Advertisements
Advertisements
प्रश्न
Solve `tan^-1 2x + tan^-1 3x = pi/4`
उत्तर
Given tan-1 (2x) + tan-1 (3x) = `pi/4`
`tan^-1 [(2x + 3x)/(1 - (2x)(3x))] = pi/4`
`tan^-1 [(5x)/(1 - 6x^2)] = pi/4`
`(5x)/(1 - 6x^2) = tan pi/4`
`(5x)/(1 - 6x^2)` = 1
⇒ 5x = 1(1 – 6x2)
⇒ 6x2 + 5x – 1 = 0
⇒ (x + 1) (6x – 1) = 0
⇒ x + 1 = 0 (or) 6x – 1 = 0
⇒ x = -1 (or) x = `1/6`
x = -1 is rejected. It doesn’t satisfies the question.
APPEARS IN
संबंधित प्रश्न
Find the principal value of `sec^(-1) (2/sqrt(3))`
Evaluate the following:
`tan^-1(tan (5pi)/6)+cos^-1{cos((13pi)/6)}`
The principal value of sin−1`(1/2)` is ______
`tan^-1(tan (7pi)/6)` = ______
Prove that:
2 tan-1 (x) = `sin^-1 ((2x)/(1 + x^2))`
Prove that:
`tan^-1 (4/3) + tan^-1 (1/7) = pi/4`
Express `tan^-1 [(cos x)/(1 - sin x)], - pi/2 < x < (3pi)/2` in the simplest form.
`("cos" 8° - "sin" 8°)/("cos" 8° + "sin" 8°)` is equal to ____________.
`"tan"^-1 sqrt3 - "sec"^-1 (-2)` is equal to ____________.
If |Z1| = |Z2| and arg (Z1) + arg (Z2) = 0, then
