Advertisements
Advertisements
प्रश्न
If `sin^-1 (2a)/(1+a^2)+sin^-1 (2b)/(1+b^2)=2tan^-1x,` Prove that `x=(a+b)/(1-ab).`
Advertisements
उत्तर
Let: a = tan z
b = tan y
Then,
`sin^-1 (2a)/(1+a^2)+sin^-1 (2b)/(1+b^2)=2tan^-1x`
`=>sin^-1 (2tanz)/(1+tan^2z)+sin^-1 (2tany)/(1+tan^2y)=2tan^-1x`
`=>sin^-1(sin2z)+sin^-1(sin2y)=2tan^-1x` `[becausesin2x=(2tanx)/(1+tan^2x)]`
`=>2z+2y=2tan^-1x`
`=>tan^-1a+tan^-1b=tan^-1x` `[becausea=tanzandb=tany]`
`=>tan^-1 (a+b)/(1-ab)=tan^-1x` `[becausetan^-1x+tan^-1y=tan^-1 (x+y)/(1-xy)]`
`=>x=(a+b)/(1-ab)`
APPEARS IN
संबंधित प्रश्न
Find the domain of definition of `f(x)=cos^-1(x^2-4)`
`sin^-1(sin (7pi)/6)`
Evaluate the following:
`cos^-1{cos (5pi)/4}`
Evaluate the following:
`cos^-1{cos ((4pi)/3)}`
Evaluate the following:
`cos^-1(cos5)`
Evaluate the following:
`tan^-1(tan (6pi)/7)`
Evaluate the following:
`tan^-1(tan1)`
Evaluate the following:
`sec^-1{sec (-(7pi)/3)}`
Evaluate the following:
`cosec^-1(cosec (13pi)/6)`
Evaluate the following:
`cot^-1(cot (19pi)/6)`
Write the following in the simplest form:
`cot^-1 a/sqrt(x^2-a^2),| x | > a`
Write the following in the simplest form:
`tan^-1(x/(a+sqrt(a^2-x^2))),-a<x<a`
Prove the following result-
`tan^-1 63/16 = sin^-1 5/13 + cos^-1 3/5`
Solve: `cos(sin^-1x)=1/6`
`4sin^-1x=pi-cos^-1x`
Solve the following equation for x:
tan−1(x + 2) + tan−1(x − 2) = tan−1 `(8/79)`, x > 0
Evaluate: `cos(sin^-1 3/5+sin^-1 5/13)`
Prove that: `cos^-1 4/5+cos^-1 12/13=cos^-1 33/65`
`tan^-1 1/4+tan^-1 2/9=1/2cos^-1 3/2=1/2sin^-1(4/5)`
`2tan^-1(1/2)+tan^-1(1/7)=tan^-1(31/17)`
Prove that
`sin{tan^-1 (1-x^2)/(2x)+cos^-1 (1-x^2)/(2x)}=1`
Solve the following equation for x:
`3sin^-1 (2x)/(1+x^2)-4cos^-1 (1-x^2)/(1+x^2)+2tan^-1 (2x)/(1-x^2)=pi/3`
If 4 sin−1 x + cos−1 x = π, then what is the value of x?
Write the principal value of \[\tan^{- 1} 1 + \cos^{- 1} \left( - \frac{1}{2} \right)\]
Write the value of \[\sec^{- 1} \left( \frac{1}{2} \right)\]
Wnte the value of the expression \[\tan\left( \frac{\sin^{- 1} x + \cos^{- 1} x}{2} \right), \text { when } x = \frac{\sqrt{3}}{2}\]
Write the value of \[\tan^{- 1} \left( \frac{1}{x} \right)\] for x < 0 in terms of `cot^-1x`
Find the value of \[2 \sec^{- 1} 2 + \sin^{- 1} \left( \frac{1}{2} \right)\]
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 =
If x < 0, y < 0 such that xy = 1, then tan−1 x + tan−1 y equals
\[\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11}\] is equal to
The domain of \[\cos^{- 1} \left( x^2 - 4 \right)\] is
The value of \[\tan\left( \cos^{- 1} \frac{3}{5} + \tan^{- 1} \frac{1}{4} \right)\]
Prove that : \[\cot^{- 1} \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = \frac{x}{2}, 0 < x < \frac{\pi}{2}\] .
Find the simplified form of `cos^-1 (3/5 cosx + 4/5 sin x)`, where x ∈ `[(-3pi)/4, pi/4]`
The period of the function f(x) = tan3x is ____________.
The value of tan `("cos"^-1 4/5 + "tan"^-1 2/3) =`
