Advertisements
Advertisements
प्रश्न
Evaluate the following:
`cos^-1{cos (5pi)/4}`
Advertisements
उत्तर
We know
`cos^-1(costheta)=thetaif 0<=theta<=pi`
We have
`cos^-1{cos (5pi)/4}=cos^-1{cos(2pi-(3pi)/4)}`
`=cos^-1{cos((3pi)/4)}`
`=(3pi)/4`
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))`
`sin^-1(sin4)`
Evaluate the following:
`cos^-1{cos (13pi)/6}`
Evaluate the following:
`tan^-1(tan pi/3)`
Evaluate the following:
`sec^-1(sec pi/3)`
Evaluate the following:
`sec^-1{sec (-(7pi)/3)}`
Evaluate the following:
`cot^-1(cot (9pi)/4)`
Write the following in the simplest form:
`tan^-1{sqrt(1+x^2)-x},x in R`
Evaluate the following:
`sin(cos^-1 5/13)`
Evaluate the following:
`tan(cos^-1 8/17)`
Evaluate the following:
`cos(tan^-1 24/7)`
Prove the following result
`cos(sin^-1 3/5+cot^-1 3/2)=6/(5sqrt13)`
Evaluate:
`sec{cot^-1(-5/12)}`
Evaluate:
`cosec{cot^-1(-12/5)}`
If `(sin^-1x)^2+(cos^-1x)^2=(17pi^2)/36,` Find x
`5tan^-1x+3cot^-1x=2x`
Solve the following equation for x:
`tan^-1 2x+tan^-1 3x = npi+(3pi)/4`
Evaluate: `cos(sin^-1 3/5+sin^-1 5/13)`
`sin^-1 63/65=sin^-1 5/13+cos^-1 3/5`
`tan^-1 1/4+tan^-1 2/9=1/2cos^-1 3/2=1/2sin^-1(4/5)`
`tan^-1 2/3=1/2tan^-1 12/5`
Show that `2tan^-1x+sin^-1 (2x)/(1+x^2)` is constant for x ≥ 1, find that constant.
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 sin (cot−1 x).
Write the range of tan−1 x.
Evaluate sin \[\left( \tan^{- 1} \frac{3}{4} \right)\]
Write the value of cos2 \[\left( \frac{1}{2} \cos^{- 1} \frac{3}{5} \right)\]
Write the value of \[\tan^{- 1} \frac{a}{b} - \tan^{- 1} \left( \frac{a - b}{a + b} \right)\]
If 4 sin−1 x + cos−1 x = π, then what is the value of x?
Write the value of \[\tan\left( 2 \tan^{- 1} \frac{1}{5} \right)\]
The set of values of `\text(cosec)^-1(sqrt3/2)`
If θ = sin−1 {sin (−600°)}, then one of the possible values of θ 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
It \[\tan^{- 1} \frac{x + 1}{x - 1} + \tan^{- 1} \frac{x - 1}{x} = \tan^{- 1}\] (−7), then the value of x is
If \[\cos^{- 1} x > \sin^{- 1} x\], then
Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .
Prove that : \[\cot^{- 1} \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = \frac{x}{2}, 0 < x < \frac{\pi}{2}\] .
