Question

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

Answer 1

Hint: To solve this question, we should know a few formulas like $\tan A=\dfrac{\sin A}{\cos A}$ and $\cot A=\dfrac{\cos A}{\sin A}$. Also, we should know the transformation from cos A to sec A and sin A to cosec A. We should also have knowledge of LCM and then we would be able to prove the required expression.

Complete step-by-step solution -
In the question, we have to prove that
$\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}=\sec A\operatorname{cosec}A+1$
To prove this expression, we will start from the left-hand side of the expression. So, we will get,
$LHS=\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}$
Now, we know that $\tan A=\dfrac{\sin A}{\cos A}$ and $\cot A=\dfrac{\cos A}{\sin A}$. By using these values, we will get LHS as
$LHS=\dfrac{\dfrac{\sin A}{\cos A}}{1-\dfrac{\cos A}{\sin A}}+\dfrac{\dfrac{\cos A}{\sin A}}{1-\dfrac{\sin A}{\cos A}}$
Now, we will take the LCM of the denominator of both the terms. So, we will get,
$LHS=\dfrac{\dfrac{\sin A}{\cos A}}{\dfrac{\sin A-\cos A}{\sin A}}+\dfrac{\dfrac{\cos A}{\sin A}}{\dfrac{\cos A-\sin A}{\cos A}}$
$LHS=\dfrac{{{\sin }^{2}}A}{\cos A\left( \sin A-\cos A \right)}+\dfrac{{{\cos }^{2}}A}{\sin A\left( \cos A-\sin A \right)}$
Now, we will take the negative sign common from (cos A – sin A).
$LHS=\dfrac{{{\sin }^{2}}A}{\cos A\left( \sin A-\cos A \right)}-\dfrac{{{\cos }^{2}}A}{\sin A\left( \sin A-\cos A \right)}$
Now, we will take LCM of both the terms, so we get,
$LHS=\dfrac{{{\sin }^{3}}A-{{\cos }^{3}}A}{\sin A\cos A\left( \sin A-\cos A \right)}$
Now, we know that ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$. So, we can write ${{\sin }^{3}}A-{{\cos }^{3}}A$ as $\left( \sin A-\cos A \right)\left( {{\sin }^{2}}A+{{\cos }^{2}}A+\sin A\cos A \right)$. Therefore, we can write LHS as,
$LHS=\dfrac{\left( \sin A-\cos A \right)\left( {{\sin }^{2}}A+{{\cos }^{2}}A+\sin A\cos A \right)}{\sin A\cos A\left( \sin A-\cos A \right)}$
Now, we know that (sin A – cos A) is common on both the numerator and denominator. So, we can cancel them out. Also, we know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$. Therefore, we will get LHS as
$LHS=\dfrac{1+\sin A\cos A}{\sin A\cos A}$
Now, we will separate the terms of the numerator with the denominator. So, we will get,
$LHS=\dfrac{1}{\sin A\cos A}+\dfrac{\sin A\cos A}{\sin A\cos A}$
Now, we know that the common terms in the numerator and denominator get canceled out. And we also know that $\dfrac{1}{\sin A}=\operatorname{cosec}A$ and $\dfrac{1}{\cos A}=\sec A$. Therefore, we will get LHS as,
$LHS=\sec A\operatorname{cosec}A+1$
LHS = RHS
Hence proved.

Note: In this question, the possible mistake one can make is at the time of taking negative sign common from (cos A – sin A), and after that, there are chances of making calculation errors because of the negative sign. Here we try to convert trigonometric expressions in a way such that our like terms are cancelled out.