Advertisements
Advertisements
Question
Solve: tan-1 (x + 1) + tan-1 (x – 1) = `tan^-1 (4/7)`
Advertisements
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
Evaluate the following:
`tan^-1(tan (5pi)/6)+cos^-1{cos((13pi)/6)}`
Find the principal value of the following: tan-1(– 1)
Prove the following:
`sin^-1(3/5) + cos^-1(12/13) = sin^-1(56/65)`
Show that `tan^-1 (1/2) + tan^-1 (2/11) = tan^-1 (3/4)`
sin[3 sin-1 (0.4)] = ______.
If 2 tan–1(cos θ) = tan–1(2 cosec θ), then show that θ = π 4, where n is any integer.
`("cos" 8° - "sin" 8°)/("cos" 8° + "sin" 8°)` is equal to ____________.
`"sin" 265° - "cos" 265°` is ____________.
What is the values of `cos^-1 (cos (7pi)/6)`
Find the principal value of `cot^-1 ((-1)/sqrt(3))`
