हिंदी

`Sin^-1 63/65=Sin^-1 5/13+Cos^-1 3/5` - Mathematics

Advertisements
Advertisements

प्रश्न

`sin^-1  63/65=sin^-1  5/13+cos^-1  3/5`

Advertisements

उत्तर

RHS

      `sin^-1  5/13+cos^-1  3/5`

`=sin^-1  5/13+sin^-1  4/5`        `[because cos^-1x=sin^-1sqrt(1-x^2)]`

`=sin^-1{5/13sqrt(1-(4/5)^2)+4/5sqrt(1-(5/13)^2)}`

`=sin^-1{5/13xx3/5+4/5xx12/13}`

`=sin^-1{15/65+48/65}`

`=sin^-1  63/65=`LHS

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Inverse Trigonometric Functions - Exercise 4.12 [पृष्ठ ८९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 4 Inverse Trigonometric Functions
Exercise 4.12 | Q 2.1 | पृष्ठ ८९

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

Solve the equation for x:sin1x+sin1(1x)=cos1x


If `cos^-1( x/a) +cos^-1 (y/b)=alpha` , prove that `x^2/a^2-2(xy)/(ab) cos alpha +y^2/b^2=sin^2alpha`


Prove that

`tan^(-1) [(sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x))]=pi/4-1/2 cos^(-1)x,-1/sqrt2<=x<=1`


If a line makes angles 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.


If `(sin^-1x)^2 + (sin^-1y)^2+(sin^-1z)^2=3/4pi^2,`  find the value of x2 + y2 + z2 


`sin^-1(sin2)`


Evaluate the following:

`cos^-1{cos  (13pi)/6}`


Evaluate the following:

`tan^-1(tan  (7pi)/6)`


Evaluate the following:

`sec^-1(sec  pi/3)`


Evaluate the following:

`\text(cosec)^-1(\text{cosec}  pi/4)`


Write the following in the simplest form:

`tan^-1{x+sqrt(1+x^2)},x in R `


Write the following in the simplest form:

`tan^-1{(sqrt(1+x^2)-1)/x},x !=0`


Write the following in the simplest form:

`sin^-1{(sqrt(1+x)+sqrt(1-x))/2},0<x<1`


Prove the following result

`cos(sin^-1  3/5+cot^-1  3/2)=6/(5sqrt13)`


Evaluate: `sin{cos^-1(-3/5)+cot^-1(-5/12)}`


If `cos^-1x + cos^-1y =pi/4,`  find the value of `sin^-1x+sin^-1y`


Solve the following equation for x:

`tan^-1(2+x)+tan^-1(2-x)=tan^-1  2/3, where  x< -sqrt3 or, x>sqrt3`


Evaluate: `cos(sin^-1  3/5+sin^-1  5/13)`


`tan^-1  1/4+tan^-1  2/9=1/2cos^-1  3/2=1/2sin^-1(4/5)`


`2tan^-1(1/2)+tan^-1(1/7)=tan^-1(31/17)`


Solve the following equation for x:

`3sin^-1  (2x)/(1+x^2)-4cos^-1  (1-x^2)/(1+x^2)+2tan^-1  (2x)/(1-x^2)=pi/3`


Solve the following equation for x:

`tan^-1((x-2)/(x-1))+tan^-1((x+2)/(x+1))=pi/4`


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 −1 < x < 0, then write the value of `sin^-1((2x)/(1+x^2))+cos^-1((1-x^2)/(1+x^2))`


Evaluate sin \[\left( \tan^{- 1} \frac{3}{4} \right)\]


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.

 


The set of values of `\text(cosec)^-1(sqrt3/2)`


Find the value of \[2 \sec^{- 1} 2 + \sin^{- 1} \left( \frac{1}{2} \right)\]


If x < 0, y < 0 such that xy = 1, then tan−1 x + tan−1 y equals

 


Let f (x) = `e^(cos^-1){sin(x+pi/3}.`
Then, f (8π/9) = 


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

 


The value of \[\tan\left( \cos^{- 1} \frac{3}{5} + \tan^{- 1} \frac{1}{4} \right)\]

 


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.


Write the value of \[\cos^{- 1} \left( - \frac{1}{2} \right) + 2 \sin^{- 1} \left( \frac{1}{2} \right)\] .


Find the real solutions of the equation
`tan^-1 sqrt(x(x + 1)) + sin^-1 sqrt(x^2 + x + 1) = pi/2`


The equation sin-1 x – cos-1 x = cos-1 `(sqrt3/2)` has ____________.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×