Advertisements
Advertisements
Question
Solve: tan-1 (x + 1) + tan-1 (x – 1) = `tan^-1 (4/7)`
Solution
tan-1 (x + 1) + tan-1 (x – 1) = `tan^-1 (4/7)`
`tan^-1 (((x + 1) + (x - 1))/(1 - (x + 1)(x - 1))) = tan^-1 (4/7)`
`tan^-1 ((2x)/(1 - (x^2 - 1))) = tan^-1 (4/7)`
`tan^-1 ((2x)/(1 - x^2 + 1)) = tan^-1 (4/7)`
`tan^-1 ((2x)/(2 - x^2)) = tan^-1 (4/7)`
∴ `(2x)/(2 - x^2) = 4/7`
`(x)/(2 - x^2) = 2/7`
⇒ 7x = 2(2 – x2)
⇒ 7x = 4 – 2x2
⇒ 2x2 + 7x – 4 = 0
⇒ (x + 4) (2x – 1) = 0
⇒ x + 4 = 0 (or) 2x – 1 = 0
⇒ x = -4 (or) x = `1/2`
x = -4 is rejected, since does not satisfies the question.
∴ x = `1/2`
APPEARS IN
RELATED QUESTIONS
Find the principal values of `sin^(-1) (-1/2)`
In ΔABC, if a = 18, b = 24, c = 30 then find the values of cosA
Prove that:
2 tan-1 (x) = `sin^-1 ((2x)/(1 + x^2))`
Solve for x `tan^-1((1 - x)/(1 + x)) = 1/2 tan^-1x, x > 0`
Show that `sin^-1 5/13 + cos^-1 3/5 = tan^-1 63/16`
`"tan"^-1 (sqrt3)`
If tan-1 (x – 1) + tan-1 x + tan-1 (x + 1) = tan-1 3x, then the values of x are ____________.
`sin[π/3 - sin^-1 (-1/2)]` is equal to:
`2"tan"^-1 ("cos x") = "tan"^-1 (2 "cosec x")`
The equation of the tangent to the curve given by x = a sin3t, y = bcos3t at a point where t = `pi/2` is
