If $A+B+C+D=2 \pi$,then $\cos A-\cos B+\cos C-\cos D=$

  • A
    $-4 \sin \frac{A+B}{2} \cos \frac{A+C}{2} \sin \frac{A+D}{2}$
  • B
    $4 \sin \frac{A+B}{2} \sin \frac{A+C}{2} \sin \frac{A+D}{2}$
  • C
    $-4 \sin \frac{A+B}{2} \sin \frac{A+C}{2} \sin \frac{A+D}{2}$
  • D
    $4 \sin \frac{A+B}{2} \cos \frac{A+C}{2} \sin \frac{A+D}{2}$

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The number of solutions of the equation $2 \cos(e^x) = 5^x + 5^{-x}$ is:

Consider the following two statements.
Statement $p$: The value of $\sin 120^\circ$ can be derived by taking $\theta = 240^\circ$ in the equation $2\sin \frac{\theta}{2} = \sqrt{1 + \sin \theta} - \sqrt{1 - \sin \theta}$.
Statement $q$: The angles $A, B, C$ and $D$ of any quadrilateral $ABCD$ satisfy the equation $\cos \left( \frac{1}{2}(A + C) \right) + \cos \left( \frac{1}{2}(B + D) \right) = 0$.
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