Question

If sin $\theta$ + cos $\theta$ = a, cos $\theta$ - sin $\theta$ = b, then sin $\theta$ (sin $\theta$ - cos $\theta$ ) + $si{n}^2\theta (si{n}^2\theta -co{s}^2\theta )+si{n}^3\theta (si{n}^3\theta -co{s}^3\theta )$+ . . .. . is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$[si{n}^2\theta +si{n}^4\theta +si{n}^6\theta +.....\infty ]-[sin\theta cos\theta +si{n}^2\theta co{s}^2\theta +si{n}^3\theta co{s}^3\theta +.....\infty ]=\frac{si{n}^2\theta }{1-si{n}^2\theta }-\frac{sin\theta cos\theta }{1-sin\theta cos\theta }=\frac{si{n}^2\theta }{co{s}^2\theta }-\frac{sin\theta cos\theta }{1-sin\theta cos\theta }but\frac{a^2-1}{2}=sin\theta cos\theta andab=co{s}^2\theta -si{n}^2\theta \Rightarrow =\frac{1-ab}{1+ab}+\frac{1-a^2}{3-a^2}$