Question

write the value of cot^2degree 30+sec^2 45 degree

please answer fast​

Answer 1

ANSWER:

cot²30° + sec²45°

We know that

cot30° = √3

sec45° = √2

= (√3)² + (√2)²

= 3 + 2

= 5

cot²30° + sec²45° = 5

More information:

→sin²θ + cos²θ = 1

sin²θ = 1 - cos²θ

sinθ = ±√1 - cos²θ

cos²θ = 1- sin²θ

cosθ = ±√1 - sin²θ

→ sec²θ - tan²θ = 1

sec²θ = 1 + tan²θ

secθ = ±√1 + tan²θ

tan²θ = sec²θ -1

tanθ = ±√sec²θ - 1

→ cosec²θ - cot²θ= 1

cosec²θ = 1 + cot²θ

cosecθ = ±√1 + cot²θ

cot²θ = cosec²θ -1

cotθ = ±√cosec²θ - 1