Question
If 2cos²A + 3sin²A = 13/5, then find the value of (sec²A - 1)
Solution
2cos²A + 3sin²A = 13/5
Or, -1cos²A + 3(cos²A + sin²A) = 13/5
Or, -cos²A + 3 = 13/5
Or, -cos²A = 13/5 - 3 = -2/5 => cos²A = 2/5
So, sec²A = 1 / (2/5) = 5/2
Required value = 5/2 - 1 = 3/2
More Trigonometry Questions
- Find the maximum value of 25 sin A + 11 cos A.
- If x = 4 cos A + 5 sin A snd y = 4 sin A – 5 cos A, then the value of x2+y2 is :
- If 17sin A = 8, where 0
- Question 4
- If √3 cosec 2x = 2, then the value of x:
- If (cos A - sin A) = √2 cos (90° - A), then find the value of cot A.
- If sin α + cos β = 2 (0˚ ≤ β < α ≤ 90˚), then sin[(2α + β)/3] =?
- if x sin 45 ˚ = y cosec 30 ˚ then find the value of
- What is the value of [(sin x + sin y) (sin x – sin y)]/[(cosx + cosy) (cosy – cosx)]?
- If sec A + tan A = 4 and sec A - tan A = (1/4), find the value of cos A.
Relevant for Exams:
