Question

lf $\sin \theta +\cos \theta =a,\cos \theta -\sin \theta =b$ and $0<\theta <\frac{7\pi }{2}$, then $\sin \theta (\sin \theta -\cos \theta )+{\sin }^2\theta ({\sin }^2\theta -{\cos }^2\theta )+{\sin }^3\theta ({\sin }^3\theta -{\cos }^3\theta )+...\infty$ is equal to

• A. $\frac{1-ab}{1+ab}$
• B. $\frac{1-a^2}{3-a^2}$
• C. $\frac{1-ab}{1+ab}+\frac{1-a^2}{3-a^2}$
• D. $\frac{1+ab}{1-ab}-\frac{a^2-1}{3-a^2}$

Answer 1

The correct option is C $\frac{1-ab}{1+ab}+\frac{1-a^2}{3-a^2}$

$\sin \theta (\sin \theta -\cos \theta )+{\sin }^2\theta ({\sin }^2\theta -{\cos }^2\theta )+{\sin }^3\theta ({\sin }^3\theta -{\cos }^3\theta )+...\infty$

$=({\sin }^2\theta +{\sin }^4\theta +{\sin }^6\theta +...)-[\sin \theta \cos \theta +(\sin \theta \cos \theta {)}^2+(\sin \theta \cos \theta {)}^3+....]$

$=\frac{{\sin }^2\theta }{1-{\sin }^2\theta }-\frac{\sin \theta \cos \theta }{1-\sin \theta \cos \theta }$

$=\frac{{\sin }^2\theta }{{\cos }^2\theta }-\frac{\sin \theta \cos \theta }{1-\sin \theta \cos \theta }$

$=\frac{2{\sin }^2\theta }{2{\cos }^2\theta }-\frac{2\sin \theta \cos \theta }{2-2\sin \theta \cos \theta }$

$=\frac{1-\cos 2\theta }{1+\cos 2\theta }+\frac{1-1-2\sin \theta \cos \theta }{2+1-1-2\sin \theta \cos \theta }$

$=\frac{1-(\cos \theta -\sin \theta \left)\right(\cos \theta +\sin \theta )}{1+(\cos \theta -\sin \theta \left)\right(\cos \theta +\sin \theta )}+\frac{1-(\sin \theta +\cos \theta {)}^2}{3-(\sin \theta +\cos \theta {)}^2}$

$=\frac{1-ab}{1+ab}+\frac{1-a^2}{3-a^2}$