मराठी

If Tan−1 X + Tan−1 Y = `Pi/4`, Then Write the Value Of X + Y + Xy. - Mathematics

Advertisements
Advertisements

प्रश्न

If tan−1 x + tan−1 y = `pi/4`,  then write the value of x + y + xy.

उत्तर

We know that 

\[\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \left( \frac{x + y}{1 - xy} \right)\]
Now,
\[\tan^{- 1} x + \tan^{- 1} y = \frac{\pi}{4}\]
\[ \Rightarrow \tan^{- 1} \left( \frac{x + y}{1 - xy} \right) = \frac{\pi}{4}\]
\[ \Rightarrow \frac{x + y}{1 - xy} = \tan\frac{\pi}{4}\]
\[ \Rightarrow \frac{x + y}{1 - xy} = 1 \]
\[ \Rightarrow x + y = 1 - xy\]
\[ \Rightarrow x + y + xy = 1\]
∴ \[x + y + xy = 1\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Inverse Trigonometric Functions - Exercise 4.15 [पृष्ठ ११७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.15 | Q 23 | पृष्ठ ११७

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Solve the following for x :

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


Solve for x:

`2tan^(-1)(cosx)=tan^(-1)(2"cosec" x)`


`sin^-1{(sin - (17pi)/8)}`


Evaluate the following:

`cos^-1{cos  (13pi)/6}`


Evaluate the following:

`cos^-1(cos5)`


Evaluate the following:

`tan^-1(tan2)`


Evaluate the following:

`cosec^-1(cosec  (3pi)/4)`


Evaluate the following:

`cosec^-1(cosec  (6pi)/5)`


Evaluate the following:

`cot^-1(cot  (4pi)/3)`


Evaluate the following:

`cot^-1(cot  (9pi)/4)`


Evaluate the following:

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


Write the following in the simplest form:

`tan^-1(x/(a+sqrt(a^2-x^2))),-a<x<a`


Evaluate the following:

`sin(sin^-1  7/25)`

 


Evaluate the following:

`sin(sec^-1  17/8)`


Evaluate the following:

`sec(sin^-1  12/13)`


Evaluate the following:

`cos(tan^-1  24/7)`


Prove the following result

`tan(cos^-1  4/5+tan^-1  2/3)=17/6`


Evaluate:

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


Evaluate:

`cos(tan^-1  3/4)`


Prove the following result:

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


Prove the following result:

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


Solve the following equation for x:

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


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


If `cos^-1  x/2+cos^-1  y/3=alpha,` then prove that  `9x^2-12xy cosa+4y^2=36sin^2a.`


`tan^-1  1/4+tan^-1  2/9=1/2cos^-1  3/2=1/2sin^-1(4/5)`


Show that `2tan^-1x+sin^-1  (2x)/(1+x^2)` is constant for x ≥ 1, find that constant.


Solve the following equation for x:

`tan^-1  1/4+2tan^-1  1/5+tan^-1  1/6+tan^-1  1/x=pi/4`


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


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


Write the value of cos\[\left( \frac{1}{2} \cos^{- 1} \frac{3}{5} \right)\]


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


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


If \[\sin^{- 1} \left( \frac{1}{3} \right) + \cos^{- 1} x = \frac{\pi}{2},\] then find x.

 


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


If \[\cos\left( \sin^{- 1} \frac{2}{5} + \cos^{- 1} x \right) = 0\], find the value of x.

 

It \[\tan^{- 1} \frac{x + 1}{x - 1} + \tan^{- 1} \frac{x - 1}{x} = \tan^{- 1}\]   (−7), then the value of x is

 


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

 


The period of the function f(x) = tan3x is ____________.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×