If sin 3A = cos (A+10°), when 3A is an acute angle, then find value of 2sec \(\frac{6A}{2}\) + 6sin2 3A - \(\frac{3}{2}\) tan2 3A.
Answer & explanation
Correct answer: option 1
If sin A = cos B, then A + B = 90°
Therefore,
⇒ 3A + A + 10° = 90°
⇒ A = 20°
Now, Put value of A :
⇒ 2 sec\(\frac{120°}{2}\) + 6 sin2(60°) - \(\frac{3}{2}\)tan2 60°
⇒ 2 (2) + 6 (\(\frac{3}{4}\)) - \(\frac{3}{2}\)(3)
⇒ 4 + \(\frac{9}{2}\) - \(\frac{9}{2}\) = 4