Advertisements
Advertisements
प्रश्न
Let f (x) = `e^(cos^-1){sin(x+pi/3}.`
Then, f (8π/9) =
पर्याय
e5π/18
e13π/18
e−2π/18
none of these
Advertisements
उत्तर
(b) e13π/18
Given: \[f\left( x \right) = e^{\cos^{- 1}} \left\{ \sin\left( x + \frac{\pi}{3} \right) \right\}\]
Then,
\[f\left( \frac{8\pi}{9} \right) = e^{\cos^{- 1}} \left\{ \sin\left( \frac{8\pi}{9} + \frac{\pi}{3} \right) \right\} \]
\[ = e^{\cos^{- 1}} \left\{ \sin\left( \frac{11\pi}{9} \right) \right\} \]
\[ = e^{\cos^{- 1}} \left\{ \cos\left( \frac{\pi}{2} + \frac{13\pi}{18} \right) \right\} \left[ \because \cos\left( \frac{\pi}{2} + \theta \right) = \sin\theta \right]\]
\[ = e^{\cos^{- 1}} \left\{ \cos\left( \frac{13\pi}{18} \right) \right\} \]
\[ = e^\frac{13\pi}{18}\]
APPEARS IN
संबंधित प्रश्न
If tan-1x+tan-1y=π/4,xy<1, then write the value of x+y+xy.
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
`sin^-1(sin pi/6)`
Evaluate the following:
`cos^-1{cos (5pi)/4}`
Evaluate the following:
`sec^-1(sec (7pi)/3)`
Evaluate the following:
`sec^-1(sec (13pi)/4)`
Evaluate the following:
`cosec^-1(cosec (13pi)/6)`
Write the following in the simplest form:
`tan^-1sqrt((a-x)/(a+x)),-a<x<a`
Write the following in the simplest form:
`tan^-1(x/(a+sqrt(a^2-x^2))),-a<x<a`
Evaluate the following:
`sin(tan^-1 24/7)`
Evaluate the following:
`sin(sec^-1 17/8)`
Prove the following result
`sin(cos^-1 3/5+sin^-1 5/13)=63/65`
Evaluate:
`cot(tan^-1a+cot^-1a)`
`sin^-1x=pi/6+cos^-1x`
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:
tan−1(x + 2) + tan−1(x − 2) = tan−1 `(8/79)`, x > 0
Solve the following equation for x:
`tan^-1(2+x)+tan^-1(2-x)=tan^-1 2/3, where x< -sqrt3 or, x>sqrt3`
Evaluate: `cos(sin^-1 3/5+sin^-1 5/13)`
Evaluate the following:
`tan 1/2(cos^-1 sqrt5/3)`
`tan^-1 1/7+2tan^-1 1/3=pi/4`
`2tan^-1 1/5+tan^-1 1/8=tan^-1 4/7`
`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)`
Show that `2tan^-1x+sin^-1 (2x)/(1+x^2)` is constant for x ≥ 1, find that constant.
Solve the following equation for x:
`cos^-1((x^2-1)/(x^2+1))+1/2tan^-1((2x)/(1-x^2))=(2x)/3`
What is the value of cos−1 `(cos (2x)/3)+sin^-1(sin (2x)/3)?`
Write the value of sin−1
\[\left( \sin( -{600}°) \right)\].
Write the value of cos2 \[\left( \frac{1}{2} \cos^{- 1} \frac{3}{5} \right)\]
Write the value of tan−1\[\left\{ \tan\left( \frac{15\pi}{4} \right) \right\}\]
Write the principal value of \[\cos^{- 1} \left( \cos\frac{2\pi}{3} \right) + \sin^{- 1} \left( \sin\frac{2\pi}{3} \right)\]
Wnte the value of the expression \[\tan\left( \frac{\sin^{- 1} x + \cos^{- 1} x}{2} \right), \text { when } x = \frac{\sqrt{3}}{2}\]
Find the value of \[2 \sec^{- 1} 2 + \sin^{- 1} \left( \frac{1}{2} \right)\]
If sin−1 x − cos−1 x = `pi/6` , then x =
If α = \[\tan^{- 1} \left( \frac{\sqrt{3}x}{2y - x} \right), \beta = \tan^{- 1} \left( \frac{2x - y}{\sqrt{3}y} \right),\]
then α − β =
It \[\tan^{- 1} \frac{x + 1}{x - 1} + \tan^{- 1} \frac{x - 1}{x} = \tan^{- 1}\] (−7), then the value of x is
If x > 1, then \[2 \tan^{- 1} x + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] is equal to
Solve for x : {xcos(cot-1 x) + sin(cot-1 x)}2 = `51/50`
Find the value of `sin^-1(cos((33π)/5))`.
