Math, asked by 1ŘãjÃřŸãN, 1 year ago

Find the value of x.

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Samu1y: Do v hv to use inverse trigonometry?

Samu1y: I am in 10 th std, so just wanted to confirm, cauz' I'm not sure

1ŘãjÃřŸãN: OK mention not

Answers

Answered by nikhilvenkatat

the answer is in the above picture

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Samu1y: no... the answer is 1/root2 or -1/root2. I am not able to upload the answer though with the steps.

Answered by Skidrow

$\tan^{ - 1}( \frac{x - 1}{x - 2} ) + \tan^{ - 1}( \frac{x {+ }1}{ x+ 2}) = \frac{\pi}{4} \\ we \: know \: that \: \tan^{ - 1}(a) + tan^{ - 1}(b) \: \\ = \tan^{ - 1}( \frac{a + b}{1 - ab} ) \\ procee \:using \: this \: identity \\ taking \: \: \: l.h.s \\ \\ = > \tan^{ - 1}( \frac{x - 1}{x - 2} ) + \tan^{ - 1}( \frac{x + 1}{ x+ 2}) \\ = \tan^{ - 1}( \frac{ \frac{x - 1}{x - 2} \: + \frac{x + 1}{x + 2} }{1 - (\frac{{x}^{2} - 1^2 }{ { {x}^{2} - 2^2 }} )} ) \\ = \tan^{ - 1}( \frac{(x + 2)(x - 1) + (x + 1)(x - 2)}{( {x}^{2} - {2}^{2} ) - ( {x}^{2} - {1}^{2}) } ) \\ = \tan^{ - 1}( \frac{2 {x}^{2} - 4}{ - 3} ) \\ now \: it \: is \: given \: that \: it \: is \: equal \: to \: \\ ( \frac{\pi}{4} ) \: \\ = > {4 - 2 {x}^{2} } \: = 3 \times \tan^{ }( \frac{\pi}{4} ) \\ = > {x}^{2} = \frac{1}{2} \\ = > x = \binom{ + }{ - } \sqrt{ \frac{1}{2} }$

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