Question

Prove the following statement: $\dfrac{\sin A}{1+\cos A}+\dfrac{1+\cos A}{\sin A}=2\operatorname{cosec}A$ .

Answer 1

We proceed from the left hand side of the statement and add the trigonometric fractions. We use trigonometric identity of sine and cosine ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ and then take 2 common to cancel out $\left( 1+\cos A \right)$ from the numerator and denominator. We use the reciprocal relation between sine and cosecant $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$ to arrive at the right hand side.

Complete step-by-step answer:
We are given the following trigonometric equation as a statement to prove.
$\dfrac{\sin A}{1+\cos A}+\dfrac{1+\cos A}{\sin A}=2\operatorname{cosec}A$
We assume that all the trigonometric angles and fractions are well defined. Let us proceed from the left hand side of the statement and add the trigonometric fractions given in sine and cosines. We have;

& \Rightarrow \dfrac{\sin A}{1+\cos A}+\dfrac{1+\cos A}{\sin A} \\\ & \Rightarrow \dfrac{{{\sin }^{2}}A+{{\left( 1+\cos A \right)}^{2}}}{\left( 1+\cos A \right)\sin A} \\\ \end{aligned}$$ We use the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ for $a=1,b=\cos A$ in the numerator of fraction in the above step to have; $$\begin{aligned} & \Rightarrow \dfrac{{{\sin }^{2}}A+1+2\cos A+{{\cos }^{2}}A}{\left( 1+\cos A \right)\sin A} \\\ & \Rightarrow \dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A+1+2\cos A}{\left( 1+\cos A \right)\sin A} \\\ \end{aligned}$$ We use the Pythagorean trigonometric identity of sine and cosine ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ for $\theta =A$ in the above step to have; $$\begin{aligned} & \Rightarrow \dfrac{1+1+2\cos A}{\left( 1+\cos A \right)\sin A} \\\ & \Rightarrow \dfrac{2+2\cos A}{\left( 1+\cos A \right)\sin A} \\\ \end{aligned}$$ We take 2 common in the numerator of fraction in the above step to have; $$\Rightarrow \dfrac{2\left( 1+\cos A \right)}{\left( 1+\cos A \right)\sin A}$$ We divide $1+\cos A$ in the numerator and denominator of the fraction in the above step to have; $$\Rightarrow \dfrac{2}{\sin A}$$ We use the reciprocal relation between sine and cosecant $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$ in the above step t have; $$\Rightarrow 2\times \dfrac{1}{\sin A}=2\operatorname{cosec}A$$ The above obtained expression is the expression in the right hand side of the statement equation. Hence the statement is proved. $$$$ **Note:** We note that in the question $\operatorname{cosec}A$ is well-defined which means $A$ cannot have measure equal to integral multiple of $\pi $ that is $A\ne n\pi $ for any arbitrary integer $n$ since cosecant is not defined for any angle $A=n\pi $. We also note that the question presumes $\cos A+1\ne 0$ which if we solve to get that $A$ cannot be an odd integral multiple of $\pi $ that is $A\ne \left( 2n+1 \right)\pi $.