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Introduction to Trigonometry Test - 44

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Introduction to Trigonometry Test - 44
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  • Question 1
    1 / -0
    The value of $$\displaystyle \tan { { 5 }^{ o } } .\tan { { 85 }^{ o } } .\tan { { 31 }^{ o } } .\tan { { 5 }9^{ o } } .\tan { { 45 }^{ o } } $$ is :
    Solution
    The value of $$\tan 5^o. \tan 85^o.\tan 31^o. \tan 59^o.\tan 45^o$$ is 
    $$=\tan (90-85)^o.\tan 85^o. \tan (90-59)^o. \tan 45^o$$
    $$=\cot 85^o.\tan 85^o. \cot 59^o. \tan 59^o. \tan 45^o$$
    $$=\tan 45^o$$
    $$=1$$
  • Question 2
    1 / -0
    Find the value of $$\displaystyle \cos { \left( { 90 }^{ o }-A \right)  } \tan { \left( { 90 }^{ o }-A \right)  } \sec { \left( { 90 }^{ o }-A \right)  } $$
    Solution
    $$\displaystyle \cos { \left( { 90 }^{ o }-A \right)  } .\tan { \left( { 90 }^{ o }-A \right)  } \sec { \left( { 90 }^{ o }-A \right)  } $$
    $$=\displaystyle \sin { A } \times \cot{ A }\times \text{cosec} A$$ ..... $$[\because \cos(90^{o} - \theta)=\sin \theta, \tan(90^{o}-\theta)=\cot \theta, \sec(90^{o}-\theta)=\text{cosec} \theta]$$
    $$=\sin A \times \dfrac{\cos A}{\sin A}\times \dfrac{1}{\sin A}$$
    $$=\dfrac{\cos A}{\sin A}=\cot A$$
    Hence, option A is correct.
  • Question 3
    1 / -0
    The value of $$\displaystyle \frac { \sin { { 60 }^{ o } }  }{ { \cos }^{ 2 }{ 45 }^{ o } } -\cot{ { 30 }^{ o } }+5\cos { { 90 }^{ o } } $$ is :
    Solution
    $$\displaystyle \frac { \sin { { 60 }^{ o } }  }{ { \cos }^{ 2 }{ 45 }^{ o } } -\cot{ { 30 }^{ o } }+5\cos { { 90 }^{ o } } $$ 
    $$=\displaystyle \frac { \frac { \sqrt { 3 }  }{ 2 }  }{ \frac { 1 }{ 2 }  } -\sqrt { 3 } +0=\sqrt { 3 } -\sqrt { 3 } =0$$
  • Question 4
    1 / -0
    If  $$\displaystyle \theta ={ 45 }^{ o }$$, then $$\displaystyle2 \sin { \theta  } cos{ \theta }$$ is :
    Solution
    Given, $$\theta=45^o$$
    Therefore, value of  $$\displaystyle2 \sin { \theta  } cos{ \theta }$$ is
    $$=\displaystyle 2*\sin  \left( { 45 }^{ o } \right) cos {45^0}$$
    $$=2*(1/2)$$
    $$=1$$
  • Question 5
    1 / -0
    If $$\displaystyle \frac { x\text{ cosec }^{ 2 }{ 30 }^{ o }{ \sec }^{ 2 }{ 45 }^{ o } }{ 8{ \cos }^{ 2 }{ 45 }^{ o }{ \sin }^{ 2 }{ 90 }^{ o } } ={ \tan }^{ 2 }{ 60 }^{ o }-{ \tan }^{ 2 }{ 45 }^{ o }$$, then $$x$$ is :
    Solution
    $$\displaystyle \frac { { x\left( 2 \right)  }^{ 2 }\times { \left( \sqrt { 2 }  \right)  }^{ 2 } }{ { 8\left( \dfrac { 1 }{ \sqrt { 2 }  }  \right)  }^{ 2 }\times { \sin }^{ 2 }\theta  } ={ \left( \sqrt { 3 }  \right)  }^{ 2 }-{ \left( 1 \right)  }^{ 2 }$$

