English

`Sin^-1 5/13+Cos^-1 3/5=Tan^-1 63/16` - Mathematics

Advertisements
Advertisements

Question

`sin^-1  5/13+cos^-1  3/5=tan^-1  63/16`

Solution

LHS = `sin^-1  5/13+cos^-1  3/5`

`=sin^-1  5/13+cos^-1  3/5`

`=sin^-1  5/13+sin^-1sqrt(1-(3/5)^2)`         `[because sin^-1x=cos^-1sqrt(1-x^2)]`

`=sin^-1  5/13+sin^-1  4/5`

`=sin^-1[5/13sqrt(1-(4/5)^2)+4/5sqrt(1-(5/13)^2)]`          `[because sin^-1x+sin^-1y=sin^-1(xsqrt(1-y^2)+ysqrt(1-x^2))]`

`=sin^-1(5/13xx3/5+4/5xx12/13)`

`=sin^-1(3/13+48/65)`

`=sin^-1(63/65)`

`=tan^-1((63/65)/sqrt(1-(63/65)^2))`     `[becausesin^-1x=tan^-1(x/sqrt(1-x^2))]`

`=tan^-1((63/65)/(16/65))`

`=tan^-1(63/16)=`  RHS

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Inverse Trigonometric Functions - Exercise 4.12 [Page 89]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 4 Inverse Trigonometric Functions
Exercise 4.12 | Q 2.2 | Page 89

RELATED QUESTIONS

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


`sin^-1(sin  (17pi)/8)`


`sin^-1(sin3)`


`sin^-1(sin4)`


Evaluate the following:

`cos^-1{cos  (5pi)/4}`


Evaluate the following:

`cos^-1(cos5)`


Evaluate the following:

`tan^-1(tan  pi/3)`


Evaluate the following:

`tan^-1(tan  (6pi)/7)`


Evaluate the following:

`sec^-1(sec  (7pi)/3)`


Evaluate the following:

`sec^-1(sec  (9pi)/5)`


Evaluate the following:

`sec^-1{sec  (-(7pi)/3)}`


Evaluate the following:

`sec^-1(sec  (25pi)/6)`


Evaluate the following:

`cosec^-1(cosec  (3pi)/4)`


Evaluate the following:

`sin(sin^-1  7/25)`

 


Evaluate the following:

`sec(sin^-1  12/13)`


Evaluate:

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


Evaluate:

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


Evaluate:

`cos(sec^-1x+\text(cosec)^-1x)`,|x|≥1


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


Solve the following equation for x:

`tan^-1  2x+tan^-1  3x = npi+(3pi)/4`


Solve the following equation for x:

tan−1`((1-x)/(1+x))-1/2` tan−1x = 0, where x > 0


Evaluate the following:

`tan{2tan^-1  1/5-pi/4}`


`4tan^-1  1/5-tan^-1  1/239=pi/4`


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`


If `sin^-1x+sin^-1y+sin^-1z=(3pi)/2,`  then write the value of x + y + z.


If x > 1, then write the value of sin−1 `((2x)/(1+x^2))` in terms of tan−1 x.


If x < 0, then write the value of cos−1 `((1-x^2)/(1+x^2))` in terms of tan−1 x.


Write the value of tan1 x + tan−1 `(1/x)`  for x < 0.


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


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


If \[\tan^{- 1} (\sqrt{3}) + \cot^{- 1} x = \frac{\pi}{2},\] find x.


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

 


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

 


\[\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11}\]  is equal to

 

 


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

 


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×