1. Tardigrade
  2. Question
  3. Mathematics
  4. Sin2 17.5° + Sin2 72.5° is equal to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

KCETKCET 2007Trigonometric Functions

Solution:

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