Advertisements
Advertisements
प्रश्न
Evaluate the following:
`sec^-1(sec (5pi)/4)`
Advertisements
उत्तर
We know that
sec-1 (sec θ) = θ, [0, π/2) ∪ (π/2, π]
We have
`sec^-1(sec (5pi)/4)=sec^-1[sec(2pi-(3pi)/4)]`
`=sec^-1[sec((3pi)/4)]`
`=(3pi)/4`
APPEARS IN
संबंधित प्रश्न
Solve the equation for x:sin−1x+sin−1(1−x)=cos−1x
If `tan^(-1)((x-2)/(x-4)) +tan^(-1)((x+2)/(x+4))=pi/4` ,find the value of x
Find the domain of `f(x) =2cos^-1 2x+sin^-1x.`
`sin^-1(sin pi/6)`
`sin^-1(sin (7pi)/6)`
`sin^-1{(sin - (17pi)/8)}`
`sin^-1(sin4)`
Evaluate the following:
`cot^-1(cot (19pi)/6)`
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{(sqrt(1+x^2)-1)/x},x !=0`
Evaluate the following:
`cot(cos^-1 3/5)`
Evaluate:
`cos{sin^-1(-7/25)}`
Evaluate:
`sec{cot^-1(-5/12)}`
Evaluate:
`sin(tan^-1x+tan^-1 1/x)` for x > 0
`sin(sin^-1 1/5+cos^-1x)=1`
Solve the following equation for x:
`tan^-1((x-2)/(x-4))+tan^-1((x+2)/(x+4))=pi/4`
Evaluate the following:
`tan{2tan^-1 1/5-pi/4}`
`2tan^-1(1/2)+tan^-1(1/7)=tan^-1(31/17)`
Find the value of the following:
`tan^-1{2cos(2sin^-1 1/2)}`
For any a, b, x, y > 0, prove that:
`2/3tan^-1((3ab^2-a^3)/(b^3-3a^2b))+2/3tan^-1((3xy^2-x^3)/(y^3-3x^2y))=tan^-1 (2alphabeta)/(alpha^2-beta^2)`
`where alpha =-ax+by, beta=bx+ay`
Write the value of
\[\cos^{- 1} \left( \frac{1}{2} \right) + 2 \sin^{- 1} \left( \frac{1}{2} \right)\].
Write the value of cos−1 (cos 350°) − sin−1 (sin 350°)
Write the value of cos−1 \[\left( \cos\frac{5\pi}{4} \right)\]
Write the value of \[\sec^{- 1} \left( \frac{1}{2} \right)\]
Wnte the value of the expression \[\tan\left( \frac{\sin^{- 1} x + \cos^{- 1} x}{2} \right), \text { when } x = \frac{\sqrt{3}}{2}\]
The set of values of `\text(cosec)^-1(sqrt3/2)`
Find the value of \[\tan^{- 1} \left( \tan\frac{9\pi}{8} \right)\]
The value of tan \[\left\{ \cos^{- 1} \frac{1}{5\sqrt{2}} - \sin^{- 1} \frac{4}{\sqrt{17}} \right\}\] is
If \[\cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = \alpha, then\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = \]
The positive integral solution of the equation
\[\tan^{- 1} x + \cos^{- 1} \frac{y}{\sqrt{1 + y^2}} = \sin^{- 1} \frac{3}{\sqrt{10}}\text{ is }\]
If α = \[\tan^{- 1} \left( \tan\frac{5\pi}{4} \right) \text{ and }\beta = \tan^{- 1} \left( - \tan\frac{2\pi}{3} \right)\] , then
\[\text{ If } u = \cot^{- 1} \sqrt{\tan \theta} - \tan^{- 1} \sqrt{\tan \theta}\text{ then }, \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) =\]
\[\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11}\] is equal to
If tan−1 3 + tan−1 x = tan−1 8, then x =
The value of \[\sin\left( 2\left( \tan^{- 1} 0 . 75 \right) \right)\] is equal to
Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .
Find the domain of `sec^(-1)(3x-1)`.
Find the value of x, if tan `[sec^(-1) (1/x) ] = sin ( tan^(-1) 2) , x > 0 `.
