Advertisements
Advertisements
प्रश्न
`sin^-1(sin (5pi)/6)`
उत्तर
We know
`sin(sin^-1theta)=theta if - pi/2<=theta<=pi/2`
We have
`sin^-1(sin (5pi)/6)=sin^-1{sin(pi-pi/6)}`
`=sin^-1(sin pi/6)`
`=pi/6`
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))`
Find the principal values of the following:
`cos^-1(tan (3pi)/4)`
Evaluate the following:
`tan^-1(tan pi/3)`
Evaluate the following:
`tan^-1(tan (6pi)/7)`
Evaluate the following:
`tan^-1(tan (7pi)/6)`
Evaluate the following:
`tan^-1(tan (9pi)/4)`
Evaluate the following:
`sec^-1(sec (2pi)/3)`
Evaluate the following:
`sec^-1(sec (7pi)/3)`
Evaluate the following:
`sec^-1(sec (9pi)/5)`
Evaluate the following:
`sec^-1(sec (13pi)/4)`
Evaluate the following:
`cosec^-1(cosec (6pi)/5)`
Prove the following result
`sin(cos^-1 3/5+sin^-1 5/13)=63/65`
Evaluate:
`cosec{cot^-1(-12/5)}`
Evaluate:
`cot(sin^-1 3/4+sec^-1 4/3)`
`sin^-1x=pi/6+cos^-1x`
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(x −1) + tan−1x tan−1(x + 1) = tan−13x
Solve the following equation for x:
cot−1x − cot−1(x + 2) =`pi/12`, x > 0
Prove that: `cos^-1 4/5+cos^-1 12/13=cos^-1 33/65`
Prove that
`tan^-1((1-x^2)/(2x))+cot^-1((1-x^2)/(2x))=pi/2`
Find the value of the following:
`cos(sec^-1x+\text(cosec)^-1x),` | x | ≥ 1
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 value of sin−1 (sin 1550°).
Write the value of tan−1\[\left\{ \tan\left( \frac{15\pi}{4} \right) \right\}\]
Write the principal value of \[\cos^{- 1} \left( \cos\frac{2\pi}{3} \right) + \sin^{- 1} \left( \sin\frac{2\pi}{3} \right)\]
Write the value of \[\cos^{- 1} \left( \cos\frac{14\pi}{3} \right)\]
Write the value of \[\tan^{- 1} \left( \frac{1}{x} \right)\] for x < 0 in terms of `cot^-1x`
Write the value of `cot^-1(-x)` for all `x in R` in terms of `cot^-1(x)`
Wnte the value of\[\cos\left( \frac{\tan^{- 1} x + \cot^{- 1} x}{3} \right), \text{ when } x = - \frac{1}{\sqrt{3}}\]
The value of tan \[\left\{ \cos^{- 1} \frac{1}{5\sqrt{2}} - \sin^{- 1} \frac{4}{\sqrt{17}} \right\}\] is
sin\[\left[ \cot^{- 1} \left\{ \tan\left( \cos^{- 1} x \right) \right\} \right]\] is equal to
\[\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11}\] is equal to
The value of \[\cos^{- 1} \left( \cos\frac{5\pi}{3} \right) + \sin^{- 1} \left( \sin\frac{5\pi}{3} \right)\] is
\[\cot\left( \frac{\pi}{4} - 2 \cot^{- 1} 3 \right) =\]
The value of \[\tan\left( \cos^{- 1} \frac{3}{5} + \tan^{- 1} \frac{1}{4} \right)\]
Prove that : \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) = \frac{\pi}{4} + \frac{1}{2} \cos^{- 1} x^2 ; 1 < x < 1\].
Find the simplified form of `cos^-1 (3/5 cosx + 4/5 sin x)`, where x ∈ `[(-3pi)/4, pi/4]`
