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Introduction to Trigonometry Test - 42

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Introduction to Trigonometry Test - 42
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  • Question 1
    1 / -0
    The value of $$\displaystyle \cot 15^{\circ}\cot 16^{\circ}\cot 17^{\circ}.....\cot 73^{\circ}\cot 74^{\circ}\cot 75^{\circ}$$ is
    Solution
    $$\cot { 15° } \cot { 16° } \cot { 17° } .........\cot { 73° } \cot { 74° } \cot { 75° } $$
    $$=\cot { \left( 90°-75° \right)  } \cot { \left( 90°-74° \right)  } \cot { \left( 90°-73° \right)  } ......\cot45^o......\cot { 73° } \cot { 74° } \cot { 75° } $$
    $$=\tan { 75° } \tan { 74° } \tan { 73° } .......\cot45^o......\cot { 73° } \cot { 74° } \cot { 75° } $$
    $$=(\tan { 75° }.\cot75^o) (\tan { 74° }.\cot74^o)( \tan { 73° }.\cot73^o) .......\cot45^o $$
    $$=1\times1\times1\times.........\times1$$
    $$=1$$
    Hence, the answer is $$1.$$
  • Question 2
    1 / -0
    Evaluate : $$\displaystyle \frac{3\cos 53^{\circ} \text{cosec}37^{\circ}}{(\cos ^{2}29^{\circ}+\cos ^{2}61^{\circ})}-3\tan ^{2}45^{\circ}$$
    Solution
    $$\displaystyle \frac{3\cos 53^{\circ} cosec 37^{\circ}}{(\cos ^{2}29^{\circ}+\cos ^{2}61^{\circ})}-3\tan ^{2}45^{\circ}$$

    = $$\displaystyle \frac{3\cos 53^{\circ} \text{cosec }(90 - 53)^{\circ}}{(\cos ^{2}29^{\circ}+\cos ^{2}(90 - 29)^{\circ})}-3\tan ^{2}45^{\circ}$$

    = $$\displaystyle \frac{3\cos 53^{\circ} \sec 53^{\circ}}{(\cos ^{2}29^{\circ}+\sin ^{2}29^{\circ})}-3\tan ^{2}45^{\circ}$$

    = $$\displaystyle \frac{3}{1}-3(1)^2$$

    = $$0$$
  • Question 3
    1 / -0
    Evaluate: $$\displaystyle \tan 7^{\circ}\tan 23^{\circ}\tan 60^{\circ}\tan 67{^{\circ}}\tan 83^{\circ}$$
    Solution
    $$\displaystyle \tan 7^{\circ}\tan 23^{\circ}\tan 60^{\circ}\tan 67{^{\circ}}\tan 83^{\circ}$$
    $$\displaystyle =\tan 7^{\circ}.\tan 23^{\circ}.\sqrt{3}.\tan (90^{\circ}-23^{\circ}).\tan (90^{\circ}-7^{\circ})$$                       $$\displaystyle \left [ \because \tan (90^{\circ}-\theta )=\cot \theta  \right ]$$
    =$$\displaystyle =\tan 7^{\circ}.\tan 23^{\circ}.\sqrt{3}.\cot 23^{\circ}.\cot 7^{\circ}$$
    $$\displaystyle \tan 7^{\circ}.\cot 7^{\circ}\tan 23^{\circ}.\cot 23^{\circ}.\sqrt{3}$$                         $$\displaystyle (\because \tan \theta .\cot \theta =1)$$
    $$\displaystyle =1\times 1\times \sqrt{3}=\sqrt{3}$$
  • Question 4
    1 / -0
    Let S = $$\displaystyle \sin ^{2}30^{\circ}+\sin ^{2}45^{\circ}+\sin ^{2}60^{\circ}$$ and 
    P = $$\displaystyle \text{cosec}^{2}45^{\circ}.\sec ^{2}30^{\circ}.\sin ^{3}90^{\circ}.\cos 60^{\circ}$$, then the correct statement is
    Solution
    $$S = \displaystyle \sin ^{2}30^{\circ}+\sin ^{2}45^{\circ}+\sin ^{2}60^{\circ}$$ and $$P = \displaystyle cosec^{2}45^{\circ}.\sec ^{2}30^{\circ}.\sin ^{3}90^{\circ}.\cos 60^{\circ}$$

    Now, $$S = \dfrac{1}{4} + \dfrac{1}{2} + \dfrac{3}{4} = \dfrac{1 + 2 + 3}{4} = \dfrac32$$

    $$P = (\sqrt 2)^2 \cdot \left(\dfrac{2}{\sqrt3}\right)^2 \cdot 1^3 \cdot \dfrac{1}{2} = \dfrac{4}{3}$$

