Question

Solve the following equation for x: $3sin^{-1}(\frac{2x}{1 + x^{2}}) - 4cos^{-1}(\frac{1 - x^{2}}{1 + x^{2}}) + 2tan^{-1}(\frac{2x}{1 - x^{2}}) = \frac{\pi}{3}$.

Answer 1

Solution :

Let x = tanA ---- ( 1 )

=> A = tan^-1x ----- ( 2 )

=> 3sin^-1{ 2x/(1+x²)}

- 4cos^-1{(1-x²)/(1+x²)}

+ 2tan^-1{2x/(1-x²)} = π/3

=>3sin^-1{2tanA/(1+tan²A)}

- 4cos^-1{(1-tan²A)/(1+tan²A)}

+ 2tan^-1{2tanA/(1-tan²A)} = π/3

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We know that ,

i ) sin2A = 2tanA/( 1 + tan² A )

ii ) cos2A = (1-tan²A)/(1+tan²A)

iii) tan2A = 2tanA/(1-tan²A)

***************************************

=> 3sin^-1{sin2A} - 4cos^-1(cos2A)

+ 2tan^-1(tan2A) = π/3

=> 2A = π/3

=> A = π/6

=> tan^-1 x = π/6 [ from ( 2 ) ]

=> x = tan ( π/6 )

=> x = 1/√3

••••

Answer 2

HELLO DEAR,



We Know:-
2tanA/(1 + tan²A) = sin2A
(1 - tan²A)/(1 + tan²A) = cos2A
2tanA/(1 - tan²A) = tan2A

now,

let x = tanA

so,
$3sin^{-1}(\frac{2x}{1 + x^{2}}) - 4cos^{-1}(\frac{1 - x^{2}}{1 + x^{2}}) + 2tan^{-1}(\frac{2x}{1 - x^{2}}) = \frac{\pi}{3}$.

=> 3sin-¹(2tanA)/(1 + tan²A) - 4cos-¹(1 - tan²A)/(1 + tan²A) + 2tan-¹(2tanA)/(1 - tan²A) = π/3

=> 3sin-¹(sin2A) - 4cos-¹(cos2A) + 2tan-¹(tan2A) = π/3

=> 3(2A) - 4(2A) + 2(2A) = π/3

=> 2A = π/3

=> A = π/6

therefore, x = tanA = tanπ\6

x = 1/√3



I HOPE IT'S HELP YOU DEAR,
THANKS