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Introduction to Trigonometry Test - 46

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Introduction to Trigonometry Test - 46
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  • Question 1
    1 / -0
    The value of tan $$1^{\circ}$$ tan $$2^{\circ}$$ tan $$3^{\circ}$$$$\times..........\times$$tan $$89^{\circ}$$ is
    Solution
    $$ \tan 1^{\circ} \tan 2^{\circ} \tan3^{\circ}\times..........\times \tan 89^{\circ}$$
    $$= (\tan 1^{\circ} \tan89^{\circ})( \tan 2^{\circ} \tan 88^{\circ}) \times....\times \tan 45^{\circ}$$
    $$= (\tan 1^{\circ} \cot1^{\circ})( \tan 2^{\circ} \cot2^{\circ})\times ...\times  \tan45^{\circ}$$      $$[\because \tan \theta=\cot (90^{\circ}-\theta)$$ and $$\tan \theta \times \cot \theta=1 ]$$
    $$=1$$
  • Question 2
    1 / -0
    value of $$ \displaystyle \dfrac{\tan ^{2}60^{\circ} -2\tan^{2}45^{\circ}+\sec ^{2}0^{\circ}}{3\sin ^{2}45^{\circ}\sin 90^{\circ}+\cos ^{2}60^{\circ}cos^{3}0^{\circ}}  $$
    Solution

    Consider the given expression,

      $$ \Rightarrow \,\dfrac{{{\tan }^{2}}{{60}^{0}}-2{{\tan }^{2}}{{45}^{0}}+{{\sec }^{2}}0}{3{{\sin }^{2}}{{45}^{0}}\sin {{90}^{0}}+{{\cos }^{2}}{{60}^{0}}{{\cos }^{3}}{{0}^{0}}} $$


     $$ \Rightarrow \,\dfrac{{{\sqrt{3}}^{2}}-2\times {{1}^{2}}+{{1}^{2}}}{3\times {{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}\times 1+{{\left( \dfrac{1}{2} \right)}^{2}}\times {{1}^{3}}} $$


     $$ \Rightarrow \,\dfrac{3-2+1}{\dfrac{3}{2}+\dfrac{1}{4}} $$

     $$ \Rightarrow \,\dfrac{2}{\dfrac{6+1}{4}} $$

     $$ \Rightarrow \,\dfrac{8}{7} $$

    Hence, this is the answer.

  • Question 3
    1 / -0
    The value of $$\displaystyle \frac { \sin { { 70 }^{ o } }  }{ \cos { { 20 }^{ o } }  } +\frac { \text{cosec }{ 20 }^{ o } }{ \sec { { 70 }^{ o } }  } -2\cos { { 70 }^{ o } } \text{cosec }{ 20 }^{ o }$$ is :
    Solution
    The value of $$\displaystyle \frac { \sin { { 70 }^{ o } }  }{ \cos { { 20 }^{ o } }  } +\frac { cosec{ 20 }^{ o } }{ \sec { { 70 }^{ o } }  } -2\cos { { 70 }^{ o } } cosec{ 20 }^{ o }$$ is

