Advertisements
Advertisements
प्रश्न
If x < 0, then write the value of cos−1 `((1-x^2)/(1+x^2))` in terms of tan−1 x.
Advertisements
उत्तर
Let x = tan y
Then,
`cos^-1((1-x^2)/(1+x^2))=cos^-1((1-tan^2y)/(1+tan^2y))`
`=cos^-1(cos2y)` `[because (1-tan^2x)/(1+tan^2)=cos2x]`
= 2y ...(1)
The value of x is negative.
So, let x = -a where a > 0.
`-a = tan y`
`=>y=tan^-1(-a)`
Now,
`cos^-1((1-x^2)/(1+x^2))=2y` [Using (1)]
`=2tan^-1(-a)`
`=-2tan^-1x` `[becausex=-a]`
APPEARS IN
संबंधित प्रश्न
Solve the following for x:
`sin^(-1)(1-x)-2sin^-1 x=pi/2`
`sin^-1(sin (5pi)/6)`
`sin^-1{(sin - (17pi)/8)}`
Evaluate the following:
`cos^-1(cos12)`
Evaluate the following:
`sec^-1(sec (7pi)/3)`
Evaluate the following:
`sin(sec^-1 17/8)`
Evaluate the following:
`sec(sin^-1 12/13)`
Evaluate the following:
`cos(tan^-1 24/7)`
Prove the following result
`tan(cos^-1 4/5+tan^-1 2/3)=17/6`
Evaluate:
`cos{sin^-1(-7/25)}`
Evaluate:
`tan{cos^-1(-7/25)}`
Evaluate:
`cosec{cot^-1(-12/5)}`
Evaluate:
`sin(tan^-1x+tan^-1 1/x)` for x > 0
Evaluate:
`cos(sec^-1x+\text(cosec)^-1x)`,|x|≥1
If `cos^-1x + cos^-1y =pi/4,` find the value of `sin^-1x+sin^-1y`
If `sin^-1x+sin^-1y=pi/3` and `cos^-1x-cos^-1y=pi/6`, find the values of x and y.
If `cot(cos^-1 3/5+sin^-1x)=0`, find the values of x.
`5tan^-1x+3cot^-1x=2x`
Prove the following result:
`sin^-1 12/13+cos^-1 4/5+tan^-1 63/16=pi`
Solve the following equation for x:
`tan^-1 2x+tan^-1 3x = npi+(3pi)/4`
Solve the following equation for x:
`tan^-1((1-x)/(1+x))-1/2 tan^-1x` = 0, where x > 0
Evaluate: `cos(sin^-1 3/5+sin^-1 5/13)`
`sin^-1 63/65=sin^-1 5/13+cos^-1 3/5`
`2sin^-1 3/5-tan^-1 17/31=pi/4`
Solve the following equation for x:
`cos^-1((x^2-1)/(x^2+1))+1/2tan^-1((2x)/(1-x^2))=(2x)/3`
For any a, b, x, y > 0, prove that:
`2/3tan^-1((3ab^2-a^3)/(b^3-3a^2b))+2/3tan^-1((3xy^2-x^3)/(y^3-3x^2y))=tan^-1 (2alphabeta)/(alpha^2-beta^2)`
`where alpha =-ax+by, beta=bx+ay`
Write the difference between maximum and minimum values of sin−1 x for x ∈ [− 1, 1].
Write the value of
\[\cos^{- 1} \left( \frac{1}{2} \right) + 2 \sin^{- 1} \left( \frac{1}{2} \right)\].
Write the value of cos−1 \[\left( \cos\frac{5\pi}{4} \right)\]
If \[\tan^{- 1} (\sqrt{3}) + \cot^{- 1} x = \frac{\pi}{2},\] find x.
Write the principal value of `sin^-1(-1/2)`
Write the principal value of \[\cos^{- 1} \left( \cos680^\circ \right)\]
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 =
2 tan−1 {cosec (tan−1 x) − tan (cot−1 x)} is equal to
If \[\cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = \alpha, then\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = \]
\[\text{ If } u = \cot^{- 1} \sqrt{\tan \theta} - \tan^{- 1} \sqrt{\tan \theta}\text{ then }, \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) =\]
\[\text{ If }\cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{\theta}{2}, \text{ then }4 x^2 - 12xy \cos\frac{\theta}{2} + 9 y^2 =\]
Prove that : \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) = \frac{\pi}{4} + \frac{1}{2} \cos^{- 1} x^2 ; 1 < x < 1\].
