Advertisements
Advertisements
प्रश्न
It \[\tan^{- 1} \frac{x + 1}{x - 1} + \tan^{- 1} \frac{x - 1}{x} = \tan^{- 1}\] (−7), then the value of x is
विकल्प
0
−2
1
2
Advertisements
उत्तर
(d) 2
We know that
\[\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \left( \frac{x + y}{1 - xy} \right)\]
\[\therefore \tan^{- 1} \left( \frac{x + 1}{x - 1} \right) + \tan^{- 1} \left( \frac{x - 1}{x} \right) = \tan^{- 1} \left( - 7 \right)\]
\[ \Rightarrow \tan^{- 1} \left( \frac{\frac{x + 1}{x - 1} + \frac{x - 1}{x}}{1 - \frac{x + 1}{x - 1} \times \frac{x - 1}{x}} \right) = \tan^{- 1} \left( - 7 \right)\]
\[ \Rightarrow \tan^{- 1} \left( \frac{\frac{x^2 + x + x^2 - 2x + 1}{x\left( x - 1 \right)}}{\frac{x^2 - x - x^2 + 1}{x\left( x - 1 \right)}} \right) = \tan^{- 1} \left( - 7 \right)\]
\[ \Rightarrow \tan^{- 1} \left( \frac{2 x^2 - x + 1}{- x + 1} \right) = \tan^{- 1} \left( - 7 \right)\]
So, we get
\[\frac{2 x^2 - x + 1}{- x + 1} = - 7\]
\[ \Rightarrow 2 x^2 - x + 1 = 7x - 7\]
\[ \Rightarrow 2 x^2 - 8x + 8 = 0\]
\[ \Rightarrow x^2 - 4x + 4 = 0\]
\[ \Rightarrow \left( x - 2 \right)^2 = 0\]
\[ \Rightarrow x = 2\]
APPEARS IN
संबंधित प्रश्न
Prove that :
`2 tan^-1 (sqrt((a-b)/(a+b))tan(x/2))=cos^-1 ((a cos x+b)/(a+b cosx))`
Solve the following for x :
`tan^(-1)((x-2)/(x-3))+tan^(-1)((x+2)/(x+3))=pi/4,|x|<1`
Evaluate the following:
`tan^-1(tan2)`
Evaluate the following:
`\text(cosec)^-1(\text{cosec} pi/4)`
Evaluate the following:
`cosec^-1(cosec (6pi)/5)`
Evaluate the following:
`cosec^-1(cosec (11pi)/6)`
Evaluate the following:
`cosec^-1{cosec (-(9pi)/4)}`
Evaluate the following:
`cot^-1(cot (19pi)/6)`
Evaluate the following:
`sin(sec^-1 17/8)`
Prove the following result
`tan(cos^-1 4/5+tan^-1 2/3)=17/6`
Solve: `cos(sin^-1x)=1/6`
Evaluate:
`sin(tan^-1x+tan^-1 1/x)` for x > 0
If `cot(cos^-1 3/5+sin^-1x)=0`, find the values of x.
`sin^-1x=pi/6+cos^-1x`
`tan^-1x+2cot^-1x=(2x)/3`
Solve the following equation for x:
`tan^-1((x-2)/(x-4))+tan^-1((x+2)/(x+4))=pi/4`
Prove that:
`2sin^-1 3/5=tan^-1 24/7`
`2sin^-1 3/5-tan^-1 17/31=pi/4`
If x < 0, then write the value of cos−1 `((1-x^2)/(1+x^2))` in terms of tan−1 x.
Write the value of
\[\cos^{- 1} \left( \frac{1}{2} \right) + 2 \sin^{- 1} \left( \frac{1}{2} \right)\].
Evaluate sin
\[\left( \frac{1}{2} \cos^{- 1} \frac{4}{5} \right)\]
Evaluate: \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]
If \[\sin^{- 1} \left( \frac{1}{3} \right) + \cos^{- 1} x = \frac{\pi}{2},\] then find x.
Write the principal value of `tan^-1sqrt3+cot^-1sqrt3`
If \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} \right)\] = α, then x2 =
sin\[\left[ \cot^{- 1} \left\{ \tan\left( \cos^{- 1} x \right) \right\} \right]\] is equal to
If x < 0, y < 0 such that xy = 1, then tan−1 x + tan−1 y equals
\[\text{ If } u = \cot^{- 1} \sqrt{\tan \theta} - \tan^{- 1} \sqrt{\tan \theta}\text{ then }, \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) =\]
\[\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11}\] is equal to
The value of \[\sin^{- 1} \left( \cos\frac{33\pi}{5} \right)\] is
If \[3\sin^{- 1} \left( \frac{2x}{1 + x^2} \right) - 4 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) + 2 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) = \frac{\pi}{3}\] is equal to
The value of \[\sin\left( 2\left( \tan^{- 1} 0 . 75 \right) \right)\] is equal to
If x > 1, then \[2 \tan^{- 1} x + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] is equal to
If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find \[\frac{dy}{dx}\] When \[\theta = \frac{\pi}{3}\] .
Find the domain of `sec^(-1)(3x-1)`.
Find the domain of `sec^(-1) x-tan^(-1)x`
Find the value of x, if tan `[sec^(-1) (1/x) ] = sin ( tan^(-1) 2) , x > 0 `.
The equation sin-1 x – cos-1 x = cos-1 `(sqrt3/2)` has ____________.
