Advertisements
Advertisements
Question
`2tan^-1 3/4-tan^-1 17/31=pi/4`
Advertisements
Solution
LHS = `2tan^-1 3/4-tan^-1 17/31`
`=tan^-1{(2xx3/4)/(1-(3/4)^2)}-tan^-1 17/31` `[because2tan^-1x=tan^-1{(2x)/(1-x^2)}]`
`=tan^-1{(3/2)/(7/16)}-tan^-1 17/31`
`=tan^-1 24/7-tan^-1 17/31`
`=tan^-1((24/7-17/31)/(1+24/7xx17/31))` `[becausetan^-1x-tan^-1y=tan^-1((x-y)/(1+xy))]`
`=tan^-1((625/217)/(625/217))`
`=tan^-1 1=pi/4=` RHS
APPEARS IN
RELATED QUESTIONS
Write the value of `tan(2tan^(-1)(1/5))`
If sin [cot−1 (x+1)] = cos(tan−1x), 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`
Find the domain of `f(x) =2cos^-1 2x+sin^-1x.`
Find the principal values of the following:
`cos^-1(-1/sqrt2)`
`sin^-1(sin4)`
Evaluate the following:
`sec^-1(sec (13pi)/4)`
Evaluate the following:
`cosec^-1(cosec (13pi)/6)`
Evaluate the following:
`cot^-1(cot pi/3)`
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(x/(a+sqrt(a^2-x^2))),-a<x<a`
Evaluate the following:
`sin(cos^-1 5/13)`
Evaluate the following:
`sec(sin^-1 12/13)`
`5tan^-1x+3cot^-1x=2x`
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 − 2) = tan−1 `(8/79)`, x > 0
Solve the following equation for x:
`tan^-1((x-2)/(x-4))+tan^-1((x+2)/(x+4))=pi/4`
Solve the following:
`sin^-1x+sin^-1 2x=pi/3`
Evaluate the following:
`sin(1/2cos^-1 4/5)`
Prove that
`tan^-1((1-x^2)/(2x))+cot^-1((1-x^2)/(2x))=pi/2`
Prove that:
`tan^-1 (2ab)/(a^2-b^2)+tan^-1 (2xy)/(x^2-y^2)=tan^-1 (2alphabeta)/(alpha^2-beta^2),` where `alpha=ax-by and beta=ay+bx.`
Write the value of cos−1 (cos 1540°).
Show that \[\sin^{- 1} (2x\sqrt{1 - x^2}) = 2 \sin^{- 1} x\]
Evaluate: \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]
Write the principal value of \[\cos^{- 1} \left( \cos\frac{2\pi}{3} \right) + \sin^{- 1} \left( \sin\frac{2\pi}{3} \right)\]
If \[\cos\left( \sin^{- 1} \frac{2}{5} + \cos^{- 1} x \right) = 0\], find the value of x.
The positive integral solution of the equation
\[\tan^{- 1} x + \cos^{- 1} \frac{y}{\sqrt{1 + y^2}} = \sin^{- 1} \frac{3}{\sqrt{10}}\text{ is }\]
\[\text{ If }\cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{\theta}{2}, \text{ then }4 x^2 - 12xy \cos\frac{\theta}{2} + 9 y^2 =\]
Let f (x) = `e^(cos^-1){sin(x+pi/3}.`
Then, f (8π/9) =
\[\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11}\] is equal to
The value of \[\sin^{- 1} \left( \cos\frac{33\pi}{5} \right)\] is
If 4 cos−1 x + sin−1 x = π, then the value of x is
It \[\tan^{- 1} \frac{x + 1}{x - 1} + \tan^{- 1} \frac{x - 1}{x} = \tan^{- 1}\] (−7), then the value of x is
In a ∆ ABC, if C is a right angle, then
\[\tan^{- 1} \left( \frac{a}{b + c} \right) + \tan^{- 1} \left( \frac{b}{c + a} \right) =\]
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 `.
