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Inverse Trigonometric Functions Test - 56

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Inverse Trigonometric Functions Test - 56
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  • Question 1
    1 / -0
    The range of $$ f\left( x \right) =\sin { =^{ -1 }x+ } \cos { =^{ -1 } } x+\tan { ^{ -1 } } x$$ is ?
    Solution

  • Question 2
    1 / -0
    The value of $$\sin ^{ -1 }{ \left[ \cot { \left( \sin ^{ -1 }{ \sqrt { \frac { 2-\sqrt { 3 }  }{ 4 }  }  } +\cos ^{ -1 }{ \frac { \sqrt { 2 }  }{ 4 } +\sec ^{ -1 }{ \sqrt { 2 }  }  }  \right)  }  \right]  } $$ is equal to
    Solution

  • Question 3
    1 / -0
    The value of $$\tan { \left\{ 4\tan { ^{ -1 }\left( \dfrac { 1 }{ 5 }  \right) -\tan { ^{ -1 }\left( \dfrac { 1 }{ 239 }  \right)  }  }  \right\}  }$$ is
    Solution

  • Question 4
    1 / -0
    The exhaustive set of values of $$'a'$$ such that $$x^{2}+ax+sing^{-1}\ (x^{2}-4x+5)+\ cos^{-1}(x^{2}-4x+5)=0$$ has at least one solution is
    Solution

  • Question 5
    1 / -0
    If $$\left |{\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)\right |\,\, < \,\dfrac{\pi }{3},\,then\,:$$ 
    Solution

  • Question 6
    1 / -0
    $$\sin^{-1}\left(a-\dfrac{a^2}{3}+\dfrac{a^3}{9}+...\right)+\cos^{-1}(1+b+b^2+...)=\dfrac{\pi}{2}$$ when?
    Solution
    $$\begin{matrix} { \sin ^{ -1 }  }\left( { \frac { a }{ { 1-\left( { -\frac { a }{ 3 }  } \right)  } }  } \right) +{ \cos ^{ -1 }  }\left( { \frac { 1 }{ { 1-b } }  } \right) =\frac { \pi  }{ 2 }  \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \because Angle\, \, are\, \, in\, \, G.P\, \, then\, \, sum\, \, of\, inite\, G.P\,  } \right]  \\ { \sin ^{ -1 }  }\left( { \frac { a }{ { 1+\frac { a }{ 3 }  } }  } \right) +{ \cos ^{ -1 }  }\left( { \frac { 1 }{ { 1-b } }  } \right) =\frac { \pi  }{ 2 } \, is\, \, \left( { \frac { a }{ { 1-r } }  } \right)  \\ i{ n^{ -1 } }\left( { \frac { { 3a } }{ { a+3 } }  } \right) +{ \cos ^{ -1 }  }\left( { \frac { 1 }{ { 1-b } }  } \right) =\frac { \pi  }{ 4 } +\frac { \pi  }{ 4 } .........\left( i \right)  \\ by\, \, inverse\, \, property\, \, if\, angle\, \, of\, \sin  \, \, \, and\, \cos  ine\, \, is\, \, same\, \, then\, \,  \\ { { Sin }^{ -1 } }x\, \, +{ \cos ^{ -1 }  }x=\frac { \pi  }{ 2 }  \\ By\, \, compearing\, \,  \\ { { Sin }^{ -1 } }\left( { \frac { { 3a } }{ { a+3 } }  } \right) =\frac { \pi  }{ 4 }  \\ { Sin }\frac { \pi  }{ 4 } =\frac { { 3a } }{ { a+3 } }  \\ \frac { 1 }{ { \sqrt { 2 }  } } =\frac { { 3a } }{ { a+3 } }  \\ a+3=3\sqrt { 2a }  \\ 3=a(3\sqrt { 2-1 }  \\ a=\frac { 3 }{ { 3\sqrt { 2-1 }  } }  \\ { \cos ^{ -1 }  }\left( { \frac { 1 }{ { 1-b } }  } \right) =\frac { \pi  }{ 4 }  \\ \cos  \frac { \pi  }{ 4 } =\frac { 1 }{ { 1-b } }  \\ \frac { 1 }{ { \sqrt { 2 }  } } =\frac { 1 }{ { 1-b } }  \\ \sqrt { 2 } =1-b \\ b=1-\sqrt { 2 }  \\  \end{matrix}$$
  • Question 7
    1 / -0
    If $$y=\cos^{-1}{\cos{4}}$$, then $$y=$$
    Solution

  • Question 8
    1 / -0
    $$3\cot^{-1}{\left(\dfrac{1}{2+\sqrt{3}}\right)}-\cot^{-1}\left(\dfrac{1}{x}\right)=\cot^{-1}\left(\dfrac{1}{3}\right)+\dfrac{\pi}{2}$$ then $$x=$$?
  • Question 9
    1 / -0
    If $$\left( \tan ^ { - 1 } x \right) ^ { 2 } + \left( \cot ^ { - 1 } x \right) ^ { 2 } = \dfrac { 5 \pi ^ { 2 } } { 8 } $$, then $$x$$ equals to
    Solution

  • Question 10
    1 / -0
    $$cos({ cos }^{ -1 }cos(\frac { 8\pi  }{ 7 } )+{ tan }^{ -1 }tan(\frac { 8\pi  }{ 7 } ))$$ has the value equal to -
    Solution

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