Advertisements
Advertisements
Question
`5tan^-1x+3cot^-1x=2x`
Advertisements
Solution
`5tan^-1x+3cot^-1x=2x`
⇒ `5tan^-1x+3(pi/2-tan^-1x)=2pi` `[becausecot^-1x=pi/2-tan^-1x]`
⇒ `5tan^-1x+(3pi)/2-3tan^-1x=2pi`
⇒ `2tan^-1x=pi/2`
⇒ `tan^-1x=pi/4`
⇒ `x=tan pi/4=1`
APPEARS IN
RELATED QUESTIONS
Find the value of the following: `tan(1/2)[sin^(-1)((2x)/(1+x^2))+cos^(-1)((1-y^2)/(1+y^2))],|x| <1,y>0 and xy <1`
If sin [cot−1 (x+1)] = cos(tan−1x), then find x.
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 `tan^(-1)((x-2)/(x-4)) +tan^(-1)((x+2)/(x+4))=pi/4` ,find the value of x
Find the principal values of the following:
`cos^-1(-sqrt3/2)`
`sin^-1(sin3)`
`sin^-1(sin4)`
`sin^-1(sin12)`
`sin^-1(sin2)`
Evaluate the following:
`tan^-1(tan pi/3)`
Evaluate the following:
`sec^-1(sec pi/3)`
Evaluate the following:
`sec^-1(sec (5pi)/4)`
Write the following in the simplest form:
`tan^-1{x+sqrt(1+x^2)},x in R `
Write the following in the simplest form:
`sin^-1{(x+sqrt(1-x^2))/sqrt2},-1<x<1`
Write the following in the simplest form:
`sin^-1{(sqrt(1+x)+sqrt(1-x))/2},0<x<1`
Evaluate the following:
`sin(sec^-1 17/8)`
Evaluate:
`tan{cos^-1(-7/25)}`
Evaluate:
`sin(tan^-1x+tan^-1 1/x)` for x < 0
Prove the following result:
`tan^-1 1/7+tan^-1 1/13=tan^-1 2/9`
Solve the following equation for x:
tan−1(x + 1) + tan−1(x − 1) = tan−1`8/31`
Solve the following equation for x:
`tan^-1((1-x)/(1+x))-1/2 tan^-1x` = 0, where x > 0
Solve the following:
`sin^-1x+sin^-1 2x=pi/3`
Solve the following:
`cos^-1x+sin^-1 x/2=π/6`
If `sin^-1 (2a)/(1+a^2)-cos^-1 (1-b^2)/(1+b^2)=tan^-1 (2x)/(1-x^2)`, then prove that `x=(a-b)/(1+ab)`
Find the value of the following:
`tan^-1{2cos(2sin^-1 1/2)}`
Prove that:
`tan^-1 (2ab)/(a^2-b^2)+tan^-1 (2xy)/(x^2-y^2)=tan^-1 (2alphabeta)/(alpha^2-beta^2),` where `alpha=ax-by and beta=ay+bx.`
Evaluate sin \[\left( \tan^{- 1} \frac{3}{4} \right)\]
Write the value of cos−1 (cos 6).
Write the principal value of \[\cos^{- 1} \left( \cos680^\circ \right)\]
If \[\cos\left( \tan^{- 1} x + \cot^{- 1} \sqrt{3} \right) = 0\] , find the value of x.
Find the value of \[\cos^{- 1} \left( \cos\frac{13\pi}{6} \right)\]
The number of solutions of the equation \[\tan^{- 1} 2x + \tan^{- 1} 3x = \frac{\pi}{4}\] is
\[\text{ If } u = \cot^{- 1} \sqrt{\tan \theta} - \tan^{- 1} \sqrt{\tan \theta}\text{ then }, \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) =\]
\[\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11}\] is equal to
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 \[\cos^{- 1} \left( - \frac{1}{2} \right) + 2 \sin^{- 1} \left( \frac{1}{2} \right)\] .
Solve for x : {xcos(cot-1 x) + sin(cot-1 x)}2 = `51/50`
The value of tan `("cos"^-1 4/5 + "tan"^-1 2/3) =`
