Question

If $\frac{{\sin }^4\alpha }{a}+\frac{{\cos }^4\alpha }{b}=\frac{1}{a+b}$, then

• A. $\frac{{\sin }^4\alpha }{a^2}=\frac{{\cos }^4\alpha }{b^2}$
• B. $\frac{{\sin }^4\alpha }{b^2}=\frac{{\cos }^4\alpha }{a^2}$
• C. $\frac{{\sin }^8\alpha }{a^3}=\frac{{\cos }^8\alpha }{b^3}=\frac{1}{(a+b{)}^3}$
• D. ${\sin }^4\alpha =\frac{a^2}{(a+b{)}^2}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\frac{{\sin }^4\alpha }{a^2}=\frac{{\cos }^4\alpha }{b^2}$
C $\frac{{\sin }^8\alpha }{a^3}=\frac{{\cos }^8\alpha }{b^3}=\frac{1}{(a+b{)}^3}$
D ${\sin }^4\alpha =\frac{a^2}{(a+b{)}^2}$

$(a+b)[\frac{{\sin }^{4}x}{a}+\frac{{\cos }^{4}x}{b}]=1$
Or
${\sin }^{4}x+{\cos }^{4}x+\frac{b}{a}{\sin }^{4}x+\frac{a}{b}{\cos }^{4}x=1$
Or
$({\sin }^{2}x+{\cos }^{2}x)-2{\sin }^{2}x.{\cos }^{2}x+\frac{b}{a}{\sin }^{4}x+\frac{a}{b}{\cos }^{4}x=1$
$1-2{\sin }^{2}x.{\cos }^{2}x+\frac{b}{a}{\sin }^{4}x+\frac{a}{b}{\cos }^{4}x=1$
Or
$-2{\sin }^{2}x.{\cos }^{2}x+\frac{b}{a}{\sin }^{4}x+\frac{a}{b}{\cos }^{4}x=0$
Or
$(\sqrt{\frac{b}{a}}{\sin }^{2}x-\sqrt{\frac{a}{b}}{\cos }^{2}x{)}^2=0$
Or
$\sqrt{\frac{b}{a}}{\sin }^{2}x-\sqrt{\frac{a}{b}}{\cos }^{2}x=0$
Or
$\sqrt{\frac{b}{a}}{\sin }^{2}x=\sqrt{\frac{a}{b}}{\cos }^{2}x$
Or
$\frac{b}{a}{\sin }^{4}x=\frac{a}{b}{\cos }^{4}x$
Or
$\frac{{\sin }^{4}x}{a^2}=\frac{{\cos }^{4}x}{b^2}=n$
Then $\frac{{\sin }^{4}x}{a}+\frac{{\cos }^{4}x}{b}=\frac{1}{(a+b)}$
$an+bn=\frac{1}{a+b}$
Or
$n=\frac{1}{(a+b{)}^2}$
Hence
$\frac{{\sin }^{4}x}{a^2}=\frac{1}{(a+b{)}^2}=n$
Or
${\sin }^{4}x=a^{2}n$ and ${\cos }^{4}x=b^{2}n$
Then
$\frac{\left({\sin }^4\right(x){)}^2}{a^3}+\frac{\left({\cos }^4\right(x){)}^2}{b^3}$
$=\frac{a^4{n}^2}{a^3}+\frac{b^4.n^2}{b^3}$
$=n^2(a+b)$
$=\frac{(a+b)}{(a+b{)}^4}$
$=\frac{1}{(a+b{)}^3}$.