Dgdrxrt

If
$$asin^2theta-bcos^2theta=a-b$$
then prove
$$acos^4 theta +bsin^4 theta=fracaba+b$$

I have tried some ways to solve the answer but didn't succeed. Such as:

$$beginalign
acos^4 theta +bsin^4 theta &= b(cos^4theta+sin^4theta)+(a-b)cos^4theta \
&=b(1-2cos^2thetasin^2theta)+(asin^2theta-bcos^2theta)cos^4theta
endalign$$

and here I'm stuck because I'll have $bcos^6theta$ and somethings like these.

Or this way:

$$beginalign
asin^2 theta -bcos^2 theta &= a-b \
Rightarrowqquad asin^2theta-b(1-sin^2theta)&=a-b \
Rightarrowqquad asin^2theta + sin^2theta &=a-b \
Rightarrowqquad sintheta(a+1)&=a
endalign$$

But when I want to use $sin^2theta=1-cos^2theta$ 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.

We are given $$asin^2theta-bcos^2theta=a-b$$

Note that the given equation implies
$$beginalign
a(1-cos^2theta)-b(1-sin^2theta)&=a-b tag1\[4pt]
bsin^2theta-acos^2theta&=0 tag2 \[4pt]
tan^2theta&=fracab tag3 \[4pt]
tantheta&=sqrtfracab tag4
endalign$$

We can construct a triangle and deduce:
$$costheta=fracsqrtbsqrta+b qquad
sintheta=fracsqrtasqrta+b tag5$$

Then, we can sub the result above into $acos^4theta+bsin^4theta$ and obtain the desired result

$$beginalign
acos^4theta+bsin^4theta
&= aleft(fracsqrtbsqrta+bright)^4+bleft(fracsqrtasqrta+bright)^4 tag6\[4pt]
&=ableft(fracb(a+b)^2+fraca(a+b)^2right) tag7\[4pt]
&=abcdot frac1a+b tag8\[4pt]
&=fracaba+b tag9
endalign$$

Hint:

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

$$iffdfrac asin^2theta=dfrac bcos^2theta=dfraca+bsin^2theta+cos^2theta$$

$sin^2theta=dfrac aa+bimpliessin^4theta=left(dfrac aa+bright)^2$

Similarly, $cos^4theta=left(dfrac ba+bright)^2$

Variation of lab bhattacharjee's answer.

From the given:
$$asin^2theta-bcos^2theta=a-b iff bsin^2 theta=acos^2 theta iff fracab=tan^2theta.$$
Use:
$$tan^2theta +1=frac1cos^2theta Rightarrow cos^2theta =frac11+tan^2theta$$
to get:
$$beginalignacos^4 theta +bsin^4 theta&=acos^4theta+acos^2thetasin^2theta=\
&=acos^2theta=\
&=acdot frac11+tan^2theta=\
&=acdot frac11+frac ab=\
&=fracaba+b.
endalign$$

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

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

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

$$au-b(1-u) = a - bimplies a(1-u)^2+bu^2 = fracaba+b.$$

Solving the first equation for $u$ we get $u = frac aa+b$, so

beginalign
a(1-u)^2 + bu^2 &= a(1-frac aa+b)^2+b(frac aa+b)^2\
&= fracab^2(a+b)^2+ fraca^2b(a+b)^2\
&= fracab(a+b)(a+b)^2\
&=fracaba+b.
endalign