Question

Prove the following identity
$\dfrac{1-\cos x}{1+\cos x}={{\left( \csc x-\cot x \right)}^{2}}$

Answer 1

Hint: Multiply numerator and denominator of LHS by 1-cosx and use the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$. Use Trigonometric identities $1-{{\cos }^{2}}x={{\sin }^{2}}x$ and $\csc x=\dfrac{1}{\sin x},\cot x=\dfrac{\cos x}{\sin x}$. Alternatively, simplify RHS by using the identity ${{\csc }^{2}}x-{{\cot }^{2}}x=1$
and then using $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$. Alternatively, you can simplify both LHS and RHS using half-angle formulae and find the relation between LHS and RHS. The half-angle formulae to be used here are $1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}$, $1+\cos x=2{{\cos }^{2}}\dfrac{x}{2}$ and $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$

“Complete step-by-step answer:”
LHS $=\dfrac{1-\cos x}{1+\cos x}$
Multiplying numerator and denominator by 1-cosx, we get
LHS $=\dfrac{1-\cos x}{1+\cos x}\times \dfrac{1-\cos x}{1-\cos x}=\dfrac{{{\left( 1-\cos x \right)}^{2}}}{\left( 1+\cos x \right)\left( 1-\cos x \right)}$
We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
Put a = 1 and b = cosx we get\[\left( 1+\cos x \right)\left( 1-\cos x \right)=1-{{\cos }^{2}}x\]
We know that $1-{{\cos }^{2}}x={{\sin }^{2}}x$
Hence we have \[\left( 1+\cos x \right)\left( 1-\cos x \right)={{\sin }^{2}}x\]
Hence we have
LHS $=\dfrac{{{\left( 1-\cos x \right)}^{2}}}{{{\sin }^{2}}x}=\left( \dfrac{1-\cos x}{\sin x} \right)$
We know that $\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$
Using the above identity, we get
LHS $={{\left( \dfrac{1}{\sin x}-\dfrac{\cos x}{\sin x} \right)}^{2}}$
We know $\csc x=\dfrac{1}{\sin x},\cot x=\dfrac{\cos x}{\sin x}$
Using the above identities, we get
LHS $={{\left( \csc x-\cot x \right)}^{2}}=$ RHS

Note: Alternate solution [1]
We know that
${{\csc }^{2}}x-{{\cot }^{2}}x=1$
Using $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
$\begin{align}
  & \Rightarrow \left( \csc x-\cot x \right)\left( \csc x+\cot x \right)=1 \\
 & \Rightarrow \csc x-\cot x=\dfrac{1}{\csc x+\cot x} \\
\end{align}$
Multiplying both sides by cosec x - cot x, we get
${{\left( \csc x-\cot x \right)}^{2}}=\dfrac{\csc x-\cot x}{\csc x+\cot x}$
Hence we have
RHS $=\dfrac{\csc x-\cot x}{\csc x+\cot x}$
Multiplying the numerator and denominator by sinx we get
RHS $=\dfrac{\csc x\sin x-\cot x\sin x}{\csc x\sin x+\cot x\sin x}$
Using (cosecx) (sinx) = 1 and (cotx)(sinx) = cosx, we get
RHS $=\dfrac{1-\cos x}{1+\cos x}=$LHS
Hence we have LHS = RHS
Alternate Solution [2]
We know that $1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}$ and $1+\cos x=2{{\cos }^{2}}\dfrac{x}{2}$
Hence we have
\[\dfrac{1-\cos x}{1+\cos x}={{\tan }^{2}}\dfrac{x}{2}\]
Using $\csc x=\dfrac{1}{\sin x},\cot x=\dfrac{\cos x}{\sin x}$ in the expression cosecx -cotx we get
$\csc x-\cot x=\dfrac{1-\cos x}{\sin x}$
We know that $1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}$ and $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$
Using the above identities, we get
$\csc x-\cot x=\dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}=\tan \dfrac{x}{2}$
Hence we have
\[\dfrac{1-\cos x}{1+\cos x}={{\tan }^{2}}\dfrac{x}{2}={{\left( \tan \dfrac{x}{2} \right)}^{2}}={{\left( \csc x-\cot x \right)}^{2}}\]
Hence we have LHS = RHS
Hence proved