MCQ
If ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \frac{{2\pi }}{3},$ then ${\cos ^{ - 1}}x + {\cos ^{ - 1}}y = $
- A$\frac{{2\pi }}{3}$
- ✓$\frac{\pi }{3}$
- C$\frac{\pi }{6}$
- D$\pi $
$ \Rightarrow \frac{\pi }{2} - {\cos ^{ - 1}}x + \frac{\pi }{2} - {\cos ^{ - 1}}y = \frac{{2\pi }}{3}$
==> ${\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \pi - \frac{{2\pi }}{3} = \frac{\pi }{3}$.
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$\int_0^1 {(1 + {{\cos }^8}x)(a{x^2} + bx + c)\,dx} = \int_0^2 {(1 + {{\cos }^8}x)(a{x^2} + bx + c)\,dx} $
Then the quadratic equation $a{x^2} + bx + c = 0$ has