Question

If ${\cos }^2θ+2{\sin }^2θ+3{\cos }^2θ+4{\sin }^2θ+⋯$ (200 terms) $=10025$, where $θ$ is acute, then the value of $\sin θ−\cos θ$ is:

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

Answer 1

Answer: (A)

Given,
${\cos }^2θ+2{\sin }^2θ+3{\cos }^2θ+4{\sin }^2θ+⋯+200\text{ terms }=$
$10025$
$⇒({\cos }^2θ+3{\cos }^2θ+5{\cos }^2θ+⋯100$ $\text{ terms })+({\sin }^2θ.$
$.+2{\sin }^2θ+⋯+100\text{ terms })=10025$
$⇒100^2{\cos }^2θ+(100^2+100){\sin }^2θ=10025$
$⇒10000({\cos }^2θ+{\sin }^2θ)+100{\sin }^2$
$θ=10025$
$⇒100{\sin }^2θ=25$
$⇒{\sin }^2θ=\frac{1}{4}$
$⇒\sin θ=\frac{1}{2}(âˆ\mu θ\text{ is acute })$
$⇒θ=30^{∘}.$
$∴\sin θ−\cos θ=\sin 30^{∘}−\cos 30^{∘}$
$=\frac{1}{2}−\frac{\sqrt{3}}{2}=\frac{1−\sqrt{3}}{2}.$