3 cos260° + 2 cot230° – 5 sin245° = ?

Consider 3 cos260° + 2 cot230° – 5 sin245° = 3 (1/2)2 + 2(√3)2 – 5(1/√2)2= 3(1/4) + 2(3) – 5(1/2)= (3/4) + 6 – (5/2)= (3 + 24 – 10)/4= 17/4

Q29 of 186 Page 350

3 cos²60° + 2 cot²30° – 5 sin²45° = ?

Consider 3 cos²60° + 2 cot²30° – 5 sin²45° = 3 (1/2)² + 2(√3)² – 5(1/√2)²

= 3(1/4) + 2(3) – 5(1/2)

= (3/4) + 6 – (5/2)

= (3 + 24 – 10)/4

= 17/4

(cos 0° + sin 30° + sin 45°) (sin 90° + cos 60° – cos 45°) = ?

sin² 30° + 4 cot² 45° – sec² 60° = ?

cos² 30° cos² 45° + 4 sec² 60° + 2cos² 90° – 2 tan² 60° = ?

If cosec θ = √10 then sec θ = ?