Question

Let S = ${\sin }^2{30}^{\circ }+{\sin }^2{45}^{\circ }+{\sin }^2{60}^{\circ }$ and

P = ${\text{cosec}}^2{45}^{\circ }.{\sec }^2{30}^{\circ }.{\sin }^3{90}^{\circ }.\cos {60}^{\circ }$, then the correct statement is

• A. $S<P$
• B. $S=P$
• C. $S\cdot P=2$
• D. $S+P>3$

Answer 1

The correct option is C $S\cdot P=2$

$S={\sin }^2{30}^{\circ }+{\sin }^2{45}^{\circ }+{\sin }^2{60}^{\circ }$ and $P=cose{c}^2{45}^{\circ }.{\sec }^2{30}^{\circ }.{\sin }^3{90}^{\circ }.\cos {60}^{\circ }$
Now, $S=\frac{1}{4}+\frac{1}{2}+\frac{3}{4}=\frac{1+2+3}{4}=\frac{3}{2}$

$P=(\sqrt{2}{)}^2\cdot {\left(\frac{2}{\sqrt{3}}\right)}^2\cdot 1^3\cdot \frac{1}{2}=\frac{4}{3}$

Hence, $SP=\frac{3}{2}\cdot \frac{4}{3}=2$