Question

Prove that $\dfrac{{\sin \theta }}{{1 - \cos \theta }} = {\rm{cosec}}\theta + \cot \theta$.

Answer 1

Here, we need to prove the given expression. We will use trigonometric ratios and trigonometric identities to simplify one of the sides of the equation, such that it is equal to the other side of the equation, and hence, prove the given equation.

Formula Used:
We will use the following formulas:
The sum of squares of the sine and cosine of an angle $x$ is equal to 1, that is ${\sin ^2}x + {\cos ^2}x = 1$.
The product of sum and difference of two numbers is given by the algebraic identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$.
The cotangent of an angle $x$ is the ratio of the cosine and sine of the angle $x$, that is $\cot x = \dfrac{{\cos x}}{{\sin x}}$.
The cosecant of an angle $x$ is the reciprocal of the sine of the angle $x$, that is ${\rm{cosec }}x = \dfrac{1}{{\sin x}}$.

Complete step-by-step answer:
We will first simplify the left hand side of the equation.
Multiplying and dividing the left hand side of the equation $\dfrac{{\sin \theta }}{{1 - \cos \theta }} = {\rm{cosec}}\theta + \cot \theta$ by $1 + \cos \theta$, we get
$\Rightarrow L.H.S. = \dfrac{{\sin \theta }}{{1 - \cos \theta }} \times \left( {\dfrac{{1 + \cos \theta }}{{1 + \cos \theta }}} \right)$
Simplifying the expression, we get
$\Rightarrow L.H.S. = \dfrac{{\sin \theta \left( {1 + \cos \theta } \right)}}{{\left( {1 - \cos \theta } \right)\left( {1 + \cos \theta } \right)}}$
The product of sum and difference of two numbers is given by the algebraic identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$.
Using the algebraic identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ to simplify the denominator of the equation, we get
$\begin{array}{l} \Rightarrow L.H.S. = \dfrac{{\sin \theta \left( {1 + \cos \theta } \right)}}{{{1^2} - {{\left( {\cos \theta } \right)}^2}}}\\\ \Rightarrow L.H.S. = \dfrac{{\sin \theta \left( {1 + \cos \theta } \right)}}{{1 - {{\cos }^2}\theta }}\end{array}$
We know that the sum of squares of the sine and cosine of an angle is equal to 1, that is ${\sin ^2}\theta + {\cos ^2}\theta = 1$.
Subtracting ${\cos ^2}\theta$ from both sides of the trigonometric identity, we get
$\Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta$
Substituting $1 - {\cos ^2}\theta = {\sin ^2}\theta$ in the equation $L.H.S. = \dfrac{{\sin \theta \left( {1 + \cos \theta } \right)}}{{1 - {{\cos }^2}\theta }}$, we get
$\Rightarrow L.H.S. = \dfrac{{\sin \theta \left( {1 + \cos \theta } \right)}}{{{{\sin }^2}\theta }}$
Simplifying the expression, we get
$\Rightarrow L.H.S. = \dfrac{{1 + \cos \theta }}{{\sin \theta }}$
Splitting the sum using the L.C.M., we get
$\Rightarrow L.H.S. = \dfrac{1}{{\sin \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }}$
Now, the cotangent of an angle $x$ is the ratio of the cosine and sine of the angle $x$, that is $\cot x = \dfrac{{\cos x}}{{\sin x}}$.
Thus, we get
$\Rightarrow \cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
The cosecant of an angle $x$ is the reciprocal of the sine of the angle $x$, that is ${\rm{cosec }}x = \dfrac{1}{{\sin x}}$.
Thus, we get
$\Rightarrow {\rm{cosec}}\theta = \dfrac{1}{{\sin \theta }}$
Substituting $\dfrac{1}{{\sin \theta }} = {\rm{cosec}}\theta$ and $\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta$ in the equation $L.H.S. = \dfrac{1}{{\sin \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }}$, we get
$\Rightarrow L.H.S. = {\rm{cosec}}\theta + \cot \theta$
We can observe that ${\rm{cosec}}\theta + \cot \theta$ is the right hand side of the given equation $\dfrac{{\sin \theta }}{{1 - \cos \theta }} = {\rm{cosec}}\theta + \cot \theta$.
Thus, we get
$\Rightarrow L.H.S. = R.H.S.$
Hence, we have proved that $\dfrac{{\sin \theta }}{{1 - \cos \theta }} = {\rm{cosec}}\theta + \cot \theta$.

