SSC CHSL Model Paper 17

© examsiri.com

Question : 61 of 100

Marks: +1, -0

The value of

\cot^2 62° - \sec^2 28°

+ \cosec^2 30° + \tan^2 60°

is equal to:

8
$10/3$
6
$16/3$

Solution:

\cot^2 62° - \sec^2 28°

+ \cosec^2 30° + \tan^2 60°

=\cot^2 62° - \sec^2 (90°-62°)

+ \cosec^2 30° + \tan^2 60°

=\cot^2 62° - \cosec^2 62°

+ \cosec^2 30° + \tan^2 60°

=-1+2^2+({√3})^2

=- 1 +4 + 3 = 6

© examsiri.com

Go to Question:

Prev Question

Next Question