Question

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

Answer 1

Hint: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:
\[\Rightarrow \dfrac{\left( 1+\cos A \right)+\sin A}{\sin A\left( 1+\cos A \right)}=\dfrac{{{\left( 1+\cos A
\right)}^{2}}+{{\sin }^{2}}A}{\sin A+\sin A\cos A}\]
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.