Question

If 7sin2 θ + 3cos2 θ = 4, then find θ and hence prove that sec θ + cosec θ = 2 +

2

3​

Answer 1

The solution below is given taking A is in 1st quadrant; such that all trigonometric function values are positive.  

i) 7sin²A + 3cos²A = 4sin²A + 3sin²A + 3cos²A = 4sin²A + 3(sin²A + cos²A) = 4sin²A + 3  

[Since by identity, sin²A + cos²A = 1]  

ii) ==> 4sin²A + 3 = 4  

==> Solving sin(A) = 1/2 [Only positive value is taken]  

iii) So cos(A) = √(1 - sin²A) = (√3)/2  

iv) Thus sec(A) + csc(A) = 1/cos(A) + 1/sin(A) = 2/√3 + 2 = 2 + (2/√3) [Proved]

Hope helps...