Note: We can also solve the problem by simplifying the right hand side of the equation.
The right hand side of the equation $\dfrac{{\sin \theta }}{{1 - \cos \theta }} = {\rm{cosec}}\theta + \cot \theta$ is ${\rm{cosec}}\theta + \cot \theta$.
Thus, we get
$\Rightarrow R.H.S. = {\rm{cosec}}\theta + \cot \theta$
The cotangent of an angle $x$ is the ratio of the cosine and sine of the angle $x$, that is $\cot x = \dfrac{{\cos x}}{{\sin x}}$.
Thus, we get
$\Rightarrow \cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
The cosecant of an angle $x$ is the reciprocal of the sine of the angle $x$, that is ${\rm{cosec }}x = \dfrac{1}{{\sin x}}$.
Thus, we get
$\Rightarrow {\rm{cosec}}\theta = \dfrac{1}{{\sin \theta }}$
Substituting $\dfrac{1}{{\sin \theta }} = {\rm{cosec}}\theta$ and $\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta$ in the equation $R.H.S. = {\rm{cosec}}\theta + \cot \theta$, we get
$\Rightarrow R.H.S. = \dfrac{1}{{\sin \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }}$
Adding the terms, we get
$\Rightarrow R.H.S. = \dfrac{{1 + \cos \theta }}{{\sin \theta }}$
Multiplying and dividing the equation by $1 - \cos \theta$, we get
$\begin{array}{l} \Rightarrow R.H.S. = \dfrac{{1 + \cos \theta }}{{\sin \theta }} \times \left( {\dfrac{{1 - \cos \theta }}{{1 - \cos \theta }}} \right)\\\ \Rightarrow R.H.S. = \dfrac{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}}{{\sin \theta \left( {1 - \cos \theta } \right)}}\end{array}$
The product of sum and difference of two numbers is given by the algebraic identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$.
Using the algebraic identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ to simplify the numerator of the equation, we get
$\begin{array}{l} \Rightarrow R.H.S. = \dfrac{{{1^2} - {{\left( {\cos \theta } \right)}^2}}}{{\sin \theta \left( {1 - \cos \theta } \right)}}\\\ \Rightarrow R.H.S. = \dfrac{{1 - {{\cos }^2}\theta }}{{\sin \theta \left( {1 - \cos \theta } \right)}}\end{array}$
We know that the sum of squares of the sine and cosine of an angle is equal to 1, that is ${\sin ^2}\theta + {\cos ^2}\theta = 1$.
Subtracting ${\cos ^2}\theta$ from both sides of the trigonometric identity, we get
$\Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta$
Substituting $1 - {\cos ^2}\theta = {\sin ^2}\theta$ in the equation $R.H.S. = \dfrac{{1 - {{\cos }^2}\theta }}{{\sin \theta \left( {1 - \cos \theta } \right)}}$, we get
$\Rightarrow R.H.S. = \dfrac{{{{\sin }^2}\theta }}{{\sin \theta \left( {1 - \cos \theta } \right)}}$
Simplifying the expression, we get
$\Rightarrow R.H.S. = \dfrac{{\sin \theta }}{{1 - \cos \theta }}$
We can observe that $\dfrac{{\sin \theta }}{{1 - \cos \theta }}$ is the left hand side of the given equation $\dfrac{{\sin \theta }}{{1 - \cos \theta }} = {\rm{cosec}}\theta + \cot \theta$.
Thus, we get
$\Rightarrow R.H.S. = L.H.S.$
Hence, we have proved that $\dfrac{{\sin \theta }}{{1 - \cos \theta }} = {\rm{cosec}}\theta + \cot \theta$.