Question

If $\tan \alpha =x+1$and role="math" localid="1646425111199" $\tan \beta =x-1,$show that

role="math" localid="1646426027733" $2cot(\alpha -b)=x^2$

Short answer

The equation $2cot(\alpha -b)=x^2$can be obtained by multiplying equations$\tan \alpha =x+1\&\tan \beta =x-1$

$$\begin{gathered} (\tan \alpha )(\tan \beta )=(x+1)(x-1) \\ (\tan \alpha )(\tan \beta )+1=x^2 \\ 2\frac{(\tan \alpha )(\tan \beta )+1}{\tan \alpha -\tan \beta }=x^2 \\ 2cot(\alpha -\beta )=x^2 \end{gathered}$$

Step-by-step solution

Step 1. Given data

Given equations are

$$\begin{gathered} \tan \alpha =x+1\cdots \left(i\right) \\ \tan \beta =x-1\cdots \left(ii\right) \end{gathered}$$

Given equation that needs to be proven is

$2cot(\alpha -b)=x^2$

Step 2. Addition of equations

Subtract equation ii from the equation i

$$\begin{gathered} \tan \alpha -\tan \beta =(x+1)-(x-1) \\ \tan \alpha -\tan \beta =x+1-x+1 \\ \tan \alpha -\tan \beta =2 \\ \frac{\tan \alpha -\tan \beta }{2}=1\cdots \left(iii\right) \end{gathered}$$

Step 3. multiplication of equations

Multiply equations i and ii

$$\begin{gathered} (\tan \alpha )(\tan \beta )=(x+1)(x-1) \\ (\tan \alpha )(\tan \beta )=x^2-1 \\ (\tan \alpha )(\tan \beta )+1=x^2 \\ \frac{(\tan \alpha )(\tan \beta )+1}{1}=x^2\cdots \left(iv\right) \end{gathered}$$

Step 4. Proof

Use Equation iii in equation iv

$$\begin{gathered} \frac{(\tan \alpha )(\tan \beta )+1}{1}=x^2 \\ \frac{(\tan \alpha )(\tan \beta )+1}{{\displaystyle \frac{\tan \alpha -\tan \beta }{2}}}=x^2 \\ \frac{2}{{\displaystyle \frac{\tan \alpha -\tan \beta }{(\tan \alpha )(\tan \beta )+1}}}=x^2 \\ \frac{2}{\tan (\alpha -\beta )}=x^2 \\ 2cot(\alpha -\beta )=x^2 \end{gathered}$$