# If cosec A=2 then find the value of 1/tanA +sin A/ 1+cos Aâ€‹

\begin{aligned} &\text { We have, } \operatorname{cosec} A=2\\ &\text { Now, } \sin A=\frac{1}{\operatorname{cosec} A}\\ &\Rightarrow \sin A=\frac{1}{2}\\ &\text { We know that } \sin ^{2} A+\cos ^{2} A=1\\ &\Rightarrow \cos ^{2} A=1-\sin ^{2} A\\ &\Rightarrow \cos A=\sqrt{( 1-\sin ^{2} A)}\\ &\Rightarrow \cos A=\sqrt{\left\{1-\left(\frac{1}{2}\right)^{2}\right\}}\\ &\Rightarrow \cos A=\sqrt{\left( 1-\frac{1}{4}\right)}\\ &\Rightarrow \cos A=\sqrt{\left( \frac{4-1}{4}\right)}\\ &\Rightarrow \cos A=\frac{\sqrt{3}}{2}\\ &\text { We know that tan } A=\frac{\sin A}{\cos A}\\ &\Rightarrow \tan A=\frac{1 / 2}{\sqrt{3} / 2}\\ &\Rightarrow \tan A=\frac{1}{\sqrt{3}} \end{aligned}

$\begin{array}{l} \frac{1}{\tan A}+\frac{\sin A}{1+\cos A}\\ \\ =\frac{1}{1 / \sqrt{3}}+\frac{1 / 2}{1+\sqrt{3} / 2} \\ \\ =\sqrt{3}+\frac{1 / 2}{(2+\sqrt{3}) / 2} \\ \\ =\sqrt{3}+\frac{1}{2+\sqrt{3}} \\ \\ =\frac{2 \sqrt{3}+3+1}{2+\sqrt{3}} \\ \\ =\frac{2 \sqrt{3}+4}{2+\sqrt{3}} \\ \\ =\frac{2 \sqrt{3}+3+1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} \\ \\ =\frac{4 \sqrt{3}-6+8-4 \sqrt{3}}{4-3} \\ \\ =8-6 \\ \\ =2 \end{array}$

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