Advertisements
Advertisements
प्रश्न
If tan-1x+tan-1y=π/4,xy<1, then write the value of x+y+xy.
Advertisements
उत्तर
Here, `tan^(−1) x+tan^(−1) y=π/4, xy < 1.`
`tan^(-1)((x+y)/(1-xy))=pi/4`
`(x+y)/(1−xy)=1`
`x+y=1−xy`
`x+y+xy=1`
Therefore, the value of x + y + xy is 1.
APPEARS IN
संबंधित प्रश्न
Write the value of `tan(2tan^(-1)(1/5))`
Prove that
`tan^(-1) [(sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x))]=pi/4-1/2 cos^(-1)x,-1/sqrt2<=x<=1`
If a line makes angles 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.
Find the principal values of the following:
`cos^-1(-sqrt3/2)`
Evaluate the following:
`cos^-1(cos5)`
Evaluate the following:
`tan^-1(tan pi/3)`
Evaluate the following:
`tan^-1(tan (9pi)/4)`
Evaluate the following:
`tan^-1(tan1)`
Evaluate the following:
`cosec^-1(cosec (3pi)/4)`
Evaluate the following:
`cot^-1(cot (19pi)/6)`
Evaluate the following:
`cot^-1{cot (-(8pi)/3)}`
Prove the following result
`cos(sin^-1 3/5+cot^-1 3/2)=6/(5sqrt13)`
Evaluate:
`sin(tan^-1x+tan^-1 1/x)` for x < 0
`4sin^-1x=pi-cos^-1x`
`sin^-1 5/13+cos^-1 3/5=tan^-1 63/16`
Solve the equation `cos^-1 a/x-cos^-1 b/x=cos^-1 1/b-cos^-1 1/a`
Evaluate the following:
`tan 1/2(cos^-1 sqrt5/3)`
`tan^-1 2/3=1/2tan^-1 12/5`
`2sin^-1 3/5-tan^-1 17/31=pi/4`
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.
If x > 1, then write the value of sin−1 `((2x)/(1+x^2))` in terms of tan−1 x.
Write the value of cos−1 (cos 350°) − sin−1 (sin 350°)
The set of values of `\text(cosec)^-1(sqrt3/2)`
Write the value of `cot^-1(-x)` for all `x in R` in terms of `cot^-1(x)`
The positive integral solution of the equation
\[\tan^{- 1} x + \cos^{- 1} \frac{y}{\sqrt{1 + y^2}} = \sin^{- 1} \frac{3}{\sqrt{10}}\text{ is }\]
The number of solutions of the equation \[\tan^{- 1} 2x + \tan^{- 1} 3x = \frac{\pi}{4}\] is
If \[\sin^{- 1} \left( \frac{2a}{1 - a^2} \right) + \cos^{- 1} \left( \frac{1 - a^2}{1 + a^2} \right) = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right),\text{ where }a, x \in \left( 0, 1 \right)\] , then, the value of x is
Write the value of \[\cos^{- 1} \left( - \frac{1}{2} \right) + 2 \sin^{- 1} \left( \frac{1}{2} \right)\] .
