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

Prove the following statement:
1−tan⁡A1+tan⁡A=cot⁡A−1cot⁡A+1\dfrac{1-\tan A}{1+\tan A}=\dfrac{\cot A-1}{\cot A+1}

Explanation

Solution

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⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } and simplify the expression. Now, divide numerator and denominator by sin A and use cos⁡θsin⁡θ=cot⁡θ\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 1−tan⁡A1+tan⁡A=cot⁡A−1cot⁡A+1\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=1−tan⁡A1+tan⁡AE=\dfrac{1-\tan A}{1+\tan A}
We know that tan⁡θ=sin⁡θcos⁡θ\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=1−sin⁡Acos⁡A1+sin⁡Acos⁡AE=\dfrac{1-\dfrac{\sin A}{\cos A}}{1+\dfrac{\sin A}{\cos A}}
By simplifying the above expression, we get,
E=cos⁡A−sin⁡Acos⁡Acos⁡A+sin⁡Acos⁡AE=\dfrac{\dfrac{\cos A-\sin A}{\cos A}}{\dfrac{\cos A+\sin A}{\cos A}}
E=(cos⁡A−sin⁡Acos⁡A).(cos⁡Acos⁡A+sin⁡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=cos⁡A−sin⁡Acos⁡A+sin⁡AE=\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=cos⁡A−sin⁡Asin⁡Acos⁡A+sin⁡Asin⁡AE=\dfrac{\dfrac{\cos A-\sin A}{\sin A}}{\dfrac{\cos A+\sin A}{\sin A}}
We can also write the above expression as,
E=cos⁡Asin⁡A−sin⁡Asin⁡Acos⁡Asin⁡A+sin⁡Asin⁡AE=\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 cos⁡θsin⁡θ=cot⁡θ\dfrac{\cos \theta }{\sin \theta }=\cot \theta . By using this in the above expression, we get,
E=cot⁡A−1cot⁡A+1E=\dfrac{\cot A-1}{\cot A+1}
E = RHS
So, we get, LHS = RHS
Hence proved.
So, we have proved that 1−tan⁡A1+tan⁡A=cot⁡A−1cot⁡A+1\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=1−tan⁡A1+tan⁡AE=\dfrac{1-\tan A}{1+\tan A}
By substituting tan⁡A=1cot⁡A\tan A=\dfrac{1}{\cot A}, we get,
E=1−1cot⁡A1+1cot⁡AE=\dfrac{1-\dfrac{1}{\cot A}}{1+\dfrac{1}{\cot A}}
E=cot⁡A−1cot⁡Acot⁡A+1cot⁡AE=\dfrac{\dfrac{\cot A-1}{\cot A}}{\dfrac{\cot A+1}{\cot A}}
By canceling the like terms, we get,
E=cot⁡A−1cot⁡A+1E=\dfrac{\cot A-1}{\cot A+1}
E = RHS
Hence proved.