Advertisements
Advertisements
प्रश्न
Evaluate the following:
`tan^-1(tan pi/3)`
Advertisements
उत्तर
We know that
`tan^-1(tantheta)=theta, -pi/2<theta<pi/2`
We have
`tan^-1(tan pi/3)=pi/3`
APPEARS IN
संबंधित प्रश्न
Find the value of the following: `tan(1/2)[sin^(-1)((2x)/(1+x^2))+cos^(-1)((1-y^2)/(1+y^2))],|x| <1,y>0 and xy <1`
If (tan−1x)2 + (cot−1x)2 = 5π2/8, 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`
`sin^-1(sin (17pi)/8)`
Evaluate the following:
`tan^-1(tan4)`
Evaluate the following:
`sec^-1(sec (9pi)/5)`
Evaluate the following:
`sec^-1(sec (25pi)/6)`
Evaluate the following:
`cot^-1(cot (19pi)/6)`
Evaluate the following:
`cosec(cos^-1 3/5)`
Prove the following result
`tan(cos^-1 4/5+tan^-1 2/3)=17/6`
Evaluate:
`tan{cos^-1(-7/25)}`
Evaluate: `sin{cos^-1(-3/5)+cot^-1(-5/12)}`
Evaluate:
`cot(tan^-1a+cot^-1a)`
`sin^-1x=pi/6+cos^-1x`
Solve the following equation for x:
tan−1(x −1) + tan−1x tan−1(x + 1) = tan−13x
Solve the following equation for x:
cot−1x − cot−1(x + 2) =`pi/12`, x > 0
Evaluate: `cos(sin^-1 3/5+sin^-1 5/13)`
Solve the following:
`cos^-1x+sin^-1 x/2=π/6`
`tan^-1 2/3=1/2tan^-1 12/5`
Find the value of the following:
`cos(sec^-1x+\text(cosec)^-1x),` | x | ≥ 1
Solve the following equation for x:
`tan^-1 1/4+2tan^-1 1/5+tan^-1 1/6+tan^-1 1/x=pi/4`
If −1 < x < 0, then write the value of `sin^-1((2x)/(1+x^2))+cos^-1((1-x^2)/(1+x^2))`
Write the range of tan−1 x.
Write the value of cos−1 (cos 350°) − sin−1 (sin 350°)
Write the value ofWrite the value of \[2 \sin^{- 1} \frac{1}{2} + \cos^{- 1} \left( - \frac{1}{2} \right)\]
Wnte the value of the expression \[\tan\left( \frac{\sin^{- 1} x + \cos^{- 1} x}{2} \right), \text { when } x = \frac{\sqrt{3}}{2}\]
Find the value of \[\cos^{- 1} \left( \cos\frac{13\pi}{6} \right)\]
If \[\cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = \alpha, then\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = \]
If tan−1 3 + tan−1 x = tan−1 8, then x =
sin \[\left\{ 2 \cos^{- 1} \left( \frac{- 3}{5} \right) \right\}\] is equal to
If θ = sin−1 {sin (−600°)}, then one of the possible values of θ is
In a ∆ ABC, if C is a right angle, then
\[\tan^{- 1} \left( \frac{a}{b + c} \right) + \tan^{- 1} \left( \frac{b}{c + a} \right) =\]
\[\cot\left( \frac{\pi}{4} - 2 \cot^{- 1} 3 \right) =\]
If \[\sin^{- 1} \left( \frac{2a}{1 - a^2} \right) + \cos^{- 1} \left( \frac{1 - a^2}{1 + a^2} \right) = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right),\text{ where }a, x \in \left( 0, 1 \right)\] , 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
The value of \[\tan\left( \cos^{- 1} \frac{3}{5} + \tan^{- 1} \frac{1}{4} \right)\]
If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find \[\frac{dy}{dx}\] When \[\theta = \frac{\pi}{3}\] .
Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .
Write the value of \[\cos^{- 1} \left( - \frac{1}{2} \right) + 2 \sin^{- 1} \left( \frac{1}{2} \right)\] .
Find the value of `sin^-1(cos((33π)/5))`.
