If 5sin2A + 3cos2A = 4, 0° < A < 90°
find the value of sin2θ + sec2θ + cotθ
Answer & explanation
Correct answer: option 1
5sin2A + 3cos2A = 4
= 2sin2A + 3sin2A + 3cos2A = 4
= 2sin2A + 1×3 = 4
= 2sin2A = 1
sinA = \(\frac{1}{\sqrt {2}}\) = 45°
⇒ (\(\frac{1}{\sqrt {2}}\))2+(\(\sqrt {2}\))2+1
= \(\frac{1}{2}\)+2+1
= \(\frac{7}{2}\)