For A = 133^circ,2cos frac{A}{2} is equal to | vedclass

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(C) Given $A = 133^\circ$,so $\frac{A}{2} = 66.5^\circ$. Since $45^\circ  \cos \frac{A}{2} > 0$.

Vedclass · Accepted Answer

$\sqrt{1 + \sin A} - \sqrt{1 - \sin A}$

Vedclass · Answer

(C) Given $A = 133^\circ$,so $\frac{A}{2} = 66.5^\circ$. Since $45^\circ  \cos \frac{A}{2} > 0$. We know that $1 + \sin A = \sin^2 \frac{A}{2} + \cos^2 \frac{A}{2} + 2\sin \frac{A}{2}\cos \frac{A}{2} = (\sin \frac{A}{2} + \cos \frac{A}{2})^2$. Thus,$\sqrt{1 + \sin A} = |\sin \frac{A}{2} + \cos \frac{A}{2}| = \sin \frac{A}{2} + \cos \frac{A}{2} \dots (i)$. Similarly,$1 - \sin A = \sin^2 \frac{A}{2} + \cos^2 \frac{A}{2} - 2\sin \frac{A}{2}\cos \frac{A}{2} = (\sin \frac{A}{2} - \cos \frac{A}{2})^2$. Thus,$\sqrt{1 - \sin A} = |\sin \frac{A}{2} - \cos \frac{A}{2}|$. Since $\sin \frac{A}{2} > \cos \frac{A}{2}$,we have $\sqrt{1 - \sin A} = \sin \frac{A}{2} - \cos \frac{A}{2} \dots (ii)$. Subtracting $(ii)$ from $(i)$: $\sqrt{1 + \sin A} - \sqrt{1 - \sin A} = (\sin \frac{A}{2} + \cos \frac{A}{2}) - (\sin \frac{A}{2} - \cos \frac{A}{2}) = 2\cos \frac{A}{2}$.

For $A = 133^\circ$,$2\cos \frac{A}{2}$ is equal to

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