English

If α = Tan − 1 ( Tan 5 π 4 ) and β = Tan − 1 ( − Tan 2 π 3 ) , Then (A) 4 α = 3 β (B) 3 α = 4 β (C) α − β = 7 π 12 (D) None of These - Mathematics

Advertisements
Advertisements

Question

If α = \[\tan^{- 1} \left( \tan\frac{5\pi}{4} \right) \text{ and }\beta = \tan^{- 1} \left( - \tan\frac{2\pi}{3} \right)\] , then

 

Options

  • 4 α = 3 β

  • 3 α = 4 β

  • α − β = `(7pi)/12`

  • none of these

MCQ
Advertisements

Solution

(a) 4 α = 3 β
We know that 

\[\tan^{- 1} \left( \tan{x} \right) = x\]
\[\therefore \alpha = \tan^{- 1} \left( \tan\frac{5\pi}{4} \right)\]
\[ = \tan^{- 1} \left\{ \tan\left( \pi + \frac{\pi}{4} \right) \right\}\]
\[ = \tan^{- 1} \left( \tan\frac{\pi}{4} \right)\]
\[ = \frac{\pi}{4}\]
and
\[\beta = \tan^{- 1} \left\{ - \tan\left( \frac{2\pi}{3} \right) \right\}\]
\[ = \tan^{- 1} \left\{ - \tan\left( \pi - \frac{\pi}{3} \right) \right\}\]
\[ = \tan^{- 1} \left\{ \tan\left( \frac{\pi}{3} \right) \right\}\]
\[ = \frac{\pi}{3}\]
\[\therefore 4\alpha = \pi\]
\[3\beta = \pi\]
∴ \[4\alpha = 3\beta\]

 

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Inverse Trigonometric Functions - Exercise 4.16 [Page 120]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 4 Inverse Trigonometric Functions
Exercise 4.16 | Q 9 | Page 120

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`


`sin^-1(sin  (5pi)/6)`


`sin^-1(sin4)`


Evaluate the following:

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


Evaluate the following:

`cos^-1{cos  (13pi)/6}`


Evaluate the following:

`cos^-1(cos4)`


Evaluate the following:

`tan^-1(tan  pi/3)`


Evaluate the following:

`tan^-1(tan  (6pi)/7)`


Evaluate the following:

`sec^-1(sec  (2pi)/3)`


Evaluate:

`sin(tan^-1x+tan^-1  1/x)` for x > 0


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


Prove the following result:

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


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


Solve the following:

`cos^-1x+sin^-1  x/2=π/6`


Solve the equation `cos^-1  a/x-cos^-1  b/x=cos^-1  1/b-cos^-1  1/a`


Evaluate the following:

`tan{2tan^-1  1/5-pi/4}`


Evaluate the following:

`sin(1/2cos^-1  4/5)`


For any a, b, x, y > 0, prove that:

`2/3tan^-1((3ab^2-a^3)/(b^3-3a^2b))+2/3tan^-1((3xy^2-x^3)/(y^3-3x^2y))=tan^-1  (2alphabeta)/(alpha^2-beta^2)`

`where  alpha =-ax+by, beta=bx+ay`


Write the difference between maximum and minimum values of  sin−1 x for x ∈ [− 1, 1].


Write the range of tan−1 x.


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


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


Write the value of cos−1 (cos 6).


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


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


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


The value of tan \[\left\{ \cos^{- 1} \frac{1}{5\sqrt{2}} - \sin^{- 1} \frac{4}{\sqrt{17}} \right\}\] is

 


If tan−1 3 + tan−1 x = tan−1 8, then x =


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

 


\[\cot\left( \frac{\pi}{4} - 2 \cot^{- 1} 3 \right) =\] 

 


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

 


Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×