Question

# Prove the following trigonometric equation: $\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}=1+\sec A\operatorname{cosec}A$

Hint: Here, first of all, simplify LHS by substituting $\cot A=\dfrac{1}{\tan A}$ everywhere. Use formulas ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and $\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ to simplify LHS. Finally, substitute $\tan A=\dfrac{\sin A}{\cos A}$ to get the answer.

Here, we have to show that $\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}=1+\sec A\operatorname{cosec}A$
Let us consider the left-hand side (LHS) of the given expression as
$E=\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}$
First of all, we take the LCM of denominators, we get,
$E=\dfrac{\tan A\left( 1-\tan A \right)+\cot A\left( 1-\cot A \right)}{\left( 1-\cot A \right)\left( 1-\tan A \right)}$
Now, by simplifying the above expression, we get
$E=\dfrac{\tan A-{{\tan }^{2}}A+\cot A-{{\cot }^{2}}A}{{{1}^{2}}-\tan A-\cot A+\cot A.\tan A}$
We know that, $\tan A=\dfrac{1}{\cot A}$
By multiplying cot A on both sides, we get,
$\cot A.\tan A=\dfrac{\cot A}{\cot A}$
Hence, $\cot A.\tan A=1$
By applying this in the above expression, we get,
$E=\dfrac{\tan A-{{\tan }^{2}}A+\cot A-{{\cot }^{2}}A}{1-\tan A-\cot A+1}$
By further simplifying the above expression, we get,
$E=\dfrac{\tan A-{{\tan }^{2}}A+\cot A-{{\cot }^{2}}A}{2-\tan A-\cot A}$
Now, by substituting $\cot A=\dfrac{1}{\tan A}$ in the above expression, we get,
$E=\dfrac{\tan A-{{\tan }^{2}}A+\dfrac{1}{\tan A}-\dfrac{1}{{{\tan }^{2}}A}}{2-\tan A-\dfrac{1}{\tan A}}$
By simplifying the above expression, we get,
$E=\dfrac{\dfrac{{{\tan }^{3}}A-{{\tan }^{4}}A+\tan A-1}{{{\tan }^{2}}A}}{\dfrac{2\tan A-{{\tan }^{2}}A-1}{\tan A}}$
By cancelling the like terms, we get,
$E=\dfrac{\dfrac{{{\tan }^{3}}A-{{\tan }^{4}}A+\tan A-1}{\tan A}}{\dfrac{2\tan A-{{\tan }^{2}}A-1}{1}}$
We can also write the above expression as,
$E=\dfrac{{{\tan }^{3}}A-{{\tan }^{4}}A+\tan A-1}{\left( \tan A \right)\left( 2\tan A-{{\tan }^{2}}A-1 \right)}$
By multiplying â€“1 on both numerator and denominator of the above expression, we get,
$E=\dfrac{{{\tan }^{4}}A-{{\tan }^{3}}A-\tan A+1}{\left( \tan A \right)\left( {{\tan }^{2}}A-2\tan A+1 \right)}$
We can also write the above expression as,
$E=\dfrac{{{\tan }^{3}}A\left( \tan A-1 \right)-1\left( \tan A-1 \right)}{\left( \tan A \right)\left( {{\tan }^{2}}A-2\tan A+1 \right)}$
By taking (tan A â€“ 1) common in the numerator, we get,
$E=\dfrac{\left( {{\tan }^{3}}A-1 \right)\left( \tan A-1 \right)}{\left( \tan A \right)\left( {{\tan }^{2}}A-2\tan A+1 \right)}$
We know that ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$. By applying this in the denominator, we get,
$E=\dfrac{\left( {{\tan }^{3}}A-1 \right)\left( \tan A-1 \right)}{\left( \tan A \right){{\left( \tan A-1 \right)}^{2}}}$
By cancelling the like terms, we get
$E=\dfrac{\left( {{\tan }^{3}}A-{{1}^{3}} \right)}{\left( \tan A \right)\left( \tan A-1 \right)}$
We know that ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$. By applying this in the numerator, we get,
$E=\dfrac{\left( \tan A-1 \right)\left( {{\tan }^{2}}A+\tan A+1 \right)}{\left( \tan A \right)\left( \tan A-1 \right)}$
By cancelling the like terms, we get
$E=\dfrac{\left( {{\tan }^{2}}A \right)+\left( \tan A \right)+1}{\left( \tan A \right)}$
We can also write the above expression as,
$E=\dfrac{{{\tan }^{2}}A}{\tan A}+\dfrac{\tan A}{\tan A}+\dfrac{1}{\tan A}$
By cancelling the like terms, we get,
$E=\tan A+1+\dfrac{1}{\tan A}$
By substituting$\tan A=\dfrac{\sin A}{\cos A}$, we get
$E=\dfrac{\sin A}{\cos A}+\dfrac{1}{\dfrac{\sin A}{\cos A}}+1$
Or, $E=\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}+1$
By simplifying the above expression, we get,
$E=\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\left( \cos A \right)\left( \sin A \right)}+1$
We know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$. By applying this in the above expression, we get,
$E=\dfrac{1}{\sin A\cos A}+1$
Since we know that $\sin A=\dfrac{1}{\operatorname{cosec}A}$ and $\cos A=\dfrac{1}{\sec A}$. We get,
$E=\dfrac{1}{\dfrac{1}{\operatorname{cosec}A}.\dfrac{1}{\sec A}}+1$
Therefore, we get,
$E=\operatorname{cosec}A.\sec A+1=LHS$
Therefore, E = LHS = RHS
Hence, we have shown that $\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}=1+\sec A\operatorname{cosec}A$

Note: Students can also solve this question by substituting $\tan A=\dfrac{\sin A}{\cos A}$ and $\cot A=\dfrac{\cos A}{\sin A}$ in the first step only. Students must take special care while taking ${{\tan }^{2}}A$ as LCM in the numerator as they often make mistakes while writing the powers of tan A after taking the LCM. Also, students should always try to resolve the expression into factors by using identities like ${{\left( a\pm b \right)}^{2}}={{a}^{2}}+{{b}^{2}}\pm 2ab$ $\left( {{a}^{3}}\pm {{b}^{3}} \right)=\left( a\pm b \right)\left( {{a}^{2}}\mp ab+{{b}^{2}} \right)$ etc.