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tan-1 x-3/x-4. +. tan-1 x+3/x+4. =. …

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tan^-1x-3/x-4. +. tan^-1 x+3/x+4. =. π/4

Posted by Vishaal Velu 7 years, 1 month ago

CBSE > Class 12 > Mathematics

1 answers

Naveen Sharma 7 years, 1 month ago

I think it is x-2 and x +2 instead of x-3 and x +3 respectively

Ans.

{tex}tan^{-1} ({x-2\over x-4})+ tan^{-1}( {x+2\over x+4}) = {\pi \over 4}{/tex}

Using{tex}tan^{-1} x + tan^{-1}y = tan^{-1} ({x+y\over 1-xy}){/tex}

=> {tex}tan^{-1} ({{x-2\over x-4} + {x+2\over x+4}\over 1-( {x-2\over x-4} \times {x+2\over x+4})}) = {\pi \over 4}{/tex}

=> {tex}tan^{-1}({2x^2-16\over -12}) = {\pi \over 4}{/tex}

{tex}=> {2x^2-16\over -12}= tan^{-1} {\pi \over 4}{/tex}

{tex}=> {2x^2-16\over -12}= 1{/tex}

{tex}=> 2x^2-16= -12{/tex}

{tex}=> 2x^2= 4 => x^2 = 2{/tex}

{tex}x = \sqrt 2{/tex}

if you use 3 answer ll be {tex}x = \sqrt {17\over 2}{/tex}

0Thank You

ANSWER

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