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.