Question

Q1.If tan a = x + 1
, tan B = x -1, show that 2 cot (a-B)=x2

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Answer 1

Step-by-step explanation:

Given,

tanA = x + 1

tanB = x - 1

Now, We know that;

$\begin{gathered}\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\\\;\\\tan(A-B)=\frac{(x+1)-(x-1)}{1+(x-1)(x+1)}\\\;\\\tan(A-B)=\frac{x+1-x+1}{1+(x^2-1))}\\\;\\\tan(A-B)=\frac{2}{1+x^2-1}\\\;\\\tan(A-B)=\frac{2}{x^2}\end{gathered}$

We know that,

$\begin{gathered}\cot(A-B)=\frac{1}{\tan(A-B)}\\\;\\\cot(A-B)=\frac{1}{\frac{2}{x^2}}\\\;\\\cot(A-B)=\frac{x^2}{2}\\\;\\2\cot(A-B)=x^2\end{gathered}$

Hence Proved.

Note:-

$\begin{gathered}1.)\;\;\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\\\;\\2.)\;\;\cot\theta=\frac{1}{\tan\theta}\\\;\\3.)\;\;(a+b)(a-b)=a^2-b^2\end{gathered}$