Tan²Ф+sec²Ф=3

can someone solve this by finding all answers as the nearest whole degree

Recall that

tan²(x) + 1 = sec²(x)

for all x ≠ (2n + 1) 90°, where n is any integer. (That is, whenever x isn't an angle that is an odd multiple of 90°.)

Then

tan²(Φ) + sec²(Φ) = 3

tan²(Φ) + (tan²(Φ) + 1) = 3

2 tan²(Φ) + 1 = 3

2 tan²(Φ) = 2

tan²(Φ) = 1

tan(Φ) = ±√1

tan(Φ) = 1   or   tan(Φ) = -1

Φ = tan⁻¹(1) + n 180°   or   Φ = tan⁻¹(-1) + n 180°

Φ = 45° + n 180°   or   Φ = -45° + n 180°

The first family of solution is the set of angles

{…, -315°, -135°, 45°, 225°, 405°, …}

and the second family is the set

{…, -405°, -225°, -45°, 135°, 315°, …}

(showing the solutions for n = -2 to n = 2)

You can condense the solution set into one family by noticing each angle is 45° plus some multiple of 90°, so that

Φ = 45° + n 90°

If you're looking for solutions in a given range, such as 0 ≤ Φ < 360°, then

Φ = 45°, 135°, 225° or 315°

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.

think about it deeply and passionately about this property

i am sorry