SSC CHSL Model Paper 15

© examsiri.com

Question : 55 of 100

Marks: +1, -0

The value of

\sec^2 28° - \cot^2 62°

+ \sin^2 60° + \cosec^2 30°

is equal to :

$7/2$
3
$19/4$
$23/4$

Solution:

\sec^2 28° - \cot^2 62°

+ \sin^2 60° + \cosec^2 30°

=\sec^2(90°-62°)-\cos^2 62°

+3/4+4

[ ∵ sec (90° - θ) = cosecθ]

=1+3/4+4=23/4

© examsiri.com

Go to Question:

Prev Question

Next Question