English

Find the real solutions of the equation tan-1x(x+1)+sin-1x2+x+1=π2 - Mathematics

Advertisements
Advertisements

Question

Find the real solutions of the equation
`tan^-1 sqrt(x(x + 1)) + sin^-1 sqrt(x^2 + x + 1) = pi/2`

Sum
Advertisements

Solution

We have `tan^-1 sqrt(x(x + 1)) + sin^-1 sqrt(x^2 + x + 1) = pi/2`

⇒ `tan^-1 sqrt(x(x +1)) = pi/2 - sin^-1 sqrt(x^2 + x + 1)`

= `cos^-1 sqrt(x^2 + x + 1)`

= `tan^-1  sqrt(-x^2 - x)/sqrt(x^2 +x + 1)`  ....(From the figure)

⇒ `sqrt(x(x + 1)) = sqrt(-x^2 - x)/sqrt(x^2 + x + 1)`

⇒ `x^2 + x` = 0

⇒ x = 0, –1

shaalaa.com
  Is there an error in this question or solution?
Chapter 2: Inverse Trigonometric Functions - Exercise [Page 36]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 2 Inverse Trigonometric Functions
Exercise | Q 7 | Page 36

RELATED QUESTIONS

​Find the principal values of the following:
`cos^-1(-sqrt3/2)`


​Find the principal values of the following:

`cos^-1(-1/sqrt2)`


`sin^-1(sin  pi/6)`


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


Evaluate the following:

`cos^-1{cos  (5pi)/4}`


Evaluate the following:

`cos^-1(cos5)`


Evaluate the following:

`tan^-1(tan  (9pi)/4)`


Evaluate the following:

`tan^-1(tan2)`


Evaluate the following:

`sec^-1(sec  (5pi)/4)`


Write the following in the simplest form:

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


Prove the following result

`tan(cos^-1  4/5+tan^-1  2/3)=17/6`


Prove the following result-

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


Evaluate:

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


Evaluate:

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


If `(sin^-1x)^2+(cos^-1x)^2=(17pi^2)/36,`  Find x


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


Prove the following result:

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


Solve the following:

`cos^-1x+sin^-1  x/2=π/6`


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


Evaluate the following:

`tan  1/2(cos^-1  sqrt5/3)`


Evaluate the following:

`sin(2tan^-1  2/3)+cos(tan^-1sqrt3)`


`2tan^-1  3/4-tan^-1  17/31=pi/4`


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


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


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


Write the value of sin−1

\[\left( \sin( -{600}°) \right)\].

 

 


Write the value of cos−1 (cos 6).


Evaluate: \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]


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


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


Write the value of \[\cos\left( \sin^{- 1} x + \cos^{- 1} x \right), \left| x \right| \leq 1\]


Wnte the value of the expression \[\tan\left( \frac{\sin^{- 1} x + \cos^{- 1} x}{2} \right), \text { when } x = \frac{\sqrt{3}}{2}\]


The set of values of `\text(cosec)^-1(sqrt3/2)`


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

 


If \[3\sin^{- 1} \left( \frac{2x}{1 + x^2} \right) - 4 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) + 2 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) = \frac{\pi}{3}\] is equal to

 


Prove that : \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) = \frac{\pi}{4} + \frac{1}{2} \cos^{- 1} x^2 ;  1 < x < 1\].


Prove that : \[\cot^{- 1} \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = \frac{x}{2}, 0 < x < \frac{\pi}{2}\] .


Find the domain of `sec^(-1) x-tan^(-1)x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×