Advertisements
Advertisements
प्रश्न
Prove the following result
`cos(sin^-1 3/5+cot^-1 3/2)=6/(5sqrt13)`
उत्तर
LHS = `cos(sin^-1 3/5+cot^-1 3/2)`
`=cos(sin^-1 3/5+tan^-1 2/3)`
`=cos[cos^-1sqrt(1-(3/5)^2)+cos^-1 1/sqrt(1+(2/3)^2)]`
`=cos(cos^-1 4/5+cos^-1 3/sqrt13)`
`=cos{cos^-1[4/5xx3/sqrt13-sqrt(1-(4/5)^2)sqrt(1-(3/sqrt13)^2]}`
`=cos{cos^-1[12/(5sqrt13)-6/(5sqrt13)]}`
`=cos{cos^-1[6/(5sqrt13)]}`
`=6/(5sqrt13)=`RHS
APPEARS IN
संबंधित प्रश्न
Show that:
`2 sin^-1 (3/5)-tan^-1 (17/31)=pi/4`
Find the principal values of the following:
`cos^-1(sin (4pi)/3)`
`sin^-1(sin (5pi)/6)`
`sin^-1(sin4)`
`sin^-1(sin12)`
Evaluate the following:
`cos^-1(cos4)`
Evaluate the following:
`cos^-1(cos12)`
Evaluate the following:
`tan^-1(tan (6pi)/7)`
Evaluate the following:
`tan^-1(tan (7pi)/6)`
Evaluate the following:
`tan^-1(tan1)`
Evaluate the following:
`sec^-1(sec (7pi)/3)`
Evaluate the following:
`sec^-1{sec (-(7pi)/3)}`
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+sqrt(1+x^2)},x in R `
Write the following in the simplest form:
`tan^-1{(sqrt(1+x^2)+1)/x},x !=0`
Write the following in the simplest form:
`sin^-1{(sqrt(1+x)+sqrt(1-x))/2},0<x<1`
Evaluate:
`sec{cot^-1(-5/12)}`
Solve the following equation for x:
`tan^-1((x-2)/(x-4))+tan^-1((x+2)/(x+4))=pi/4`
Solve the following equation for x:
`tan^-1(2+x)+tan^-1(2-x)=tan^-1 2/3, where x< -sqrt3 or, x>sqrt3`
`sin^-1 63/65=sin^-1 5/13+cos^-1 3/5`
`4tan^-1 1/5-tan^-1 1/239=pi/4`
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)`
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`
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.`
If x < 0, then write the value of cos−1 `((1-x^2)/(1+x^2))` in terms of tan−1 x.
Write the value of cos−1 \[\left( \tan\frac{3\pi}{4} \right)\]
Write the value of cos \[\left( 2 \sin^{- 1} \frac{1}{2} \right)\]
Write the value of sin−1 \[\left( \cos\frac{\pi}{9} \right)\]
Write the value of \[\tan\left( 2 \tan^{- 1} \frac{1}{5} \right)\]
Write the value of \[\tan^{- 1} \left\{ 2\sin\left( 2 \cos^{- 1} \frac{\sqrt{3}}{2} \right) \right\}\]
The value of tan \[\left\{ \cos^{- 1} \frac{1}{5\sqrt{2}} - \sin^{- 1} \frac{4}{\sqrt{17}} \right\}\] is
2 tan−1 {cosec (tan−1 x) − tan (cot−1 x)} is equal to
If α = \[\tan^{- 1} \left( \tan\frac{5\pi}{4} \right) \text{ and }\beta = \tan^{- 1} \left( - \tan\frac{2\pi}{3} \right)\] , then
\[\text{ If }\cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{\theta}{2}, \text{ then }4 x^2 - 12xy \cos\frac{\theta}{2} + 9 y^2 =\]
The value of \[\cos^{- 1} \left( \cos\frac{5\pi}{3} \right) + \sin^{- 1} \left( \sin\frac{5\pi}{3} \right)\] is
If θ = sin−1 {sin (−600°)}, then one of the possible values of θ is
If x > 1, then \[2 \tan^{- 1} x + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] is equal to
If y = sin (sin x), prove that \[\frac{d^2 y}{d x^2} + \tan x \frac{dy}{dx} + y \cos^2 x = 0 .\]
The equation sin-1 x – cos-1 x = cos-1 `(sqrt3/2)` has ____________.
