Advertisements
Advertisements
प्रश्न
Find the value of the following:
`tan^-1{2cos(2sin^-1 1/2)}`
उत्तर
Let `sin^-1 1/2=y`
Then,
`siny=1/2`
`thereforetan^-1{2cos(2sin^-1 1/2)}=tan^-1{2cos2y}`
`=tan^-1(2(1-2sin^2y))` `[becausecos2x=1-2sin^2x]`
`=tan^-1{2(1-2xx1/4)}` `[becausesiny=1/2]`
`=tan^-1{2xx1/2}`
`=tan^-1 1`
`=pi/4`
`thereforetan^-1{2cos(2sin^-1 1/2)}=pi/4`
APPEARS IN
संबंधित प्रश्न
If sin [cot−1 (x+1)] = cos(tan−1x), then find x.
If `tan^(-1)((x-2)/(x-4)) +tan^(-1)((x+2)/(x+4))=pi/4` ,find the value of x
Find the principal values of the following:
`cos^-1(-1/sqrt2)`
`sin^-1{(sin - (17pi)/8)}`
`sin^-1(sin4)`
Evaluate the following:
`cos^-1(cos5)`
Evaluate the following:
`cot^-1{cot ((21pi)/4)}`
Write the following in the simplest form:
`sin{2tan^-1sqrt((1-x)/(1+x))}`
Evaluate the following:
`tan(cos^-1 8/17)`
Evaluate the following:
`cot(cos^-1 3/5)`
Evaluate:
`cos{sin^-1(-7/25)}`
Evaluate:
`cos(sec^-1x+\text(cosec)^-1x)`,|x|≥1
If `sin^-1x+sin^-1y=pi/3` and `cos^-1x-cos^-1y=pi/6`, find the values of x and y.
`5tan^-1x+3cot^-1x=2x`
Find the value of `tan^-1 (x/y)-tan^-1((x-y)/(x+y))`
Solve the following equation for x:
tan−1(x −1) + tan−1x tan−1(x + 1) = tan−13x
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: `cos(sin^-1 3/5+sin^-1 5/13)`
`tan^-1 2/3=1/2tan^-1 12/5`
`2tan^-1 1/5+tan^-1 1/8=tan^-1 4/7`
Write the value of `sin^-1((-sqrt3)/2)+cos^-1((-1)/2)`
If `sin^-1x+sin^-1y+sin^-1z=(3pi)/2,` then write the value of x + y + z.
If x > 1, then write the value of sin−1 `((2x)/(1+x^2))` in terms of tan−1 x.
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 value of cos\[\left( 2 \sin^{- 1} \frac{1}{3} \right)\]
Write the principal value of \[\cos^{- 1} \left( \cos\frac{2\pi}{3} \right) + \sin^{- 1} \left( \sin\frac{2\pi}{3} \right)\]
The set of values of `\text(cosec)^-1(sqrt3/2)`
Write the value of `cot^-1(-x)` for all `x in R` in terms of `cot^-1(x)`
If sin−1 x − cos−1 x = `pi/6` , then x =
If α = \[\tan^{- 1} \left( \tan\frac{5\pi}{4} \right) \text{ and }\beta = \tan^{- 1} \left( - \tan\frac{2\pi}{3} \right)\] , then
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 =
In a ∆ ABC, if C is a right angle, then
\[\tan^{- 1} \left( \frac{a}{b + c} \right) + \tan^{- 1} \left( \frac{b}{c + a} \right) =\]
\[\cot\left( \frac{\pi}{4} - 2 \cot^{- 1} 3 \right) =\]
Prove that : \[\cot^{- 1} \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = \frac{x}{2}, 0 < x < \frac{\pi}{2}\] .
If \[\tan^{- 1} \left( \frac{1}{1 + 1 . 2} \right) + \tan^{- 1} \left( \frac{1}{1 + 2 . 3} \right) + . . . + \tan^{- 1} \left( \frac{1}{1 + n . \left( n + 1 \right)} \right) = \tan^{- 1} \theta\] , then find the value of θ.
Find the domain of `sec^(-1) x-tan^(-1)x`