English

Evaluate: `Cot{Sec^-1(-13/5)}` - Mathematics

Advertisements
Advertisements

Question

Evaluate:

`cot{sec^-1(-13/5)}`

Advertisements

Solution

`cot{sec^-1(-13/5)}=cot{sec^-1(pi-13/5)}`

`=-cot{sec^-1(13/5)}`

`=-cot{tan^-1(sqrt(1-(5/13)^3)/(5/13))}`

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

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

`=-5/12`

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Inverse Trigonometric Functions - Exercise 4.09 [Page 58]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 4 Inverse Trigonometric Functions
Exercise 4.09 | Q 1.3 | Page 58

RELATED QUESTIONS

 

Prove that :

`2 tan^-1 (sqrt((a-b)/(a+b))tan(x/2))=cos^-1 ((a cos x+b)/(a+b cosx))`

 

`sin^-1(sin12)`


Evaluate the following:

`cos^-1{cos(-pi/4)}`


Evaluate the following:

`sec^-1(sec  pi/3)`


Evaluate the following:

`sec^-1{sec  (-(7pi)/3)}`


Evaluate the following:

`cot^-1(cot  (4pi)/3)`


Write the following in the simplest form:

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


Evaluate the following:

`sin(sin^-1  7/25)`

 


Solve: `cos(sin^-1x)=1/6`


Evaluate:

`cos{sin^-1(-7/25)}`


Evaluate:

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


Prove the following result:

`tan^-1  1/7+tan^-1  1/13=tan^-1  2/9`


Solve the following equation for x:

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


Prove that:

`2sin^-1  3/5=tan^-1  24/7`


`2tan^-1  1/5+tan^-1  1/8=tan^-1  4/7`


`4tan^-1  1/5-tan^-1  1/239=pi/4`


If `sin^-1  (2a)/(1+a^2)+sin^-1  (2b)/(1+b^2)=2tan^-1x,` Prove that  `x=(a+b)/(1-ab).`


Find the value of the following:

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


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:

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


Write the value of sin−1

\[\left( \sin( -{600}°) \right)\].

 

 


Write the value of cos\[\left( 2 \sin^{- 1} \frac{1}{3} \right)\]


Write the value of cos1 (cos 350°) − sin−1 (sin 350°)


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


Write the value of cos−1 \[\left( \cos\frac{5\pi}{4} \right)\]


Show that \[\sin^{- 1} (2x\sqrt{1 - x^2}) = 2 \sin^{- 1} x\]


Write the value of \[\sin^{- 1} \left( \frac{1}{3} \right) - \cos^{- 1} \left( - \frac{1}{3} \right)\]


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


Write the value of \[\cos^{- 1} \left( \cos\frac{14\pi}{3} \right)\]


If  \[\cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = \alpha, then\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = \]


The positive integral solution of the equation
\[\tan^{- 1} x + \cos^{- 1} \frac{y}{\sqrt{1 + y^2}} = \sin^{- 1} \frac{3}{\sqrt{10}}\text{ is }\]


The number of solutions of the equation \[\tan^{- 1} 2x + \tan^{- 1} 3x = \frac{\pi}{4}\] 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

 


If \[\sin^{- 1} \left( \frac{2a}{1 - a^2} \right) + \cos^{- 1} \left( \frac{1 - a^2}{1 + a^2} \right) = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right),\text{ where }a, x \in \left( 0, 1 \right)\] , then, the value of x is

 


Prove that : \[\cot^{- 1} \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = \frac{x}{2}, 0 < x < \frac{\pi}{2}\] .


If 2 tan−1 (cos θ) = tan−1 (2 cosec θ), (θ ≠ 0), then find the value of θ.


Find the value of x, if tan `[sec^(-1) (1/x) ] = sin ( tan^(-1) 2) , x > 0 `.


Find the value of `sin^-1(cos((33π)/5))`.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×