Question

If $co{s}^2\theta +2si{n}^2\theta +3co{s}^2\theta +4si{n}^2\theta +......\left(200\right)terms=10025$ , where $\theta$ is an acute angle , then the value of $sin\theta -cos\theta$ is

• A. $\frac{1-\sqrt{3}}{2}$
• B. $\frac{1+\sqrt{3}}{2}$
• C. $\frac{\sqrt{3}-1}{2}$
• D. $0$

Answer 1

The correct option is A $\frac{1-\sqrt{3}}{2}$

Given,

${\cos }^2\theta +2{\sin }^2\theta +3{\cos }^2\theta +4{\sin }^2\theta +.......+\left(200\right)$ terms

$=({\cos }^2\theta +3{\cos }^2\theta +......100terms)+(2{\sin }^2\theta +4{\sin }^2\theta +.......100terms)$

$=\frac{100}{2}[{\cos }^2\theta +(100-1\left)2{\cos }^2\theta \right]+\frac{100}{2}[2{\sin }^2\theta +(100-1\left)2{\sin }^2\theta \right]$ .......... $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$=50[2{\cos }^2\theta +198{\cos }^2\theta ]+50[4{\sin }^2\theta +198{\sin }^2\theta ]$

$⟹50[2{\cos }^2\theta +198{\cos }^2\theta +4{\sin }^2\theta +198{\sin }^2\theta ]=10025$

$⟹198({\cos }^2\theta +{\sin }^2\theta )+2({\cos }^2\theta +{\sin }^2\theta )+2{\sin }^2\theta =\frac{401}{2}$

$⟹198+2+2{\sin }^2\theta =\frac{401}{2}$

$⟹2{\sin }^2\theta =\frac{1}{2}$

$⟹{\sin }^2\theta =\frac{1}{4}$

$⟹\sin \theta =\pm \frac{1}{2}$

$⟹\therefore \theta ={30}^{\circ }$

given,

$\sin \theta -\cos \theta$

$=\sin 30-\cos 30$

$=\frac{1}{2}-\frac{\sqrt{3}}{2}$

$=\frac{1-\sqrt{3}}{2}$