हिंदी

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

Advertisements
Advertisements

प्रश्न

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

उत्तर

LHS = `2tan^-1(1/2)+tan^-1(1/7)`

`=tan^-1{(2xx1/2)/(1-(1/2)^2)}+tan^-1  1/7`     `[because2tan^-1x=tan^-1{(2x)/(1-x^2)}]`

`=tan^-1{1/(3/4)}+tan^-1  1/7`

`=tan^-1  4/3+tan^-1  1/7`

`=tan^-1((4/3+1/7)/(1-4/3xx1/7))`       `[because tan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))]`

`=tan^-1((31/21)/(17/21))`

`=tan^-1  31/17=`RHS

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Inverse Trigonometric Functions - Exercise 4.14 [पृष्ठ ११५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 4 Inverse Trigonometric Functions
Exercise 4.14 | Q 2.09 | पृष्ठ ११५

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

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`


Prove that

`tan^(-1) [(sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x))]=pi/4-1/2 cos^(-1)x,-1/sqrt2<=x<=1`


 

If `tan^(-1)((x-2)/(x-4)) +tan^(-1)((x+2)/(x+4))=pi/4` ,find the value of x

 

`sin^-1(sin  (5pi)/6)`


Evaluate the following:

`tan^-1(tan  (7pi)/6)`


Evaluate the following:

`sec^-1{sec  (-(7pi)/3)}`


Evaluate the following:

`cosec^-1(cosec  (6pi)/5)`


Evaluate the following:

`cot^-1{cot  ((21pi)/4)}`


Write the following in the simplest form:

`cot^-1  a/sqrt(x^2-a^2),|  x  | > a`


Write the following in the simplest form:

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


Write the following in the simplest form:

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


Write the following in the simplest form:

`tan^-1sqrt((a-x)/(a+x)),-a<x<a`


Write the following in the simplest form:

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


Evaluate the following:

`sin(cos^-1  5/13)`


Evaluate the following:

`tan(cos^-1  8/17)`


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


Evaluate:

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


Evaluate: 

`cot(sin^-1  3/4+sec^-1  4/3)`


Solve the following equation for x:

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


Evaluate: `cos(sin^-1  3/5+sin^-1  5/13)`


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


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


Evaluate the following:

`tan{2tan^-1  1/5-pi/4}`


Evaluate the following:

`sin(1/2cos^-1  4/5)`


`tan^-1  1/7+2tan^-1  1/3=pi/4`


`2sin^-1  3/5-tan^-1  17/31=pi/4`


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`


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


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


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


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 value of  `cot^-1(-x)`  for all `x in R` in terms of `cot^-1(x)`


2 tan−1 {cosec (tan−1 x) − tan (cot1 x)} is equal to


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

 


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

 


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


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


The value of tan `("cos"^-1  4/5 + "tan"^-1  2/3) =`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×