Advertisements
Advertisements
प्रश्न
Evaluate:
`sec{cot^-1(-5/12)}`
Advertisements
उत्तर
`sec{cot^-1(-5/12)}=sec{pi-cot^-1(5/12)}`
`=-sec{cot^-1(5/12)}`
`=-sec{cos^-1[1/(1+(12/5)^2)]}`
`=-sec{cos^-1(5/13)}`
`=-sec{sec^-1
(13/5)}`
`=-13/5`
APPEARS IN
संबंधित प्रश्न
Write the value of `tan(2tan^(-1)(1/5))`
Show that:
`2 sin^-1 (3/5)-tan^-1 (17/31)=pi/4`
`sin^-1(sin pi/6)`
`sin^-1(sin (7pi)/6)`
`sin^-1(sin (13pi)/7)`
`sin^-1(sin3)`
Evaluate the following:
`tan^-1(tan (6pi)/7)`
Evaluate the following:
`tan^-1(tan1)`
Evaluate the following:
`sec^-1(sec (7pi)/3)`
Evaluate:
`cos(tan^-1 3/4)`
`sin^-1x=pi/6+cos^-1x`
Prove the following result:
`tan^-1 1/4+tan^-1 2/9=sin^-1 1/sqrt5`
Solve the following equation for x:
`tan^-1 2x+tan^-1 3x = npi+(3pi)/4`
Solve the following equation for x:
`tan^-1 x/2+tan^-1 x/3=pi/4, 0<x<sqrt6`
Solve the following equation for x:
`tan^-1((x-2)/(x-4))+tan^-1((x+2)/(x+4))=pi/4`
Sum the following series:
`tan^-1 1/3+tan^-1 2/9+tan^-1 4/33+...+tan^-1 (2^(n-1))/(1+2^(2n-1))`
Evaluate: `cos(sin^-1 3/5+sin^-1 5/13)`
`sin^-1 63/65=sin^-1 5/13+cos^-1 3/5`
`tan^-1 1/7+2tan^-1 1/3=pi/4`
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`
If x > 1, then write the value of sin−1 `((2x)/(1+x^2))` in terms of tan−1 x.
Write the value of tan−1x + tan−1 `(1/x)`for x > 0.
Write the value of tan−1 x + tan−1 `(1/x)` for x < 0.
What is the value of cos−1 `(cos (2x)/3)+sin^-1(sin (2x)/3)?`
Write the value of sin−1
\[\left( \sin( -{600}°) \right)\].
If \[\tan^{- 1} (\sqrt{3}) + \cot^{- 1} x = \frac{\pi}{2},\] find x.
Write the value of \[\tan^{- 1} \left\{ 2\sin\left( 2 \cos^{- 1} \frac{\sqrt{3}}{2} \right) \right\}\]
Write the principal value of `tan^-1sqrt3+cot^-1sqrt3`
Write the value of \[\sec^{- 1} \left( \frac{1}{2} \right)\]
Write the value of \[\tan^{- 1} \left( \frac{1}{x} \right)\] for x < 0 in terms of `cot^-1x`
If \[\cos\left( \tan^{- 1} x + \cot^{- 1} \sqrt{3} \right) = 0\] , find the value of x.
If \[\cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = \alpha, then\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = \]
\[\text{ If } u = \cot^{- 1} \sqrt{\tan \theta} - \tan^{- 1} \sqrt{\tan \theta}\text{ then }, \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) =\]
\[\text{ If }\cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{\theta}{2}, \text{ then }4 x^2 - 12xy \cos\frac{\theta}{2} + 9 y^2 =\]
If tan−1 3 + tan−1 x = tan−1 8, then x =
The value of \[\sin^{- 1} \left( \cos\frac{33\pi}{5} \right)\] is
sin \[\left\{ 2 \cos^{- 1} \left( \frac{- 3}{5} \right) \right\}\] is equal to
If \[\tan^{- 1} \left( \frac{1}{1 + 1 . 2} \right) + \tan^{- 1} \left( \frac{1}{1 + 2 . 3} \right) + . . . + \tan^{- 1} \left( \frac{1}{1 + n . \left( n + 1 \right)} \right) = \tan^{- 1} \theta\] , then find the value of θ.
The period of the function f(x) = tan3x is ____________.
