 QUESTION

# Prove that $\dfrac{{\tan {\text{A}}}}{{1 - \cot {\text{A}}}} + \dfrac{{\cot {\text{A}}}}{{1 - \tan {\text{A}}}} = 1 + \left( {\sec {\text{A}}} \right)\left( {{\text{cosecA}}} \right)$.

Hint- Here, we will proceed by converting all the trigonometric functions given in the LHS of the equation which we needed to prove and then taking the LCM. After this, we will use the formula ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$ in order to get the RHS of the equation we needed to prove.

To prove: $\dfrac{{\tan {\text{A}}}}{{1 - \cot {\text{A}}}} + \dfrac{{\cot {\text{A}}}}{{1 - \tan {\text{A}}}} = 1 + \left( {\sec {\text{A}}} \right)\left( {{\text{cosecA}}} \right)$
Since, $\tan {\text{A}} = \dfrac{{\sin {\text{A}}}}{{\cos {\text{A}}}}$ and $\cot {\text{A}} = \dfrac{{\cos {\text{A}}}}{{\sin {\text{A}}}}$

Taking the LHS of the equation we need to prove and using the above formulas, we have
$\Rightarrow \dfrac{{\tan {\text{A}}}}{{1 - \cot {\text{A}}}} + \dfrac{{\cot {\text{A}}}}{{1 - \tan {\text{A}}}} = \dfrac{{\left( {\dfrac{{\sin {\text{A}}}}{{\cos {\text{A}}}}} \right)}}{{1 - \left( {\dfrac{{\cos {\text{A}}}}{{\sin {\text{A}}}}} \right)}} + \dfrac{{\left( {\dfrac{{\cos {\text{A}}}}{{\sin {\text{A}}}}} \right)}}{{1 - \left( {\dfrac{{\sin {\text{A}}}}{{\cos {\text{A}}}}} \right)}}$

By taking separately the LCM of the two terms given in the denominator of the RHS of the above equation as sinA and cosA respectively, we get

$\Rightarrow \dfrac{{\tan {\text{A}}}}{{1 - \cot {\text{A}}}} + \dfrac{{\cot {\text{A}}}}{{1 - \tan {\text{A}}}} = \dfrac{{\left( {\dfrac{{\sin {\text{A}}}}{{\cos {\text{A}}}}} \right)}}{{\left( {\dfrac{{\sin {\text{A}} - \cos {\text{A}}}}{{\sin {\text{A}}}}} \right)}} + \dfrac{{\left( {\dfrac{{\cos {\text{A}}}}{{\sin {\text{A}}}}} \right)}}{{\left( {\dfrac{{\cos {\text{A}} - \sin {\text{A}}}}{{\cos {\text{A}}}}} \right)}} \\ \Rightarrow \dfrac{{\tan {\text{A}}}}{{1 - \cot {\text{A}}}} + \dfrac{{\cot {\text{A}}}}{{1 - \tan {\text{A}}}} = \dfrac{{\left( {\sin {\text{A}}} \right)\left( {\sin {\text{A}}} \right)}}{{\left( {\sin {\text{A}} - \cos {\text{A}}} \right)\left( {\cos {\text{A}}} \right)}} + \dfrac{{\left( {\cos {\text{A}}} \right)\left( {\cos {\text{A}}} \right)}}{{\left( {\cos {\text{A}} - \sin {\text{A}}} \right)\left( {\sin {\text{A}}} \right)}} \\ \Rightarrow \dfrac{{\tan {\text{A}}}}{{1 - \cot {\text{A}}}} + \dfrac{{\cot {\text{A}}}}{{1 - \tan {\text{A}}}} = \dfrac{{{{\left( {\sin {\text{A}}} \right)}^2}}}{{\left( {\sin {\text{A}} - \cos {\text{A}}} \right)\left( {\cos {\text{A}}} \right)}} - \dfrac{{{{\left( {\cos {\text{A}}} \right)}^2}}}{{\left( {\sin {\text{A}} - \cos {\text{A}}} \right)\left( {\sin {\text{A}}} \right)}} \\$

By taking the LCM of the terms given in the RHS of the above equation as (sinA)(cosA)(sinA - cosA), we get

$\Rightarrow \dfrac{{\tan {\text{A}}}}{{1 - \cot {\text{A}}}} + \dfrac{{\cot {\text{A}}}}{{1 - \tan {\text{A}}}} = \dfrac{{{{\left( {\sin {\text{A}}} \right)}^2}\left( {\sin {\text{A}}} \right) - {{\left( {\cos {\text{A}}} \right)}^2}\left( {\cos {\text{A}}} \right)}}{{\left( {\sin {\text{A}}} \right)\left( {\cos {\text{A}}} \right)\left( {\sin {\text{A}} - \cos {\text{A}}} \right)}} \\ \Rightarrow \dfrac{{\tan {\text{A}}}}{{1 - \cot {\text{A}}}} + \dfrac{{\cot {\text{A}}}}{{1 - \tan {\text{A}}}} = \dfrac{{{{\left( {\sin {\text{A}}} \right)}^3} - {{\left( {\cos {\text{A}}} \right)}^3}}}{{\left( {\sin {\text{A}}} \right)\left( {\cos {\text{A}}} \right)\left( {\sin {\text{A}} - \cos {\text{A}}} \right)}} \\$

