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.

4

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