==> $\frac{{{{(1 - \cos 2A)}^2}}}{{4a}} + \frac{{{{(1 + \cos 2A)}^2}}}{{4b}} = \frac{1}{{a + b}}$
==> $b(a + b)(1 - 2\cos 2A + {\cos ^2}2A)$
$ + a(a + b)(1 + 2\cos 2A + {\cos ^2}2A) = 4ab$
==>$\{ b(a + b) + a(a + b)\} {\cos ^2}2A + 2(a + b)(a - b)\cos 2A$
$ + a(a + b) + b(a + b) - 4ab = 0$
==> ${(a + b)^2}{\cos ^2}2A + 2(a + b)(a - b)\cos 2A + {(a - b)^2} = 0$
==> ${\{ (a + b)\cos 2A + (a - b)\} ^2} = 0$or $\cos 2A = \frac{{b - a}}{{b + a}}$
Hence, $\frac{{{{\sin }^8}A}}{{{a^3}}} + \frac{{{{\cos }^8}A}}{{{b^3}}} = \frac{{{{(1 - \cos 2A)}^4}}}{{16{a^3}}} + \frac{{{{(1 + \cos 2A)}^4}}}{{16{b^3}}}$
$ = \frac{1}{{16{a^3}}}{\left[ {1 - \frac{{b - a}}{{b + a}}} \right]^4} + \frac{1}{{16{b^3}}}{\left[ {1 + \frac{{b - a}}{{b + a}}} \right]^4}$
$ = \frac{{16{a^4}}}{{16{a^3}{{(b + a)}^4}}} + \frac{{16{b^4}}}{{16{b^3}{{(b + a)}^4}}}$
$ = \frac{1}{{{{(b + a)}^4}}}(a + b) = \frac{1}{{{{(a + b)}^3}}}$
Trick : Put $A = {90^o}$, then
$\frac{{{{\sin }^4}A}}{a} + \frac{{{{\cos }^4}A}}{b} = \frac{1}{{a + b}}$==>$\frac{1}{a} = \frac{1}{{a + b}} \Rightarrow b = 0$
$\therefore \,\,\,\,\frac{{{{\sin }^8}A}}{{{a^3}}} + \frac{{{{\cos }^8}A}}{{{b^3}}} = \frac{1}{{{a^3}}}$
which is given by option $(a)$
Note : Students can check this question for other values of $A$ also.
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Another ellipse $E _2$ passing through the point $(0,4)$ circumscribes the rectangle $R$.. The eccentricity of the ellipse $E _2$ is