If operatorname{cosec} theta + cot theta = fr | vedclass

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(B) Given $\operatorname{cosec} 	heta + \cot 	heta = \frac{1}{3}$.Since $\operatorname{cosec} 	heta - \cot 	heta = \frac{1}{\operatorname{cosec} \

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$2^{	ext{nd}}$ quadrant

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(B) Given $\operatorname{cosec} 	heta + \cot 	heta = \frac{1}{3}$.Since $\operatorname{cosec} 	heta - \cot 	heta = \frac{1}{\operatorname{cosec} 	heta + \cot 	heta}$,we have $\operatorname{cosec} 	heta - \cot 	heta = 3$.Adding the two equations: $(\operatorname{cosec} 	heta + \cot 	heta) + (\operatorname{cosec} 	heta - \cot 	heta) = \frac{1}{3} + 3 = \frac{10}{3}$.$2 \operatorname{cosec} 	heta = \frac{10}{3}$ $\Rightarrow \operatorname{cosec} 	heta = \frac{5}{3}$ $\Rightarrow \sin 	heta = \frac{3}{5}$.Subtracting the equations: $(\operatorname{cosec} 	heta + \cot 	heta) - (\operatorname{cosec} 	heta - \cot 	heta) = \frac{1}{3} - 3 = -\frac{8}{3}$.$2 \cot 	heta = -\frac{8}{3}$ $\Rightarrow \cot 	heta = -\frac{4}{3}$ $\Rightarrow \cos 	heta = -\frac{4}{5}$.Since $\sin 	heta > 0$ and $\cos 	heta < 0$,$	heta$ lies in the $2^{	ext{nd}}$ quadrant.

If $\operatorname{cosec} \theta + \cot \theta = \frac{1}{3}$,then $\theta$ lies in the

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