For A = 30°, find the value of : $\frac{-3sin^22A+2sec^2A -tan\frac{3A}{2}}{\frac{1}{3}sin3A}$
Answer & explanation
Correct answer: option 1
$\frac{-3sin^22A+2sec^2A -tan\frac{3A}{2}}{\frac{1}{3}sin3A}$
= $\frac{-3sin^260º+2sec^230º -tan\frac{90º}{2}}{\frac{1}{3}sin90º}$
= \(\frac{- 3× 3/4 + 2×(2/√3)² - 1 }{ 1/3}\)
= \(\frac{- 9/4 + 8/3 - 1 }{ 1/3}\)
= \(\frac{(-27+32-12)/12 }{ 1/3}\)
= \(\frac{-7/12 }{ 1/3}\)
= \(\frac{-7 }{ 4}\)
−3sin22A+2sec2A−tan3A213sin3A
−3sin22A+2sec2A−tan3A213sin3A