Advertisements
Advertisements
Question
Evaluate the following:
`tan^-1(tan (6pi)/7)`
Advertisements
Solution
We know that
`tan^-1(tantheta)=theta, -pi/2<theta<pi/2`
We have
`tan^-1(tan (6pi)/7)=tan^-1[tan(pi+pi/7)]`
`=tan^-1[tan(-pi/7)]`
`=-pi/7`
APPEARS IN
RELATED QUESTIONS
If (tan−1x)2 + (cot−1x)2 = 5π2/8, then find x.
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`
`sin^-1(sin (17pi)/8)`
`sin^-1(sin3)`
`sin^-1(sin12)`
Evaluate the following:
`tan^-1(tan1)`
Evaluate the following:
`tan^-1(tan12)`
Evaluate the following:
`sec^-1(sec (13pi)/4)`
Evaluate the following:
`cot^-1{cot (-(8pi)/3)}`
Evaluate: `sin{cos^-1(-3/5)+cot^-1(-5/12)}`
Evaluate:
`cos(sec^-1x+\text(cosec)^-1x)`,|x|≥1
If `cos^-1x + cos^-1y =pi/4,` find the value of `sin^-1x+sin^-1y`
`sin^-1x=pi/6+cos^-1x`
Prove the following result:
`tan^-1 1/4+tan^-1 2/9=sin^-1 1/sqrt5`
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 the following:
`tan{2tan^-1 1/5-pi/4}`
Evaluate the following:
`sin(1/2cos^-1 4/5)`
`2tan^-1 1/5+tan^-1 1/8=tan^-1 4/7`
Prove that
`tan^-1((1-x^2)/(2x))+cot^-1((1-x^2)/(2x))=pi/2`
Prove that
`sin{tan^-1 (1-x^2)/(2x)+cos^-1 (1-x^2)/(2x)}=1`
If `sin^-1 (2a)/(1+a^2)+sin^-1 (2b)/(1+b^2)=2tan^-1x,` Prove that `x=(a+b)/(1-ab).`
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 `sin^-1x+sin^-1y+sin^-1z=(3pi)/2,` then write the value of x + y + z.
Evaluate sin
\[\left( \frac{1}{2} \cos^{- 1} \frac{4}{5} \right)\]
Write the value of cos−1 (cos 6).
Evaluate: \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]
If x < 0, y < 0 such that xy = 1, then write the value of tan−1 x + tan−1 y.
The set of values of `\text(cosec)^-1(sqrt3/2)`
Find the value of \[2 \sec^{- 1} 2 + \sin^{- 1} \left( \frac{1}{2} \right)\]
\[\text{ If }\cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{\theta}{2}, \text{ then }4 x^2 - 12xy \cos\frac{\theta}{2} + 9 y^2 =\]
If 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 =
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 value of x, if tan `[sec^(-1) (1/x) ] = sin ( tan^(-1) 2) , x > 0 `.
The period of the function f(x) = tan3x is ____________.
The value of tan `("cos"^-1 4/5 + "tan"^-1 2/3) =`
