If sin t + cos t = frac{1}{5},then tan frac{t | vedclass

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(D) We know that $\sin t = \frac{2 	an(t/2)}{1 + 	an^2(t/2)}$ and $\cos t = \frac{1 - 	an^2(t/2)}{1 + 	an^2(t/2)}$.Let $y = 	an(t/2)$. Then the e

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Both $(b)$ and $(c)$

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(D) We know that $\sin t = \frac{2 	an(t/2)}{1 + 	an^2(t/2)}$ and $\cos t = \frac{1 - 	an^2(t/2)}{1 + 	an^2(t/2)}$.Let $y = 	an(t/2)$. Then the equation becomes $\frac{2y}{1+y^2} + \frac{1-y^2}{1+y^2} = \frac{1}{5}$.Multiplying by $5(1+y^2)$,we get $5(2y + 1 - y^2) = 1 + y^2$.$10y + 5 - 5y^2 = 1 + y^2$.$6y^2 - 10y - 4 = 0$.Dividing by $2$,we get $3y^2 - 5y - 2 = 0$.Factoring the quadratic equation: $(3y + 1)(y - 2) = 0$.Thus,$y = 2$ or $y = -\frac{1}{3}$.

If $\sin t + \cos t = \frac{1}{5}$,then $\tan \frac{t}{2}$ is equal to:

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