Question

$sin^{-1}\bigg(sin\frac{2π}{3}\bigg)+cos^{-1}\bigg(cos\frac{7π}{6}\bigg)+tan^{-1}\bigg(tan\frac{3π}{4}\bigg)$ is equal to:

• A. $\frac{11π}{12}$
• B. $\frac{17π}{12}$
• C. $\frac{31π}{12}$
• D. $-\frac{3π}{4}$

Answer 1

Answer: (A)

$\frac{11π}{12}$

Answer 2

$sin^{-1}\bigg(\frac{\sqrt3}{2}\bigg)+cos^{-1}\bigg(\frac{-\sqrt3}{2}\bigg)+tan^{-1}(-1)$

= $\frac{π}{3}+\frac{5π}{6}-\frac{π}{4}$

= $\frac{4π+10π-3π}{12}$

=$\frac{11π}{12}$

Hence, the correct option is (A): $\frac{11π}{12}$