मराठी

Write the Value of Sin (Cot−1 X). - Mathematics

Advertisements
Advertisements

प्रश्न

Write the value of sin (cot−1 x).

उत्तर

We know

\[\cot^{- 1} x = \tan^{- 1} \frac{1}{x}\]

Now, we have

\[\sin\left( \cot^{- 1} x \right) = \sin\left( \tan^{- 1} \frac{1}{x} \right)\]
\[ = \sin\left[ \sin^{- 1} \left( \frac{\frac{1}{x}}{\sqrt{1 + \frac{1}{x^2}}} \right) \right] \left[ \because \tan^{- 1} x = \sin^{- 1} \left( \frac{x}{\sqrt{1 + x^2}} \right) \right]\]
\[ = \sin\left[ \sin^{- 1} \left( \frac{\frac{1}{x}}{\frac{\sqrt{x^2 + 1}}{x}} \right) \right]\]
\[ = \sin\left( \sin^{- 1} \frac{1}{\sqrt{x^2 + 1}} \right)\]
\[ = \frac{1}{\sqrt{x^2 + 1}} \left[ \because \sin\left( \sin^{- 1} x = x \right) \right]\]

Hence, 

\[\sin\left( \cot^{- 1} x \right) = \frac{1}{\sqrt{x^2 - 1}}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Inverse Trigonometric Functions - Exercise 4.15 [पृष्ठ ११७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.15 | Q 10 | पृष्ठ ११७

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Write the value of `tan(2tan^(-1)(1/5))`


 

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(-sqrt3/2)`


​Find the principal values of the following:

`cos^-1(-1/sqrt2)`


Evaluate the following:

`cos^-1{cos  ((4pi)/3)}`


Evaluate the following:

`cos^-1{cos  (13pi)/6}`


Evaluate the following:

`tan^-1(tan4)`


Evaluate the following:

`cosec^-1(cosec  (13pi)/6)`


Evaluate the following:

`cosec^-1{cosec  (-(9pi)/4)}`


Evaluate the following:

`cot^-1(cot  (9pi)/4)`


Write the following in the simplest form:

`sin{2tan^-1sqrt((1-x)/(1+x))}`


Evaluate the following:

`sin(tan^-1  24/7)`


Evaluate the following:

`cosec(cos^-1  3/5)`


Evaluate:

`sec{cot^-1(-5/12)}`


Evaluate:

`sin(tan^-1x+tan^-1  1/x)` for x < 0


Evaluate:

`sin(tan^-1x+tan^-1  1/x)` for x > 0


`5tan^-1x+3cot^-1x=2x`


Solve the following equation for x:

tan−1(x + 1) + tan−1(x − 1) = tan−1`8/31`


`(9pi)/8-9/4sin^-1  1/3=9/4sin^-1  (2sqrt2)/3`


Prove that:

`2sin^-1  3/5=tan^-1  24/7`


`2tan^-1  3/4-tan^-1  17/31=pi/4`


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


Find the value of the following:

`tan^-1{2cos(2sin^-1  1/2)}`


Find the value of the following:

`cos(sec^-1x+\text(cosec)^-1x),` | x | ≥ 1


Solve the following equation for x:

`2tan^-1(sinx)=tan^-1(2sinx),x!=pi/2`


Write the value of cos−1 (cos 6).


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

 


If 4 sin−1 x + cos−1 x = π, then what is the value of x?


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


Write the value of \[\cos\left( \sin^{- 1} x + \cos^{- 1} x \right), \left| x \right| \leq 1\]


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


If \[\cos\left( \sin^{- 1} \frac{2}{5} + \cos^{- 1} x \right) = 0\], find the value of x.

 

If sin−1 − cos−1 x = `pi/6` , then x = 


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 =


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


If 4 cos−1 x + sin−1 x = π, then the value of x is

 


If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find \[\frac{dy}{dx}\] When  \[\theta = \frac{\pi}{3}\] .


Find the simplified form of `cos^-1 (3/5 cosx + 4/5 sin x)`, where x ∈ `[(-3pi)/4, pi/4]`


tanx is periodic with period ____________.


The value of sin `["cos"^-1 (7/25)]` is ____________.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×