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Trig help!!!?
I do not understand how to solve this when there is more than one trigonometric ratio. I know that sec is positive in quadrants 1&4 and negative in 2&3. Csc is positive in 1&2 and negative in 3&4
Sec t=square root3 csc t
1. Answer must be in 0<t,2pi
2. Then solve symbolically
TIA!
2 Answers
- Some BodyLv 72 years agoFavorite Answer
sec t = √3 csc t
I recommend rewriting in terms of sine and cosine:
1 / cos t = √3 / sin t
Cross multiply:
sin t = √3 cos t
Divide by cosine:
tan t = √3
Solve:
t = π/3 + kπ
Between 0 and 2π, t = π/3 or 4π/3.
- ?Lv 72 years ago
sec t = √3 csc t
// Square both sides to get rid of radical
[sec t]² = [√3 csc t]²
sec² t = 3 csc² t
...1.............3
------- = --------
cos² t....sin² t
// Substitute the trig identity sin² t = (1 - cos²t)
// and cross multiply
...1.............3
------- = -------------
cos² t....(1 - cos² t)
3 cos² t = (1 - cos² t)
// Combine all the cos² t terms
3 cos² t + cos² t = 1
4 cos² t = 1
cos² t = 1/4
// Take the square root
cos t = ±½
When cos t = +½,
t = cos⁻¹ (½)
t = 60° = π/3
When cos t = -½,
t = cos⁻¹ (½)
t = 120° = 2π/3
// So your 2 solutions are
t = 60° = π/3 and t = 120° = 2π/3................ANS
