Question

Evaluate 5 cos^2 60° + 4 sec^2 30° - tan^2 45°/ sin^2 30° + cos^2 30° + cos^2 90°

Answer 1

Question : 5 cos^2 60° + 4 sec^2 30° - tan^2 45°/ sin^2 30° + cos^2 30° + cos^2 90°

Solution:
cos 60° = 1/2 => cos^2 60° = 1/4 => 5*cos^2 60° = 5/4

sec 30° = 2/√3 => sec^2 30° = 4/3 => 4*sec^2 30° = 16/3

tan 45° = 1 => tan^2 45° = 1

sin 30° = 1/2 => sin^2 30° = 1/4

cos 30° = √3/2 => cos^2 30° = 3/4

cos 90° = 0 => cos^2 90° = 0


[Q.1] (5 cos^2 60° + 4 sec^2 30° - tan^2 45°)/ (sin^2 30° + cos^2 30° + cos^2 90°)

(5/4 + 16/3 - 1)/ (1/4 + 3/4 + 0)
=> (15+64-12)/12
=> 67/12

[Q.2] (5 cos^2 60° + 4 sec^2 30°) - {tan^2 45°/ (sin^2 30° + cos^2 30° + cos^2 90°)}

(5/4 + 16/3)-{1/(1/4 + 3/4 + 0)}
=> {(15+64)/12} - (1)
=> (79/12)-(1)
=> (79 - 12)/12
=> 67/12

NOTE : We can also substitute sin^2 30° + cos^2 30° = 1 ( Since sin^2 A + cos^2 A = 1)

[Q.3] (5 cos^2 60° + 4 sec^2 30° - tan^2 45°)/ (sin^2 30° + cos^2 30° + cos^2 90°)

(5/4 + 16/3 - 1)/ (1 + 0)
=> (15+64-12)/12
=> 67/12


[Q.4] (5 cos^2 60° + 4 sec^2 30° - tan^2 45°)/ (sin^2 30° + cos^2 30°) + cos^2 90°

(5/4 + 16/3 -1)/(1/4 + 3/4) + 0
=>(5/4 + 16/3 - 1)/ (1 + 0)
=> (15+64-12)/12
=> 67/12