Question

If $\frac{si{n}^4\theta }{a}+\frac{co{s}^4\theta }{b}=\frac{1}{a+b},then$

• A. $co{s}^2\theta =\frac{b}{a+b}$
• B. $si{n}^2\theta =\frac{a}{a+b}$
• C. ${\cos }^6\theta +{\sin }^6\theta =\frac{a^2-ab+b^2}{a^2+2ab+b^2}$
• D. $\frac{co{s}^8\theta }{b^3}+\frac{si{n}^8\theta }{a^3}=\frac{1}{(a+b{)}^3}$

Answer 1

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

The correct options are
A $co{s}^2\theta =\frac{b}{a+b}$
B $si{n}^2\theta =\frac{a}{a+b}$
C ${\cos }^6\theta +{\sin }^6\theta =\frac{a^2-ab+b^2}{a^2+2ab+b^2}$
D $\frac{co{s}^8\theta }{b^3}+\frac{si{n}^8\theta }{a^3}=\frac{1}{(a+b{)}^3}$

Let $si{n}^2\theta =t$
$\therefore \frac{si{n}^4\theta }{a}+\frac{co{s}^4\theta }{b}=\frac{1}{a+b}$
$\frac{t^2}{a}+\frac{(1-t{)}^2}{b}=\frac{1}{a+b}$
$\frac{t^2}{a}+\frac{t^2-2t+1}{b}=\frac{1}{a+b}$

$b{t}^2+a{t}^2-2at+a=\frac{ab}{a+b}$
$(a+b)t^2-2at+\frac{a^2}{a+b}=0$

$t=\frac{2a\pm \sqrt{4a^2-4a^2}}{2(a+b)}$
$\therefore si{n}^2\theta =t=\frac{a}{a+b}$ ....................(1)
$\therefore co{s}^2\theta =\frac{b}{a+b}$ ......................(2)
Now using (1) and (2) we can also deduce the result for option C and option D.