Math, asked by alizaaish864, 9 months ago

(11) If tan a = x +1 and tan B = x - 1, then prove that : 2 cot (a - b) = x square

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Answered by BrainlyIAS

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 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

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

amitkumar44481: Great :-)

Answered by BrainlyNisha001

$\huge\mathfrak\blue{Answer:-}$

$<font color = white >$

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.✌️☺️

$1. \: \tan(a - b) = \frac{ \tan \: a - \tan \: b}{1 + \tan \: a \tan \: b } \\ 2. \: \cot \: 0 = \frac{1}{ \tan \: 0} \\ 3. \: (a + b)(a - b) = {a}^{2} - {b}^{2}$

____________________♡

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(11) If tan a = x +1 and tan B = x - 1, then prove that : 2 cot (a - b) = x square​

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Hence proved.✌️☺️

(11) If tan a = x +1 and tan B = x - 1, then prove that : 2 cot (a - b) = x square