Question

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;

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

We know that,

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

Hence Proved.

Note:-

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

Answer 2

Answer

2 cot (a - b) = x²

Given

$\bullet \;\; \rm tan\ a=x+1\\\\\bullet \;\; \rm tan\ b=x-1$

To Prove

$\bullet \;\; \rm 2\ cot(a-b)=x^2$

Formula to be Applied

$\bullet \;\; \rm tan(x-y)=\dfrac{tan\ x-tan\ y}{1+tan\ x.tan\ y}\\\\\bullet \;\; \rm (x+y)(x-y)=x^2-y^2\\\\\bullet \;\; \rm tan\ x=\dfrac{1}{cot\ x}$

Solution

$\rm tan(a-b)=\dfrac{tan\ a-tan\ b}{1+tan\ a.tan\ b}\\\\\to\ \rm tan(a-b)=\dfrac{(x+1)-(x-1)}{1+(x+1)(x-1)}\\\\\to\ \rm tan(a-b)=\dfrac{2}{1+x^2-1}\\\\\to\ \rm tan(a-b)=\dfrac{2}{x^2}\\\\\to\ \rm \dfrac{1}{cot(a-b)}=\dfrac{2}{x^2}\\\\\to\ \rm x^2=2\ cot(a-b)\\\\\to\ \rm 2\ cot(a-b)=x^2\\\\ \rm Hence\ proved$