Advertisements
Advertisements
प्रश्न
If sin [cot−1 (x+1)] = cos(tan−1x), then find x.
Advertisements
उत्तर
If sin [cot−1 (x+1)] = cos(tan−1x), then find x.
`=>sin{sin^(-1) (1/(sqrt(1+(1+x)^2)))}`
`=cos{cos^(-1)(1/sqrt(1+x^2))} [because cot^(-1)=sin^(-1)1/sqrt(1+x^2) and tan^(-1)x=cos^(-1)(1/sqrt(1+x^2))]`
`⇒1/sqrt(1+(x+1)^2)=1/sqrt(1+x^2)`
`⇒1/sqrt(2+x^2+2x)=1/sqrt(1+x^2)`
`⇒sqrt(1+x2)=sqrt(x^2+2x+2)`
Squaring both sides, we get
⇒1+x2=x2+2x+2
⇒2x+2=1
⇒x=−1/2
APPEARS IN
संबंधित प्रश्न
Solve the following for x:
`sin^(-1)(1-x)-2sin^-1 x=pi/2`
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
Evaluate the following:
`cos^-1(cos5)`
Evaluate the following:
`tan^-1(tan pi/3)`
Evaluate the following:
`sec^-1(sec (2pi)/3)`
Write the following in the simplest form:
`tan^-1{sqrt(1+x^2)-x},x in R`
Write the following in the simplest form:
`tan^-1(x/(a+sqrt(a^2-x^2))),-a<x<a`
Write the following in the simplest form:
`sin^-1{(x+sqrt(1-x^2))/sqrt2},-1<x<1`
Evaluate the following:
`sin(cos^-1 5/13)`
Evaluate the following:
`sec(sin^-1 12/13)`
Evaluate the following:
`cot(cos^-1 3/5)`
Evaluate:
`cot(sin^-1 3/4+sec^-1 4/3)`
Find the value of `tan^-1 (x/y)-tan^-1((x-y)/(x+y))`
Solve the following equation for x:
`tan^-1 x/2+tan^-1 x/3=pi/4, 0<x<sqrt6`
`sin^-1 5/13+cos^-1 3/5=tan^-1 63/16`
`2sin^-1 3/5-tan^-1 17/31=pi/4`
`2tan^-1(1/2)+tan^-1(1/7)=tan^-1(31/17)`
Find the value of the following:
`cos(sec^-1x+\text(cosec)^-1x),` | x | ≥ 1
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−1 \[\left( \cos\frac{5\pi}{4} \right)\]
Show that \[\sin^{- 1} (2x\sqrt{1 - x^2}) = 2 \sin^{- 1} x\]
Wnte the value of\[\cos\left( \frac{\tan^{- 1} x + \cot^{- 1} x}{3} \right), \text{ when } x = - \frac{1}{\sqrt{3}}\]
Find the value of \[\cos^{- 1} \left( \cos\frac{13\pi}{6} \right)\]
Find the value of \[\tan^{- 1} \left( \tan\frac{9\pi}{8} \right)\]
If \[\cos^{- 1} x > \sin^{- 1} x\], then
If tan−1 (cot θ) = 2 θ, then θ =
If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find \[\frac{dy}{dx}\] When \[\theta = \frac{\pi}{3}\] .
Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .
The period of the function f(x) = tan3x is ____________.
