Q. The value of $sin^{2} 5 1^{\circ} + sin^{2} 3 9^{\circ}$ is

9908 201 KCETKCET 2020Trigonometric Functions Report Error

A

1

39%

B

0

24%

C

$sin 1 2^{\circ}$

20%

D

$cos 1 2^{\circ}$

18%

Solution:

$sin^{2} 5 1^{\circ} + sin^{2} 3 9^{\circ}$
$= cos^{2} 3 9^{\circ} + sin^{2} 3 9^{\circ}$
$= 1$