# Prove the following identity

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

Last updated date: 29th Mar 2023

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Answer

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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

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

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