Question

Prove the following statement:
\[\dfrac{1-\tan A}{1+\tan A}=\dfrac{\cot A-1}{\cot A+1}\]

Answer 1

Hint: First of all, consider the LHS of the given equation and convert the whole expression in terms of sin A and cos A by using the formula \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] and simplify the expression. Now, divide numerator and denominator by sin A and use \[\dfrac{\cos \theta }{\sin \theta }=\cot \theta \] to prove the desired result.
Complete step-by-step answer:
In this question, we have to prove that \[\dfrac{1-\tan A}{1+\tan A}=\dfrac{\cot A-1}{\cot A+1}\]. Let us consider the LHS of the equation given in the question.
\[E=\dfrac{1-\tan A}{1+\tan A}\]
We know that \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. So by substituting the value of tan A in terms of sin A and cos A in the above expression, we get,
\[E=\dfrac{1-\dfrac{\sin A}{\cos A}}{1+\dfrac{\sin A}{\cos A}}\]
By simplifying the above expression, we get,
\[E=\dfrac{\dfrac{\cos A-\sin A}{\cos A}}{\dfrac{\cos A+\sin A}{\cos A}}\]
\[E=\left( \dfrac{\cos A-\sin A}{\cos A} \right).\left( \dfrac{\cos A}{\cos A+\sin A} \right)\]
Now, by canceling the like terms of the above expression, we get,
\[E=\dfrac{\cos A-\sin A}{\cos A+\sin A}\]
By dividing the numerator and denominator of the above expression by sin A, we get,
\[E=\dfrac{\dfrac{\cos A-\sin A}{\sin A}}{\dfrac{\cos A+\sin A}{\sin A}}\]
We can also write the above expression as,
\[E=\dfrac{\dfrac{\cos A}{\sin A}-\dfrac{\sin A}{\sin A}}{\dfrac{\cos A}{\sin A}+\dfrac{\sin A}{\sin A}}\]
Now, we know that \[\dfrac{\cos \theta }{\sin \theta }=\cot \theta \]. By using this in the above expression, we get,
\[E=\dfrac{\cot A-1}{\cot A+1}\]
E = RHS
So, we get, LHS = RHS
Hence proved.
So, we have proved that \[\dfrac{1-\tan A}{1+\tan A}=\dfrac{\cot A-1}{\cot A+1}\].

Note: Students can also solve this question in the following way. Let us consider the LHS of the given equation.
\[E=\dfrac{1-\tan A}{1+\tan A}\]
By substituting \[\tan A=\dfrac{1}{\cot A}\], we get,
\[E=\dfrac{1-\dfrac{1}{\cot A}}{1+\dfrac{1}{\cot A}}\]
\[E=\dfrac{\dfrac{\cot A-1}{\cot A}}{\dfrac{\cot A+1}{\cot A}}\]
By canceling the like terms, we get,
\[E=\dfrac{\cot A-1}{\cot A+1}\]
E = RHS
Hence proved.