Question

If ${\sin }^2{30}^{\circ }+{\sin }^2{60}^{\circ }+3({\cos }^2{70}^{\circ }+{\cos }^2{20}^{\circ })=4K,thenthevalueofKis$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A 1

Given: ${\sin }^2{30}^{\circ }+{\sin }^2{60}^{\circ }+3({\cos }^2{70}^{\circ }+{\cos }^2{20}^{\circ })=4K$

$\therefore 4K={\sin }^2{30}^{\circ }+{\sin }^2{60}^{\circ }+3({\cos }^2{70}^{\circ }+{\cos }^2{20}^{\circ })$

$={\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2+3({\sin }^2{20}^{\circ }+{\cos }^2{20}^{\circ })$
$(\because \sin ({90}^{\circ }-\theta )=\cos \theta )$

$=\frac{1}{4}+\frac{3}{4}+3\times 1$

$(\because {\sin }^2\theta +{\cos }^2\theta =1)$

= 1 + 3

$\Rightarrow$ 4K = 4

$\Rightarrow$ K = 1

Hence, the correct answer is option (a).