(Sin^2 ; 17.5^{\circ} + Sin^2 ; 72.5^{\circ}) is equal to | Mathematics | SolveeIt

Question

$Sin^2 \; 17.5^{\circ} + Sin^2 \; 72.5^{\circ}$ is equal to

$tan^2 45^\circ$

$Cos^2 90^\circ$

$Sin^2 \; 45^\circ$

$Cos^ 2 \; 30^\circ$

Answer

$tan^2 45^\circ$

Explanation

Solution

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

Mathematics Question on Trigonometric Functions

Solution