Using the formula ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$, we get

$\Rightarrow \dfrac{{\tan {\text{A}}}}{{1 - \cot {\text{A}}}} + \dfrac{{\cot {\text{A}}}}{{1 - \tan {\text{A}}}} = \dfrac{{\left( {\sin {\text{A}} - \cos {\text{A}}} \right)\left[ {{{\left( {\sin {\text{A}}} \right)}^2} + {{\left( {\cos {\text{A}}} \right)}^2} + \left( {\sin {\text{A}}} \right)\left( {\cos {\text{A}}} \right)} \right]}}{{\left( {\sin {\text{A}}} \right)\left( {\cos {\text{A}}} \right)\left( {\sin {\text{A}} - \cos {\text{A}}} \right)}}$
By cancelling (sinA - cosA) from the numerator and the denominator of the RHS of the above equation, we get

$\Rightarrow \dfrac{{\tan {\text{A}}}}{{1 - \cot {\text{A}}}} + \dfrac{{\cot {\text{A}}}}{{1 - \tan {\text{A}}}} = \dfrac{{\left[ {{{\left( {\sin {\text{A}}} \right)}^2} + {{\left( {\cos {\text{A}}} \right)}^2} + \left( {\sin {\text{A}}} \right)\left( {\cos {\text{A}}} \right)} \right]}}{{\left( {\sin {\text{A}}} \right)\left( {\cos {\text{A}}} \right)}}$

Using the identity ${\left( {\sin {\text{A}}} \right)^2} + {\left( {\cos {\text{A}}} \right)^2} = 1$ in the above equation, we get

$\Rightarrow \dfrac{{\tan {\text{A}}}}{{1 - \cot {\text{A}}}} + \dfrac{{\cot {\text{A}}}}{{1 - \tan {\text{A}}}} = \dfrac{{1 + \left( {\sin {\text{A}}} \right)\left( {\cos {\text{A}}} \right)}}{{\left( {\sin {\text{A}}} \right)\left( {\cos {\text{A}}} \right)}} \\ \Rightarrow \dfrac{{\tan {\text{A}}}}{{1 - \cot {\text{A}}}} + \dfrac{{\cot {\text{A}}}}{{1 - \tan {\text{A}}}} = \dfrac{1}{{\left( {\sin {\text{A}}} \right)\left( {\cos {\text{A}}} \right)}} + \dfrac{{\left( {\sin {\text{A}}} \right)\left( {\cos {\text{A}}} \right)}}{{\left( {\sin {\text{A}}} \right)\left( {\cos {\text{A}}} \right)}} \\ \Rightarrow \dfrac{{\tan {\text{A}}}}{{1 - \cot {\text{A}}}} + \dfrac{{\cot {\text{A}}}}{{1 - \tan {\text{A}}}} = \left( {\dfrac{1}{{\sin {\text{A}}}}} \right)\left( {\dfrac{1}{{\cos {\text{A}}}}} \right) + 1 \\$

Using the formulas ${\text{cosecA}} = \dfrac{1}{{\sin {\text{A}}}}$ and $\sec {\text{A}} = \dfrac{1}{{\cos {\text{A}}}}$ in the above equation, we get

$\Rightarrow \dfrac{{\tan {\text{A}}}}{{1 - \cot {\text{A}}}} + \dfrac{{\cot {\text{A}}}}{{1 - \tan {\text{A}}}} = \left( {{\text{cosecA}}} \right)\left( {\sec {\text{A}}} \right) + 1 \\ \Rightarrow \dfrac{{\tan {\text{A}}}}{{1 - \cot {\text{A}}}} + \dfrac{{\cot {\text{A}}}}{{1 - \tan {\text{A}}}} = \left( {\sec {\text{A}}} \right)\left( {{\text{cosecA}}} \right) + 1 \\$

The above equation is the same equation which we needed to prove.

Note- In this particular problem, we have used the basic definitions of tangent trigonometric function, cotangent trigonometric function, secant trigonometric function and cosine trigonometric function as given by $\tan {\text{A}} = \dfrac{{\sin {\text{A}}}}{{\cos {\text{A}}}}$, $\cot {\text{A}} = \dfrac{{\cos {\text{A}}}}{{\sin {\text{A}}}}$, $\sec {\text{A}} = \dfrac{1}{{\cos {\text{A}}}}$ and ${\text{cosecA}} = \dfrac{1}{{\sin {\text{A}}}}$.