    $$\displaystyle \frac { 8x }{ 8\times \dfrac { 1 }{ 2 } \times 1 } =3-1=2$$
    $$\Rightarrow \displaystyle 2x=2$$$$\Rightarrow \displaystyle x=1$$
  • Question 6
    1 / -0
    Evaluate: $$\displaystyle 3{\cot }^{ 2 }{ 60 }^{ o }+{ \sec }^{ 4 }{ 45 }^{ o }-{ \tan }^{ 2 }{ 60 }^{ o }$$
    Solution
    Given that: $$3\cot ^{ 2 }{ { 60^o }  } +\sec ^{ 4 }{  { 45 ^o}  } -\tan ^{ 2 }{ { 60^\circ }  } $$
     $$=3\cot ^{ 2 }{ \left(  { 90 ^o- }  { 30 ^o }  \right)  } +\sec ^{ 4 }{{ 45 ^\circ }  } -\tan ^{ 2 }{  { 60^o }  } $$

     $$=3\times( \dfrac { 1 }{ \sqrt { 3 }  } )^2-(\sqrt { 3 })^2 +{ \left( \sqrt { 2 }  \right)  }^{ 2 }$$
     $$=1 -3+2$$
     $$=0$$
  • Question 7
    1 / -0
    Evaluate: $$\displaystyle \sin { { 40 }^{ o } } .\sec{ { 50 }^{ o } }-\cfrac { \tan { { 40 }^{ o } }  }{ \cot { { 50 }^{ o } }  } +1$$
    Solution
    $$\displaystyle \sin { { 40 }^{ o } } \times \sec { \left( { 90 }^{ o }-{ 40 }^{ o } \right) -\frac { \tan { { 40 }^{ o } }  }{ \cot { \left( { 90 }^{ o }-{ 40 }^{ o } \right)  }  }  } +1$$

    $$\displaystyle =\sin { { 40 }^{ o } } \times cosec{ 40 }^{ o }-\frac { \tan { { 40 }^{ o } }  }{ \tan { { 40 }^{ o } }  } +1$$
    $$\displaystyle =1-1+1=1$$
  • Question 8
    1 / -0
    The value of  $$\displaystyle { \cos }^{ 2 }\left( { 90 }^{ o }-\theta  \right) +{ \cos }^{ 2 }\theta $$ is :
    Solution
    The value of $$\cos^2(90^o-\theta)+\cos^2\theta$$ is
    $$=\sin^2\theta+\cos^2\theta$$
    $$=1$$
  • Question 9
    1 / -0
    The value of $$\displaystyle \tan { \theta  }. \tan { \left( { 90 }^{ o }-\theta  \right)  } +\cos { \theta  } .\text{cosec}\left( { 90 }^{ o }-\theta  \right) $$ is
    Solution
    We know $$\tan(90-\theta)=\cot\theta$$
    and $$\text{cosec}(90-\theta)=\sec\theta$$
    Also $$\tan\theta=\dfrac{1}{\cot\theta}$$,
    $$\cos\theta=\dfrac{1}{\sec\theta}$$
    So, $$\tan \theta. \tan (90^o-\theta)+\cos \theta . \text{cosec} (90^o-\theta)$$
    $$=\tan\theta\cot\theta+\cos\theta\sec\theta$$
    $$=1+1=2$$
  • Question 10
    1 / -0
    $$\displaystyle \frac { \sec { \theta  }  }{ \text{cosec }\left( { 90 }^{ o }-\theta  \right)  } -\frac { \sin { \theta  }  }{ \cos { \left( { 90 }^{ o }-\theta  \right)  }  } +\cos { { 0 }^{ o } } $$ is equal to :
    Solution
    We need to find value of $$\dfrac {\sec \theta}{\text {cosec }(90^o-\theta)}-\dfrac {\sin \theta}{\cos (90^o-\theta)}+\cos 0^o$$
    $$=\dfrac {\sec \theta}{\sec \theta}-\dfrac {\sin \theta}{\sin\theta}+1$$
    $$=1-1+1$$
    $$=1$$
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