Question

$Si{n}^{2}17.5^{\circ }+Si{n}^{2}72.5^{\circ }$ is equal to

• A. $ta{n}^{2}45^{\circ }$
• B. $Co{s}^{2}90^{\circ }$
• C. $Si{n}^{2}45^{\circ }$
• D. $Co{s}^{2}30^{\circ }$

Answer 1

Answer: (A)

We have, ${\sin }^{2}17.5^{\circ }+{\sin }^{2}72.5^{\circ }$
$={\sin }^{2}17.5^{\circ }+{\sin }^2(90^{\circ }-17.5^{\circ })$
$={\sin }^{2}17.5^{\circ }+{\cos }^{2}17.5^{\circ }$
$=1=1^2$
$={\tan }^{2}45^{\circ }$