    $$\displaystyle =\frac { \sin { { 70 }^{ o } }  }{ \cos { \left( { 90 }^{ o }-{ 70 }^{ o } \right)  }  } +\frac { cosec{ 20 }^{ o } }{ \sec { \left( { 90 }^{ o }-{ 20 }^{ o } \right)  }  } -\frac { 2\cos { { 70 }^{ o } }  }{ \sin { { 20 }^{ o } }  } $$
    $$\displaystyle =\frac { \sin { { 70 }^{ o } }  }{ \sin { { 70 }^{ o } }  } +\frac { cosec{ 20 }^{ o } }{ cosec{ 20 }^{ o } } -\frac { 2\cos { { 70 }^{ o } }  }{ \sin { \left( { 90 }^{ o }-{ 70 }^{ o } \right)  }  } $$
    $$\displaystyle \left[ \because \quad \cos { \left( { 90 }^{ o }-\theta  \right)  } \sin { \theta  } and\quad \sec { \left( { 90 }^{ o }-\theta  \right) =cosec\theta  }  \right] $$
    $$\displaystyle =1+1-\frac { 2\cos { { 70 }^{ o } }  }{ \cos { { 70 }^{ o } }  } =2-2=0$$
  • Question 4
    1 / -0
    The value of $$1+\cot^2{A}$$ is
    Solution
    Using the identity $$\sin^2 A + \cos^2 A = 1$$
    Divide both the sides by $$\sin^2 A$$
    $$\displaystyle \cfrac {\sin^2 A}{\sin^2 A} + \cfrac {\cos^2 A}{\sin^2 A} $$ = $$\displaystyle \cfrac {1}{\sin^2 A}$$
    i.e. $$ 1 +  \cot^2 A = \text{cosec}^2 A$$
  • Question 5
    1 / -0
    The value of $$\displaystyle \sec { { 41 }^{ o } } \sin { { 49 }^{ o }+ } \cos { { 49 }^{ o } } \text{cosec }{ 41 }^{ o }$$ is :
    Solution
    $$\displaystyle \sec { { 41 }^{ o } } \sin { { 49 }^{ o }+ } \cos { { 49 }^{ o } } cosec{ 41 }^{ o }$$
    $$\displaystyle =\frac { \sin { { 49 }^{ o } }  }{ \cos { { 41 }^{ o } }  } +\frac { \cos { { 49 }^{ o } }  }{ \sin { { 41 }^{ o } }  } $$
    $$\displaystyle =\frac { \sin { { 49 }^{ o } }  }{ \cos { \left( { 90 }^{ o }-{ 49 }^{ o } \right)  }  } +\frac { \cos { { 49 }^{ o } }  }{ \sin { \left( { 90 }^{ o }-{ 49 }^{ o } \right)  }  } $$
    $$\displaystyle =\frac { \sin { { 49 }^{ o } }  }{ \sin { { 49 }^{ o } }  } +\frac { \cos { { 49 }^{ o } }  }{ \cos { { 49 }^{ o } }  } $$
    $$\displaystyle =1+1=2$$
  • Question 6
    1 / -0
    The value of $$\displaystyle \frac { 2\cos { { 67 }^{ o } }  }{ \sin { { 23 }^{ o } }  } -\frac { \tan { { 40 }^{ o } }  }{ \cot { { 50 }^{ o } }  } -\sin { { 90 }^{ o } } $$ is :
    Solution
    $$\displaystyle \frac { 2\cos { { 67 }^{ o } }  }{ \sin { { 23 }^{ o } }  } +\frac { \tan { { 40 }^{ o } }  }{ \cot { { 50 }^{ o } }  } -\sin { { 90 }^{ o } } $$
    $$\displaystyle =\frac { 2\cos { { 67 }^{ o } }  }{ \cos { \left( { 90 }^{ o }-{ 23 }^{ o } \right)  }  } -\frac { \tan { { 40 }^{ o } }  }{ \tan { \left( { 90 }^{ o }-{ 50 }^{ o } \right)  }  } -\sin { { 90 }^{ o } } $$
    $$\displaystyle =\frac { 2\cos { { 67 }^{ o } }  }{ \cos { { 67 }^{ o } }  } -\frac { \tan { { 40 }^{ o } }  }{ \tan { { 40 }^{ o } }  } -\sin { { 90 }^{ o } } $$
    $$\displaystyle =2-1-1=0$$
  • Question 7
    1 / -0
    The value of   $$\displaystyle \sin { \theta  } \cos { \theta  } -\frac { \sin { \theta  } \cos { \left( { 90 }^{ o }-\theta  \right)  } \cos { \theta  }  }{ \sec { \left( { 90 }^{ o }-\theta  \right)  }  } -\frac { \cos { \theta  } \sin { \left( { 90 }^{ o }-\theta  \right)  } \sin { \theta  }  }{ \text{cosec }\left( { 90 }^{ o }-\theta  \right)  } $$ is :
    Solution
    Let $$A = \displaystyle \sin { \theta  } \cos { \theta  } -\frac { \sin { \theta  } \cos { \left( { 90 }^{ o }-\theta  \right)  } \cos { \theta  }  }{ \sec { \left( { 90 }^{ o }-\theta  \right)  }  } -\frac { \cos { \theta  } \sin { \left( { 90 }^{ o }-\theta  \right)  } \sin { \theta  }  }{ \text{cosec }\left( { 90 }^{ o }-\theta  \right)  } $$
    Using the identities  $$ \cos({90}^{o}-\theta) =\sin{\theta}, \sin( {90}^{o}-\theta) =\cos{\theta}, \sec({90}^{o}-\theta)=\text{cosec }\theta, \text{cosec}({90}^{o}-\theta) =\sec{\theta} $$, we get

    $$A =\sin { \theta  } \cos { \theta  } -\dfrac { \sin { \theta  } \sin { \theta  } \cos { \theta  }  }{ \text{cosec }\theta  } -\dfrac { \cos { \theta  } \cos { \theta  } \sin { \theta  }  }{ \sec { \theta  }  } $$ 

