Advertisements
Advertisements
Question
Solve the following equation for x:
`tan^-1 2x+tan^-1 3x = npi+(3pi)/4`
Solution
We know
`tan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))`
`therefore tan^-1 2x+tan^-1 3x=npi+(3pi)/4`
⇒ `tan^-1((2x+3x)/(1-2x xx3x))=npi+(3pi)/4`
⇒ `(5x)/(1-6x^2)=tan(npi+(3pi)/4)`
⇒ `(5x)/(1-6x^2)=-1`
⇒ `5x=-1+6x^2`
⇒ `6x^2-5x-1=0`
⇒ `(6x+1)(x-1)=0`
⇒ `x=-1/6` [As x=1 is not satisfying the equation]
APPEARS IN
RELATED QUESTIONS
Solve for x:
`2tan^(-1)(cosx)=tan^(-1)(2"cosec" x)`
Solve the following for x:
`sin^(-1)(1-x)-2sin^-1 x=pi/2`
If sin [cot−1 (x+1)] = cos(tan−1x), then find x.
Find the domain of definition of `f(x)=cos^-1(x^2-4)`
Find the domain of `f(x) =2cos^-1 2x+sin^-1x.`
Evaluate the following:
`tan^-1(tan pi/3)`
Evaluate the following:
`tan^-1(tan1)`
Evaluate the following:
`sec^-1(sec (5pi)/4)`
Evaluate the following:
`sec^-1(sec (7pi)/3)`
Evaluate the following:
`\text(cosec)^-1(\text{cosec} pi/4)`
Evaluate the following:
`cosec^-1(cosec (3pi)/4)`
Evaluate the following:
`cosec^-1(cosec (13pi)/6)`
Evaluate the following:
`cot^-1(cot (4pi)/3)`
Evaluate the following:
`cot^-1{cot ((21pi)/4)}`
Write the following in the simplest form:
`tan^-1{sqrt(1+x^2)-x},x in R `
Evaluate:
`cos(tan^-1 3/4)`
`tan^-1x+2cot^-1x=(2x)/3`
Solve the following equation for x:
tan−1(x + 2) + tan−1(x − 2) = tan−1 `(8/79)`, x > 0
`(9pi)/8-9/4sin^-1 1/3=9/4sin^-1 (2sqrt2)/3`
Solve the following:
`sin^-1x+sin^-1 2x=pi/3`
Evaluate the following:
`tan{2tan^-1 1/5-pi/4}`
`sin^-1 4/5+2tan^-1 1/3=pi/2`
Solve the following equation for x:
`tan^-1((x-2)/(x-1))+tan^-1((x+2)/(x+1))=pi/4`
If `sin^-1x+sin^-1y+sin^-1z=(3pi)/2,` then write the value of x + y + z.
Write the value of tan−1 x + tan−1 `(1/x)` for x < 0.
Write the value of cos−1 (cos 1540°).
Write the value of cos2 \[\left( \frac{1}{2} \cos^{- 1} \frac{3}{5} \right)\]
Write the value of tan−1\[\left\{ \tan\left( \frac{15\pi}{4} \right) \right\}\]
Write the value ofWrite the value of \[2 \sin^{- 1} \frac{1}{2} + \cos^{- 1} \left( - \frac{1}{2} \right)\]
If \[\cos\left( \tan^{- 1} x + \cot^{- 1} \sqrt{3} \right) = 0\] , find the value of x.
If \[\cos\left( \sin^{- 1} \frac{2}{5} + \cos^{- 1} x \right) = 0\], find the value of x.
If \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} \right)\] = α, then x2 =
sin\[\left[ \cot^{- 1} \left\{ \tan\left( \cos^{- 1} x \right) \right\} \right]\] is equal to
If α = \[\tan^{- 1} \left( \tan\frac{5\pi}{4} \right) \text{ and }\beta = \tan^{- 1} \left( - \tan\frac{2\pi}{3} \right)\] , then
If α = \[\tan^{- 1} \left( \frac{\sqrt{3}x}{2y - x} \right), \beta = \tan^{- 1} \left( \frac{2x - y}{\sqrt{3}y} \right),\]
then α − β =
\[\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11}\] is equal to
The value of \[\sin^{- 1} \left( \cos\frac{33\pi}{5} \right)\] is
The value of \[\tan\left( \cos^{- 1} \frac{3}{5} + \tan^{- 1} \frac{1}{4} \right)\]
If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find \[\frac{dy}{dx}\] When \[\theta = \frac{\pi}{3}\] .
