Question

Prove $\dfrac{1+\cos A}{\sin A}+\dfrac{\sin A}{1+\cos A}=2\text{cosec}A$

Answer 1

To find the proof, we first need to find the LCM of the LHS and after that we need to find the value in terms of $cos$ and $sin$ to convert into $cosec$ and then equate both the LHS and RHS together to form $2\text{cosec}A$ on both LHS and RHS.

Complete solution step by step:
First let us find the LCM of the LHS by finding the product of the denominator as:
$\Rightarrow \dfrac{1+\cos A}{\sin A}+\dfrac{\sin A}{1+\cos A}$
Forming the LCM as $\dfrac{\left( 1+\cos A \right)+\sin A}{\sin A\left( 1+\cos A \right)}$.
Now finding the fractions, we get the value of the LHS as:

\right)}^{2}}+{{\sin }^{2}}A}{\sin A+\sin A\cos A}$$ Now converting the value of $${{\left( 1+\cos A \right)}^{2}}$$ into [\1+2\cos A+{{\cos }^{2}}A\], we get the numerator as: $$\Rightarrow \dfrac{1+2\cos A+{{\cos }^{2}}A+{{\sin }^{2}}A}{\sin A+\sin A\cos A}$$ Converting the value of $${{\cos }^{2}}A+{{\sin }^{2}}A$$ into 1. $$\Rightarrow \dfrac{1+2\cos A+1}{\sin A+\sin A\cos A}$$ $$\Rightarrow \dfrac{2+2\cos A}{\sin A+\sin A\cos A}$$ Taking the value of $$\sin A$$ common we get the value of the denominator as $$\sin A\left( 1+\cos A \right)$$. We get the rest of the equation as: $$\Rightarrow \dfrac{2\left( 1+\cos A \right)}{\sin A\left( 1+\cos A \right)}=\dfrac{2}{\sin A}$$ $$\Rightarrow \dfrac{2}{\sin A}=2\text{cosec}A$$ **Therefore, the value of the LHS of the equation is $$2\text{cosec}A$$ which is equal to RHS.** Hence, proved. **Note:** Trigonometric values like $${{\cos }^{2}}A+{{\sin }^{2}}A=1$$ and $$\dfrac{1}{\sin A}=\text{cosec}A$$ should be known when solving question as it makes the question solving easier and faster.