Question

If $\tan \alpha =x+1$ and $\tan \beta =x-1$ , show that $2\cot \left( \alpha -\beta \right)={{x}^{2}}$ .

Answer 1

Hint: Try to simplify the left-hand side of the equation given in the question by using the property that $\cot A=\dfrac{1}{\tan A}$ , followed by applying the formula of tan(A-B).

Complete step-by-step answer:
Starting with the left-hand side of the equation that is given in the question.
We know, $\cot A=\dfrac{1}{\tan A}$ . So, applying this to the left-hand side of the equation in the question, we get
$2\cot \left( \alpha -\beta \right)$
$=\dfrac{2}{\tan \left( \alpha -\beta \right)}$
Now we know that $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$ . On using this in our expression, we get
$=\dfrac{2\left( 1+\tan \alpha \tan \beta \right)}{\tan \alpha -tan\beta }$
Now it is given in the question that $\tan \alpha =x+1$ and $\tan \beta =x-1$ . So, when we put these values in our expression, we get
$=\dfrac{2\left( 1+\left( 1+x \right)\left( x-1 \right) \right)}{1+x-\left( x-1 \right)}$
Now, we will simplify the above expression using the formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ . On doing so, we get
$=\dfrac{2\left( 1+{{x}^{2}}-1 \right)}{1+x-x+1}$
$=\dfrac{2{{x}^{2}}}{2}$
$={{x}^{2}}$
The left-hand side of the equation given in the question is equal to the right-hand side of the equation. Hence, we can say that we have proved the equation given in the question.

Note: Be careful about the calculation and the signs while opening the brackets. The general mistake that a student can make is 1+x-(x-1)=1+x-x-1. Also, be careful about the signs in the formula of tan(A-B). Whenever you are dealing with an expression having cotangents, secant, and cosecant involved, it is better to convert it to an equivalent expression in terms of sine, cosine, and tangent, as most of the formulas we know are valid for sine, cosine, and tangents only.