Question

How do you prove $\dfrac{1+\cos x}{\sin x}+\dfrac{\sin x}{1+\cos x}=\csc x$ ?

Answer 1

Hint: We begin from left hand side of the given statement by adding the two fractional trigonometric expressions by the working rule to add fractions$\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}$ and then use Pythagorean trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to simplify. We finally use the reciprocal relation between sine and cosine $\csc \theta =\dfrac{1}{\sin \theta }$ to arrive at the right hand side.

Complete step-by-step solution:
We are given the following statement to prove.
\[\dfrac{1+\cos x}{\sin x}+\dfrac{\sin x}{1+\cos x}=\csc x\]
We begin from left hand side using the working rule for adding fraction $\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}$ for $a=1+\cos x=d,b=\sin x=c$ have;
\[\begin{align}
  & \dfrac{1+\cos x}{\sin x}+\dfrac{\sin x}{1+\cos x} \\
 & \Rightarrow \dfrac{\left( 1+\cos x \right)\left( 1+\cos x \right)+\sin x\cdot \sin x}{\sin x\left( 1+\cos x \right)} \\
 & \Rightarrow \dfrac{{{\left( 1+\cos x \right)}^{2}}+\sin x\cdot \sin x}{\sin x\left( 1+\cos x \right)} \\
\end{align}\]
We use the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ in the numerator of the above step for $a=1,b=\cos x$to have;
\[\begin{align}
  & \Rightarrow \dfrac{1+{{\cos }^{2}}x+2\cdot 1\cdot \cos x+{{\sin }^{2}}x}{\sin x\left( 1+\cos x \right)} \\
 & \Rightarrow \dfrac{1+{{\sin }^{2}}x+{{\cos }^{2}}x+2\cos x}{\sin x\left( 1+\cos x \right)} \\
\end{align}\]
We know from Pythagorean trigonometric identity that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ for all values of $\theta $. We use this identity in numerator the above step for $\theta =x$to have
\[\begin{align}
  & \Rightarrow \dfrac{1+1+2\cos x}{\sin x\left( 1+\cos x \right)} \\
 & \Rightarrow \dfrac{2+2\cos x}{\sin x\left( 1+\cos x \right)} \\
 & \Rightarrow \dfrac{2\left( 1+\cos x \right)}{\sin x\left( 1+\cos x \right)} \\
\end{align}\]
We cancel out the term $1+\cos x$ in the numerator and denominator of the above step to have;
\[\Rightarrow \dfrac{2}{\sin x}\]
We use the reciprocal relation of sine and cosecant $\csc \theta =\dfrac{1}{\sin \theta }$ for $\theta =x$ in the above step to have
 \[\Rightarrow \dfrac{2}{\sin x}=2\csc x\]
The above expression is on the right hand side of the statement of proof. Hence the given statement is proved.
Alternative Method: We shall alternatively solve using cosine and sine double angle formulas. We know that the double angle formula of cosine $1+\cos \theta =2{{\cos }^{2}}\left( \dfrac{\theta }{2} \right)$. We know from the sine double angle formula that $\sin 2\theta =2\sin \theta \cos \theta $. We use this formula in the left hand side of the given expression and then proceed to have;
\[\begin{align}
  & \dfrac{2{{\cos }^{2}}\left( \dfrac{x}{2} \right)}{2\sin \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)}+\dfrac{2\sin \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)}{2{{\cos }^{2}}\left( \dfrac{x}{2} \right)} \\
 & \Rightarrow \dfrac{\cos \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)}{\sin \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)}+\dfrac{\sin \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)}{\cos \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)} \\
\end{align}\]
We cancel out the term $\cos \left( \dfrac{x}{2} \right)$ in the above step to have;
\[\begin{align}
  & \Rightarrow \dfrac{\cos \left( \dfrac{x}{2} \right)}{\sin \left( \dfrac{x}{2} \right)}+\dfrac{\sin \left( \dfrac{x}{2} \right)}{\cos \left( \dfrac{x}{2} \right)} \\
 & \Rightarrow \dfrac{{{\cos }^{2}}\left( \dfrac{x}{2} \right)+{{\sin }^{2}}\left( \dfrac{x}{2} \right)}{\sin \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)} \\
\end{align}\]
We use the Pythagorean trigonometric identity that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ for $\theta =\dfrac{x}{2}$ in the above step to have;
\[\begin{align}
  & \Rightarrow \dfrac{1}{\sin \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)} \\
 & \Rightarrow \dfrac{2}{2\sin \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)} \\
\end{align}\]
We use the sine double angle formula $\sin 2\theta =2\sin \theta \cos \theta $ for $\theta =\dfrac{x}{2}$ in the denominator of above step to have;
\[\begin{align}
  & \Rightarrow \dfrac{2}{\sin \left( 2\times \dfrac{x}{2} \right)} \\
 & \Rightarrow \dfrac{2}{\sin x} \\
 & \Rightarrow 2\times \dfrac{1}{\sin x} \\
\end{align}\]
We use the reciprocal relation of sine and cosecant $\csc \theta =\dfrac{1}{\sin \theta }$ for $\theta =x$ in the above step to have
\[\Rightarrow 2\times \csc x=2\csc x\]
The above expression is on the right hand side of the statement of proof. Hence the given statement is proved.

Note: We note that an equation is called identity when the expression is true for all values of parameters $x$. We see that in the given expression assumes conditions $\sin x\ne 0\Rightarrow x=n\pi $ and $1+\cos x\ne 0\Rightarrow c\ne \left( 2n+1 \right)\pi $ where $n$ is any integer. So the given equation is not an identity. We need to be clear of the confusion between formulas $1+\cos \theta ={{\cos }^{2}}\left( \dfrac{\theta }{2} \right)$ and $1-\cos \theta ={{\sin }^{2}}\left( \dfrac{\theta }{2} \right)$ by remembering that in most formulas negative sign is associated with sine and positive sign associated with cosine.