If θ is an acute angle such that sec²θ = 3, then the value of tan²θ-cosec²θ/tan²θ+cosec²θ is
A. 4/7
B. 3/7
C. 2/7
D. 1/7
Answers
-1 or -1/2 depending on the brackets
Step-by-step explanation:
Given: sec²θ = 3 where θ is an acute angle
sec θ = √3
sin θ = 1/√3
tan²θ - cosec²θ / tan²θ + cosec²θ = sin²θ/Cos²θ - (1/cos²θ / sin²θ/cos²θ) + 1 /cos²θ
= sin²θ/Cos²θ - 1/sin²θ + 1 /cos²θ
= sin²θ/(1-sin²θ) - 1/sin²θ + 1/(1-sin²θ)
= 1/3 / (1 - 1/3) - 1/(1/3) + 1/(1-1/3)
= (1/3 / 2/3) - 3 + 1/(2/3)
= 1/2 - 3 +3/2
= (1 - 6 +3)/ 2
= -1
(tan²θ - cosec²θ) / (tan²θ + cosec²θ) = (sin²θ/Cos²θ - 1/cos²θ) / (sin²θ/cos²θ + 1 /cos²θ)
= (sin²θ-1/Cos²θ) / (sin²θ + 1) /cos²θ
= (sin²θ - 1) / (sin²θ + 1)
= 1/3 - 1 / 1/3 +1
= -2/3 / 4/3
= -2/4
= -1/2
Note: I have arranged the question in two possible format and given the solution for each. None of the choices match the answer. So please post correct format of the question with proper brackets.