    $$\displaystyle =\sin { \theta  } \cos { \theta  }- { \sin }^{ 3 }\theta \cos { \theta  } -{ cos }^{ 3 }\theta \sin { \theta  } $$ ..... $$\displaystyle \left[ \because \frac { 1 }{ \text{cosec }\theta  } =\sin { \theta}, \frac { 1 }{ \sec { \theta  }  } =\cos { \theta  }  \right] $$

    $$\displaystyle =\sin { \theta  } \cos { \theta  } -\sin { \theta  } \cos { \theta  } \left( { \sin }^{ 2 }\theta +{ cos }^{ 2 }\theta  \right) $$

    $$\displaystyle =\sin { \theta  } \cos { \theta  } -\sin { \theta  } \cos { \theta  } \left( 1 \right) $$

    $$\displaystyle =\sin { \theta  } \cos { \theta  } -\sin { \theta  } \cos { \theta  } =0$$
  • Question 8
    1 / -0
    If $$\displaystyle \sec 2A=\text{cosec } \left ( A-42^{\circ} \right )$$ where $$2A$$ is acute angle, then value of $$A$$ is
    Solution
    Given, $$\displaystyle \sec  2A=cosec \left ( A-42^{\circ} \right )$$
    $$\displaystyle \left [ \because \sec \theta = cosec \left ( 90^{\circ}-\theta  \right ) \right ]$$
    $$\displaystyle \Rightarrow cosec \left ( 90^{\circ}-2A \right )= cosec \left ( A-42^{\circ} \right )$$
    $$\displaystyle\Rightarrow  \left ( 90^{\circ}-2A \right )=\left ( A-42^{\circ} \right )$$
    $$\displaystyle \Rightarrow 90^{\circ}+42^{\circ}=A+2A$$
    $$\displaystyle \Rightarrow 132^{\circ}=3A$$
    $$\displaystyle \Rightarrow A=44^{\circ}$$
  • Question 9
    1 / -0
    The value of $$ \displaystyle  \tan 1^{\circ}\tan 2^{\circ}\tan 3^{\circ}.....\tan 89^{\circ}  $$ is 
    Solution
    Given,
    $$\tan 1^0\tan 2^0 \tan 3^0......\tan 89^0$$

    $$=\tan (90^0-1^0)\tan (90^0-2^0)\tan (90^0-3^0)......\tan 89^0$$

    $$=\cot 89^0 \cot 88^0\cot 87^0......\tan 89^0$$  [$$\because \tan(90^0-\theta)=\cot\theta)$$]

    $$=\cot 89^0\tan 89^0 \cot 88^0\tan 88^0......\tan 45^0$$ [$$because \tan\theta \times \cot\theta=1 \quad \tan45^\circ=1$$]

    $$=1$$
  • Question 10
    1 / -0
    Find the value of 
    $$\displaystyle \frac{\sin ^{3}\theta +\cos ^{3}\theta }{\sin \theta +\cos \theta }+\frac{\left ( \cos ^{3}\theta -\sin ^{3}\theta  \right )}{\cos \theta -\sin \theta }$$ 
    Solution
    To find the value of  $$\displaystyle \frac{\sin ^{3}\theta +\cos ^{3}\theta }{\sin \theta +\cos \theta }+\frac{ \cos ^{3}\theta -\sin ^{3}\theta}{\cos \theta -\sin \theta }$$ 

    Using the formulae,
    $$a^3+b^3=(a+b)(a^2+b^2-ab)$$
    and $$a^3-b^3=(a-b)(a^2+b^2+ab)$$

    $$\displaystyle \frac{\sin ^{3}\theta +\cos ^{3}\theta }{\sin \theta +\cos \theta }+\frac{ \cos ^{3}\theta -\sin ^{3}\theta }{\cos \theta -\sin \theta }$$

    $$=\displaystyle \frac{(\sin \theta +\cos \theta)(\sin^2\theta+\cos^2\theta-\sin\theta\cos\theta) }{\sin \theta +\cos \theta }+\frac{(\cos \theta-\sin \theta )(\sin^2\theta+\cos^2\theta+\sin\theta\cos\theta) }{\cos \theta -\sin \theta }$$

    $$=(1-\sin\theta \cos\theta)+(1+\sin\theta \cos\theta)$$

    $$=1-\sin\theta \cos\theta+1+\sin\theta \cos\theta$$

    $$=2$$

    Hence, the required answer is $$2$$.
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