Proving trigonometrical identity from given identity.

Question

Given $$\dfrac{\sin^4 A}{a}+\dfrac{\cos^4 A}{b}=\dfrac{1}{a+b}$$ prove that:$$\dfrac{\sin^8 A}{a^3}+\dfrac{\cos^8 A}{b^3}=\dfrac{1}{(a+b)^3}$$

Using given, I proved: $b\sin^2 A=a\cos^2 A$ Help me proceeding after this.

Answer 1

You are almost finished. From where you left off, let us say $$b \sin^2 A =a \cos^2 A \iff \frac{\sin^2 A}{a}=\frac{\cos^2 A}{b}=c$$ For some $c$. Thus $$\sin^2 A+\cos^2 A=1 \iff ac+bc=1 \iff c=\frac{1}{a+b} \tag{1}$$ Now, note $$\dfrac{\sin^8 A}{a^3}+\dfrac{\cos^8 A}{b^3}=\frac{a^4c^4}{a^3}+\frac{b^4c^4}{b^3}=ac^4+bc^4=c^4(a+b)=\frac{1}{(a+b)^3}$$ From $(1)$. We are done.

Answer 2

HINT:

Using $\cos^2A=1-\sin^2A,$ set $\sin^2A=u$ to form a Quadratic equation in $u$ to find $u=\dfrac a{a+b}$

OR where you have left of,

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

Answer 3

$\dfrac{\sin^4A}{a}+\dfrac{\left(1-sin^2A\right)^2}{b}-\dfrac{1}{a+b}=0$

$\sin^4A\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{2\sin^2A}{b}+\dfrac{1}{b}-\dfrac{1}{a+b}=0$

$\left(\sin^2A-\dfrac{a}{a+b}\right)^2=0$

$\sin^2A=\dfrac{a}{a+b}$

$\cos^2A=\dfrac{b}{a+b}$

You can take it up from here.