Advertisements
Advertisements
प्रश्न
Prove the following result
`tan(cos^-1 4/5+tan^-1 2/3)=17/6`
उत्तर
LHS=`tan(cos^-1 4/5+tan^-1 2/3)=tan(tan^-1 sqrt(1-(4/5)^2)/(4/5)+tan^-1 2/3)` `[thereforecos^-1x=tan^-1(sqrt(1-x^2)/x)]`
`=tan(tan^-1 3/4+tan^-1 2/3)`
`=tan[tan^-1((3/4+2/3)/(1-3/4xx2/3))]` `[thereforetan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))]`
`=tan[tan^-1((17/12)/(6/12))`
`=tan[tan^-1 17/6]`
`=17/6=`RHS
APPEARS IN
संबंधित प्रश्न
`sin^-1(sin4)`
`sin^-1(sin2)`
Evaluate the following:
`cos^-1(cos4)`
Evaluate the following:
`cos^-1(cos12)`
Evaluate the following:
`sec^-1{sec (-(7pi)/3)}`
Evaluate the following:
`cosec^-1{cosec (-(9pi)/4)}`
Write the following in the simplest form:
`tan^-1{sqrt(1+x^2)-x},x in R `
Evaluate the following:
`sin(sec^-1 17/8)`
Evaluate:
`cot(tan^-1a+cot^-1a)`
If `cos^-1x + cos^-1y =pi/4,` find the value of `sin^-1x+sin^-1y`
If `cot(cos^-1 3/5+sin^-1x)=0`, find the values of x.
`sin^-1 63/65=sin^-1 5/13+cos^-1 3/5`
`sin^-1 4/5+2tan^-1 1/3=pi/2`
`2sin^-1 3/5-tan^-1 17/31=pi/4`
Show that `2tan^-1x+sin^-1 (2x)/(1+x^2)` is constant for x ≥ 1, find that constant.
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 range of tan−1 x.
Write the value of cos−1 (cos 1540°).
Write the value of cos−1 (cos 350°) − sin−1 (sin 350°)
Write the value of cos2 \[\left( \frac{1}{2} \cos^{- 1} \frac{3}{5} \right)\]
Write the value of cos−1 \[\left( \cos\frac{5\pi}{4} \right)\]
Show that \[\sin^{- 1} (2x\sqrt{1 - x^2}) = 2 \sin^{- 1} x\]
What is the principal value of `sin^-1(-sqrt3/2)?`
Write the principal value of `sin^-1(-1/2)`
Write the principal value of \[\tan^{- 1} 1 + \cos^{- 1} \left( - \frac{1}{2} \right)\]
Write the principal value of `tan^-1sqrt3+cot^-1sqrt3`
Write the value of \[\cos^{- 1} \left( \cos\frac{14\pi}{3} \right)\]
The value of tan \[\left\{ \cos^{- 1} \frac{1}{5\sqrt{2}} - \sin^{- 1} \frac{4}{\sqrt{17}} \right\}\] is
If sin−1 x − cos−1 x = `pi/6` , then x =
The number of solutions of the equation \[\tan^{- 1} 2x + \tan^{- 1} 3x = \frac{\pi}{4}\] is
If x < 0, y < 0 such that xy = 1, then tan−1 x + tan−1 y equals
In a ∆ ABC, if C is a right angle, then
\[\tan^{- 1} \left( \frac{a}{b + c} \right) + \tan^{- 1} \left( \frac{b}{c + a} \right) =\]
Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .
If 2 tan−1 (cos θ) = tan−1 (2 cosec θ), (θ ≠ 0), then find the value of θ.
If \[\tan^{- 1} \left( \frac{1}{1 + 1 . 2} \right) + \tan^{- 1} \left( \frac{1}{1 + 2 . 3} \right) + . . . + \tan^{- 1} \left( \frac{1}{1 + n . \left( n + 1 \right)} \right) = \tan^{- 1} \theta\] , then find the value of θ.
Write the value of \[\cos^{- 1} \left( - \frac{1}{2} \right) + 2 \sin^{- 1} \left( \frac{1}{2} \right)\] .
The value of sin `["cos"^-1 (7/25)]` is ____________.
