Question

If $\frac{{\sin }^4\theta }{a}+\frac{{\cos }^4\theta }{b}=\frac{1}{a+b}$, then which one of the following is incorrect?

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

Answer 1

The correct option is B $\frac{{\sin }^4\theta }{b^2}=\frac{{\cos }^4\theta }{a^2}$

We have $\frac{{\sin }^4\theta }{a}+\frac{{\cos }^4\theta }{b}=\frac{1}{a+b}$
$(a+b)(\frac{{\sin }^4\theta }{a}+\frac{{\cos }^4\theta }{b})={({\sin }^2\theta +{\cos }^2\theta )}^2$
$\Rightarrow {\sin }^4\theta +{\cos }^4\theta \frac{b}{a}{\sin }^4\theta +\frac{a}{b}{\cos }^4\theta ={\cos }^4\theta +{\cos }^4\theta +2{\sin }^2\theta {\cos }^2\theta$
$\Rightarrow {(\sqrt{\frac{b}{a}}{\sin }^2\theta -\sqrt{\frac{a}{b}}{\cos }^2\theta )}^2=0$
$\Rightarrow b{\sin }^2\theta =a{\cos }^2\theta$
$\frac{{\sin }^2\theta }{a}=\frac{{\cos }^2\theta }{b}=\frac{{\sin }^2\theta +{\cos }^2\theta }{a+b}$
$\Rightarrow {\sin }^2\theta =\frac{a}{a+b};{\cos }^2\theta =\frac{a}{a+b}$
clearly, these values satisfy options (a),(c),(d)