1{1+\tan^2\theta}$$ to get: $$\begin{align}a\cos^4 \theta +b\sin^4 \theta&=a\cos^4\theta+a\cos^2\theta\sin^2\theta=\\ &=a\cos^2\theta=\\ &=a\cdot \frac1{1+\tan^2\theta}=\\ &=a\cdot \frac1{1+\frac ab}=\\ &=\frac{ab}{a+b}. \end{align}$$

Answer 4

You made an error in your second attempt: $$a\sin^2\theta-b(1-\sin^2\theta)=a-b$$ doesn't imply $$a\sin^2\theta + \sin^2\theta =a-b,$$ but $$a\sin^2\theta -b + b\sin^2\theta = a - b,$$ which would then give you $\sin^2\theta = \frac a{a+b}$ and $\cos^2\theta = 1 - \sin^2\theta = \frac b{a+b}$. You can square it and substitute into the identity you want to prove.

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I'll use the same idea, but with substitution $u = \sin^2\theta$.

Then, $\cos^2\theta = 1 - \sin^2\theta = 1- u$ and we want to prove

$$au-b(1-u) = a - b\implies a(1-u)^2+bu^2 = \frac{ab}{a+b}.$$

Solving the first equation for $u$ we get $u = \frac a{a+b}$, so

\begin{align} a(1-u)^2 + bu^2 &= a(1-\frac a{a+b})^2+b(\frac a{a+b})^2\\ &= \frac{ab^2}{(a+b)^2}+ \frac{a^2b}{(a+b)^2}\\ &= \frac{ab(a+b)}{(a+b)^2}\\ &=\frac{ab}{a+b}. \end{align}