If 90^circ

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(D) Given $\sin A = \frac{4}{5}$ and $90^\circ < A < 180^\circ$ (Second Quadrant).Since $\sin A$ is positive and $\cos A$ is negative in the second qu

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$2$

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(D) Given $\sin A = \frac{4}{5}$ and $90^\circ Since $\sin A$ is positive and $\cos A$ is negative in the second quadrant,$\cos A = -\sqrt{1 - \sin^2 A} = -\sqrt{1 - (4/5)^2} = -\sqrt{1 - 16/25} = -\sqrt{9/25} = -3/5$.Now,$	an A = \frac{\sin A}{\cos A} = \frac{4/5}{-3/5} = -4/3$.Using the half-angle formula: $	an A = \frac{2 	an(A/2)}{1 - 	an^2(A/2)}$.Let $P = 	an(A/2)$. Then $-4/3 = \frac{2P}{1 - P^2}$.$-4(1 - P^2) = 6P \implies -4 + 4P^2 = 6P \implies 4P^2 - 6P - 4 = 0$.Dividing by $2$: $2P^2 - 3P - 2 = 0$.Factoring the quadratic: $(2P + 1)(P - 2) = 0$.So,$P = -1/2$ or $P = 2$.Since $90^\circ Therefore,$	an(A/2) = 2$.

If $90^\circ < A < 180^\circ$ and $\sin A = \frac{4}{5}$,then $\tan \frac{A}{2}$ is equal to

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