Advertisements
Advertisements
Question
Solve for x:
`2tan^(-1)(cosx)=tan^(-1)(2"cosec" x)`
Advertisements
Solution
`2tan^(-1)(cosx)=tan^(-1)(2"cosec" x)`
`=>tan^(-1)((2cosx)/(1-cos^2x))=tan^(-1)(2"cosec" x) ""[because 2tan^(-1)x=tan^(-1)(2x/(1-x^2))]`
`=>(2cosx)/(sin^2x) = 2"cosec" x`
`=>(cosx)/(sin^2x) = 1/sinx`
`=>(sinx)/(cosx) = 1`
`=>tanx = 1`
`=> x=pi/4`
APPEARS IN
RELATED QUESTIONS
Find the principal values of the following:
`cos^-1(sin (4pi)/3)`
`sin^-1(sin (5pi)/6)`
`sin^-1(sin3)`
Evaluate the following:
`sec^-1(sec (13pi)/4)`
Evaluate the following:
`cot^-1(cot (9pi)/4)`
Evaluate the following:
`cot^-1(cot (19pi)/6)`
Write the following in the simplest form:
`tan^-1sqrt((a-x)/(a+x)),-a<x<a`
Evaluate the following:
`cosec(cos^-1 3/5)`
Evaluate the following:
`tan(cos^-1 8/17)`
Evaluate:
`sin(tan^-1x+tan^-1 1/x)` for x > 0
`sin^-1 63/65=sin^-1 5/13+cos^-1 3/5`
`sin^-1 5/13+cos^-1 3/5=tan^-1 63/16`
`2tan^-1 3/4-tan^-1 17/31=pi/4`
`2tan^-1(1/2)+tan^-1(1/7)=tan^-1(31/17)`
Solve the following equation for x:
`tan^-1((2x)/(1-x^2))+cot^-1((1-x^2)/(2x))=(2pi)/3,x>0`
If −1 < x < 0, then write the value of `sin^-1((2x)/(1+x^2))+cos^-1((1-x^2)/(1+x^2))`
Write the value of cos\[\left( 2 \sin^{- 1} \frac{1}{3} \right)\]
Evaluate sin \[\left( \tan^{- 1} \frac{3}{4} \right)\]
Write the value of cos2 \[\left( \frac{1}{2} \cos^{- 1} \frac{3}{5} \right)\]
Write the value of sin \[\left\{ \frac{\pi}{3} - \sin^{- 1} \left( - \frac{1}{2} \right) \right\}\]
The set of values of `\text(cosec)^-1(sqrt3/2)`
If \[\cos\left( \sin^{- 1} \frac{2}{5} + \cos^{- 1} x \right) = 0\], find the value of x.
Find the value of \[\tan^{- 1} \left( \tan\frac{9\pi}{8} \right)\]
Let f (x) = `e^(cos^-1){sin(x+pi/3}.`
Then, f (8π/9) =
If tan−1 3 + tan−1 x = tan−1 8, then x =
The value of sin \[\left( \frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8} \right)\] is
The value of \[\tan\left( \cos^{- 1} \frac{3}{5} + \tan^{- 1} \frac{1}{4} \right)\]
Find the domain of `sec^(-1)(3x-1)`.
The value of sin `["cos"^-1 (7/25)]` is ____________.
The value of tan `("cos"^-1 4/5 + "tan"^-1 2/3) =`
