Solve for x Cos-1-x2-1 -x2-1-Tan-1-2 x -x2-1-2 - - 3

Here , cos 1 ( x2 – 1 / x2 + 1) + tan 1( 2x/ x2 – 1) = 2π/3 LHS = cos 1 ( x2 – 1 / x2 + 1) + tan 1( 2x/ x2 – 1) Dividing throughout by x2 , cos 1 ( 1 – (1/ ...

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Byju's Answer

Standard XII

Mathematics

Odd Extension of a Function

Solve for x C...

Question

Solve for x

Cos^-1[(x²-1)/(x²+1)] + Tan^-1[2x/(x²-1)] = 2π/3

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Solution

Here , cos^-1 ( x² – 1 / x² + 1) + tan^-1( 2x/ x² – 1) = 2π/3
LHS = cos^-1 ( x² – 1 / x² + 1) + tan^-1( 2x/ x² – 1)
Dividing throughout by x² ,
cos^-1 ( 1 – (1/x²) /( 1 + ( 1/x²) ) + tan^-1( 2/x / 1 – (1/x²) )
= cos^-1( 1 - (1/x)² / 1 + (1/x)² ) + tan^-1 ( 2 (1/x) / 1 – (1/x)² )
=2tan^-1(1/x) + 2tan^-1(1/x) [ using 2tan^-1x = cos^-1( 1 – x² / 1 + x² ) = tan^-1( 2x / 1 – x²) ]
=4tan^-1(1/x)
RHS = 2 π/3
So 4 tan^-1( 1/x ) = 2 π/3
Or tan^-1(1/x) = π/6
Or 1/x = 1/√3
Or x = √3

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Solve for x Cos​​​​-1[(x​​​​​​2​​​-1)/(x​​​2+1)] + Tan​-1[2x/(x​​​​​​2​​​​​-1)] = 2π/3 ​​​​​​

Solve for x

Cos^-1[(x²-1)/(x²+1)] + Tan^-1[2x/(x²-1)] = 2π/3