English

Evaluate the Following: `Sec^-1(Sec (7pi)/3)` - Mathematics

Advertisements
Advertisements

Question

Evaluate the following:

`sec^-1(sec  (7pi)/3)`

Advertisements

Solution

We know that

sec-1 (sec θ) = θ,    [0, π/2) ∪ (π/2, π]

 We have 

`sec^-1(sec  (7pi)/3)=sec^-1[sec(2pi+pi/3)]`

`=sec^-1[sec(pi/3)]`

`=pi/3`

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Inverse Trigonometric Functions - Exercise 4.07 [Page 42]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 4 Inverse Trigonometric Functions
Exercise 4.07 | Q 4.4 | Page 42

RELATED QUESTIONS

Find the value of the following: `tan(1/2)[sin^(-1)((2x)/(1+x^2))+cos^(-1)((1-y^2)/(1+y^2))],|x| <1,y>0 and xy <1`


Solve the following for x :

`tan^(-1)((x-2)/(x-3))+tan^(-1)((x+2)/(x+3))=pi/4,|x|<1`


If (tan1x)2 + (cot−1x)2 = 5π2/8, then find x.


If tan-1x+tan-1y=π/4,xy<1, then write the value of x+y+xy.


`sin^-1(sin  (7pi)/6)`


`sin^-1(sin  (5pi)/6)`


Evaluate the following:

`cos^-1{cos  (5pi)/4}`


Evaluate the following:

`cos^-1(cos4)`


Evaluate the following:

`tan^-1(tan4)`


Evaluate the following:

`sec^-1(sec  (25pi)/6)`


Evaluate the following:

`cosec^-1(cosec  (3pi)/4)`


Evaluate the following:

`cosec^-1{cosec  (-(9pi)/4)}`


Evaluate the following:

`cot^-1{cot (-(8pi)/3)}`


Write the following in the simplest form:

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


Write the following in the simplest form:

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


Evaluate the following:

`cosec(cos^-1  3/5)`


Evaluate:

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


Evaluate: 

`cot(sin^-1  3/4+sec^-1  4/3)`


`sin(sin^-1  1/5+cos^-1x)=1`


`sin^-1x=pi/6+cos^-1x`


Prove the following result:

`tan^-1  1/4+tan^-1  2/9=sin^-1  1/sqrt5`


Solve the following equation for x:

`tan^-1  2x+tan^-1  3x = npi+(3pi)/4`


Solve the following equation for x:

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


Solve the following:

`cos^-1x+sin^-1  x/2=π/6`


`4tan^-1  1/5-tan^-1  1/239=pi/4`


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`


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


Write the value of sin1 (sin 1550°).


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


Write the value of tan1\[\left\{ \tan\left( \frac{15\pi}{4} \right) \right\}\]


If x < 0, y < 0 such that xy = 1, then write the value of tan1 x + tan−1 y.


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 \[\sec^{- 1} \left( \frac{1}{2} \right)\]


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


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


If θ = sin−1 {sin (−600°)}, then one of the possible values of θ is

 


In a ∆ ABC, if C is a right angle, then
\[\tan^{- 1} \left( \frac{a}{b + c} \right) + \tan^{- 1} \left( \frac{b}{c + a} \right) =\]

 

 


If tan−1 (cot θ) = 2 θ, then θ =

 


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

 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×