हिंदी

Sin { 2 Cos − 1 ( − 3 5 ) } is Equal to (A) 6 25 (B) 24 25 (C) 4 5 (D) − 24 25 - Mathematics

Advertisements
Advertisements

प्रश्न

sin \[\left\{ 2 \cos^{- 1} \left( \frac{- 3}{5} \right) \right\}\]  is equal to

 

विकल्प

  • `6/25`

  • `24/25`

  • `4/5`

  • `-24/25`

MCQ

उत्तर

(d) `-24/25`

Let \[\cos^{- 1} \left( - \frac{3}{5} \right) = x, 0 \leq x \leq \pi\]
Then,
`cosx=-3/5`
\[\therefore \sin{x} = \sqrt{1 - \cos^2 x} = \sqrt{1 - \left( - \frac{3}{5} \right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5}\]
Now,
\[\sin\left\{ 2 \cos^{- 1} \left( - \frac{3}{5} \right) \right\} = \sin\left( 2x \right)\]
\[ = 2\sin{x} \cos{x}\]
\[ = 2 \times \frac{4}{5} \times \frac{- 3}{5}\]
\[ = - \frac{24}{25}\]



shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Inverse Trigonometric Functions - Exercise 4.16 [पृष्ठ १२१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 4 Inverse Trigonometric Functions
Exercise 4.16 | Q 21 | पृष्ठ १२१

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

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


Solve for x:

`2tan^(-1)(cosx)=tan^(-1)(2"cosec" x)`


 

If `tan^(-1)((x-2)/(x-4)) +tan^(-1)((x+2)/(x+4))=pi/4` ,find the value of x

 

Find the domain of `f(x)=cos^-1x+cosx.`


`sin^-1(sin  (13pi)/7)`


Evaluate the following:

`cos^-1(cos4)`


Evaluate the following:

`tan^-1(tan12)`


Evaluate the following:

`cosec^-1(cosec  (13pi)/6)`


Evaluate the following:

`cosec^-1{cosec  (-(9pi)/4)}`


Evaluate the following:

`cot^-1(cot  (4pi)/3)`


Evaluate the following:

`sin(sec^-1  17/8)`


Evaluate the following:

`cot(cos^-1  3/5)`


Prove the following result

`tan(cos^-1  4/5+tan^-1  2/3)=17/6`


Prove the following result

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


`tan^-1x+2cot^-1x=(2x)/3`


Prove the following result:

`sin^-1  12/13+cos^-1  4/5+tan^-1  63/16=pi`


Evaluate the following:

`tan{2tan^-1  1/5-pi/4}`


`tan^-1  2/3=1/2tan^-1  12/5`


`sin^-1  4/5+2tan^-1  1/3=pi/2`


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


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


Solve the following equation for x:

`tan^-1  1/4+2tan^-1  1/5+tan^-1  1/6+tan^-1  1/x=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`


Prove that `2tan^-1(sqrt((a-b)/(a+b))tan  theta/2)=cos^-1((a costheta+b)/(a+b costheta))`


Prove that:

`tan^-1  (2ab)/(a^2-b^2)+tan^-1  (2xy)/(x^2-y^2)=tan^-1  (2alphabeta)/(alpha^2-beta^2),`   where `alpha=ax-by  and  beta=ay+bx.`


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 range of tan−1 x.


Write the value of cos−1 \[\left( \tan\frac{3\pi}{4} \right)\]


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


Write the value of sin−1 \[\left( \cos\frac{\pi}{9} \right)\]


If \[\sin^{- 1} \left( \frac{1}{3} \right) + \cos^{- 1} x = \frac{\pi}{2},\] then find x.

 


What is the principal value of `sin^-1(-sqrt3/2)?`


Write the principal value of `sin^-1(-1/2)`


Find the value of \[\tan^{- 1} \left( \tan\frac{9\pi}{8} \right)\]


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 tan−1 (cot θ) = 2 θ, then θ =

 


The domain of  \[\cos^{- 1} \left( x^2 - 4 \right)\] is

 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×