Advertisements
Advertisements
प्रश्न
`sin^-1(sin3)`
Advertisements
उत्तर
We know
`sin(sin^-1theta)=theta if - pi/2<=theta<=pi/2`
We have
`sin^-1(sin3)=sin^-1{sin(pi-3)}`
= π - 3
APPEARS IN
संबंधित प्रश्न
If `cos^-1( x/a) +cos^-1 (y/b)=alpha` , prove that `x^2/a^2-2(xy)/(ab) cos alpha +y^2/b^2=sin^2alpha`
Prove that :
`2 tan^-1 (sqrt((a-b)/(a+b))tan(x/2))=cos^-1 ((a cos x+b)/(a+b cosx))`
If sin [cot−1 (x+1)] = cos(tan−1x), then find x.
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{cos(-pi/4)}`
Evaluate the following:
`cos^-1{cos (13pi)/6}`
Evaluate the following:
`cos^-1(cos12)`
Evaluate the following:
`tan^-1(tan pi/3)`
Evaluate the following:
`sec^-1(sec (9pi)/5)`
Evaluate the following:
`cot^-1(cot (19pi)/6)`
Evaluate the following:
`sin(tan^-1 24/7)`
Evaluate the following:
`tan(cos^-1 8/17)`
Prove the following result
`cos(sin^-1 3/5+cot^-1 3/2)=6/(5sqrt13)`
Evaluate:
`cot(tan^-1a+cot^-1a)`
Evaluate:
`cos(sec^-1x+\text(cosec)^-1x)`,|x|≥1
If `cot(cos^-1 3/5+sin^-1x)=0`, find the values of x.
Solve the following equation for x:
`tan^-1((1-x)/(1+x))-1/2 tan^-1x` = 0, where x > 0
Solve the following equation for x:
cot−1x − cot−1(x + 2) =`pi/12`, x > 0
Solve the following equation for x:
`tan^-1((x-2)/(x-4))+tan^-1((x+2)/(x+4))=pi/4`
`sin^-1 63/65=sin^-1 5/13+cos^-1 3/5`
Solve the following:
`sin^-1x+sin^-1 2x=pi/3`
Solve the following:
`cos^-1x+sin^-1 x/2=π/6`
If `cos^-1 x/2+cos^-1 y/3=alpha,` then prove that `9x^2-12xy cosa+4y^2=36sin^2a.`
Evaluate the following:
`tan 1/2(cos^-1 sqrt5/3)`
Prove that:
`2sin^-1 3/5=tan^-1 24/7`
`tan^-1 1/4+tan^-1 2/9=1/2cos^-1 3/2=1/2sin^-1(4/5)`
Prove that
`sin{tan^-1 (1-x^2)/(2x)+cos^-1 (1-x^2)/(2x)}=1`
Write the value of cos−1 (cos 1540°).
Write the value of cos\[\left( 2 \sin^{- 1} \frac{1}{3} \right)\]
If tan−1 x + tan−1 y = `pi/4`, then write the value of x + y + xy.
If \[\sin^{- 1} \left( \frac{1}{3} \right) + \cos^{- 1} x = \frac{\pi}{2},\] then find x.
Write the principal value of \[\cos^{- 1} \left( \cos680^\circ \right)\]
The value of tan \[\left\{ \cos^{- 1} \frac{1}{5\sqrt{2}} - \sin^{- 1} \frac{4}{\sqrt{17}} \right\}\] is
\[\text{ If } u = \cot^{- 1} \sqrt{\tan \theta} - \tan^{- 1} \sqrt{\tan \theta}\text{ then }, \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) =\]
\[\text{ If }\cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{\theta}{2}, \text{ then }4 x^2 - 12xy \cos\frac{\theta}{2} + 9 y^2 =\]
sin \[\left\{ 2 \cos^{- 1} \left( \frac{- 3}{5} \right) \right\}\] is equal to
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 the domain of `sec^(-1) x-tan^(-1)x`
The value of tan `("cos"^-1 4/5 + "tan"^-1 2/3) =`
