If theta does not lie in the second quadrant | vedclass

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Question

(B) Given $	an 	heta = \frac{-3}{4}$. Since $	an 	heta < 0$ and $	heta$ is not in the second quadrant,$	heta$ must be in the fourth quadrant.In

Vedclass · Accepted Answer

$\frac{-97}{75}$

Vedclass · Answer

(B) Given $	an 	heta = \frac{-3}{4}$. Since $	an 	heta In the fourth quadrant,$\sin 	heta = \frac{-3}{5}$ and $\cos 	heta = \frac{4}{5}$.Using the half-angle formula,$	an \frac{	heta}{2} = \frac{1 - \cos 	heta}{\sin 	heta} = \frac{1 - 4/5}{-3/5} = \frac{1/5}{-3/5} = \frac{-1}{3}$.Also,$\sin 2 	heta = 2 \sin 	heta \cos 	heta = 2 	imes (\frac{-3}{5}) 	imes (\frac{4}{5}) = \frac{-24}{25}$.Therefore,$	an \frac{	heta}{2} + \sin 2 	heta = \frac{-1}{3} + (\frac{-24}{25}) = \frac{-25 - 72}{75} = \frac{-97}{75}$.

If $\theta$ does not lie in the second quadrant and $\tan \theta = \frac{-3}{4}$,then $\tan \frac{\theta}{2} + \sin 2 \theta =$

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