    Hence, $$SP = \dfrac{3}{2} \cdot \dfrac{4}{3} = 2$$
  • Question 5
    1 / -0
    Evaluate: $$\cfrac { \sin { \theta  } \cos { \theta  } \sin { \left( { 90 }^{ o }-\theta  \right)  }  }{ \cos { \left( { 90 }^{ o }-\theta  \right)  }  } +\cfrac { \cos { \theta  } \sin { \theta  } \cos { \left( { 90 }^{ o }-\theta  \right)  }  }{ \sin { \left( { 90 }^{ o }-\theta  \right)  }  } +\cfrac { \sin ^{ 2 }{ { 27 }^{ o } } +\sin ^{ 2 }{ { 63 }^{ o } }  }{ \cos ^{ 2 }{ { 40 }^{ o } } +\cos ^{ 2 }{ { 50 }^{ o } }  } $$
    Solution
    $$\cfrac { \sin { \theta  } \cos { \theta  } \sin { \left( { 90 }^{ o }-\theta  \right)  }  }{ \cos { \left( { 90 }^{ o }-\theta  \right)  }  } =\cfrac { \sin { \theta  } \cos { \theta  } \cos { \theta  }  }{ \sin { \theta  }  } ....(i)$$
    $$\cfrac { \cos { \theta  } \sin { \theta  } \cos { \left( { 90 }^{ o }-\theta  \right)  }  }{ \sin { \left( { 90 }^{ o }-\theta  \right)  }  } =\cfrac { \cos { \theta  } \sin { \theta  } \sin { \theta  }  }{ \cos { \theta  }  } ......(ii)\quad $$
    $$\cfrac { \sin ^{ 2 }{ { 27 }^{ o } } +\sin ^{ 2 }{ { 63 }^{ o } }  }{ \cos ^{ 2 }{ { 40 }^{ o } } +\cos ^{ 2 }{ { 50 }^{ o } }  } $$
    $$=\cfrac { \sin ^{ 2 }{ { 27 }^{ o } } +\sin ^{ 2 }{ \left( { 90 }^{ o }-{ 27 }^{ o } \right)  }  }{ \cos ^{ 2 }{ { 40 }^{ o } } +\cos ^{ 2 }{ \left( { 90 }^{ o }-{ 40 }^{ o } \right)  }  } $$
    $$=\cfrac { \sin ^{ 2 }{ { 27 }^{ o }\cos ^{ 2 }{ { 27 }^{ o } }  }  }{ \cos ^{ 2 }{ { 40 }^{ o } } +\sin ^{ 2 }{ { 40 }^{ o } }  } =\cfrac { 1 }{ 1 } $$
    Using $$(i), (ii), (iii)$$ we get
    $$\sin { \theta  } \sin { \theta  } +\cos { \theta  } \cos { \theta  } +1$$
    $$\sin ^{ 2 }{ \theta  } +\cos ^{ 2 }{ \theta  } +1=1+1=2$$
  • Question 6
    1 / -0
    If $$\cos (\alpha+\beta)=0$$, then $$\sin (\alpha-\beta)$$, can be reduced to
    Solution
    $$\textbf{Step 1 : Use the triginometric identities for angle transformation}$$
                     $$\text{Given ,}$$ $$\cos(\alpha +\beta )=0$$ 
                     $$\Rightarrow \cos (\alpha +\beta )=\cos90\quad \quad \quad \quad \boldsymbol{[\because \cos 90^\circ=0]}$$ 
                     $$\Rightarrow \alpha +\beta =90$$
                     $$\Rightarrow\alpha =90-\beta$$ 
                     $$\therefore\ sin (\alpha -\beta )\\ =\sin(90-2\beta )=\cos2\beta\ \ \  \quad \quad \quad \boldsymbol{[\because \sin(90-\theta )=\cos\theta]} $$

    $$\textbf{Hence , option(B) is the correct answer}$$
  • Question 7
    1 / -0
    Evaluate: $$\sin { \left( { 50 }^{ o }+\theta  \right)  } -\cos { \left( { 40 }^{ o }-\theta  \right)  } +\tan {1}^{o} \tan {10}^{o} \tan {20}^{o} \tan {70}^{o} \tan {80}^{o} \tan {89}^{o}$$
    Solution
    $$\sin { \left( { 50 }^{ o }+\theta  \right)  } -\cos { \left( { 40 }^{ o }-\theta  \right)  } +\tan {1}^{o} \tan {10}^{o} \tan {20}^{o} \tan {70}^{o} \tan {80}^{o} \tan {89}^{o}$$
    $$=\sin { \left[ { 90 }^{ o }-{ 40 }^{ o }+\theta  \right]  } -\cos { \left( { 40 }^{ o }-\theta  \right)  } +(\tan { { 1 }^{ o } } \tan {{89}^{o}})(\tan{{10}^{o}}\tan{{80}^{o}})(\tan {{20}^{o}}\tan {{70}^{o}})$$
    $$=\sin {\left[{90}^{o}-({40}^{o}-\theta) \right]} -\cos {\left({40}^{o}-\theta \right)} +\tan {({90}^{o}} -{ 89 }^{ o })\tan { { 89 }^{ o } } \tan { { (90 }^{ o } } -{ 80 }^{ o })\tan { { 80}^{ o } } \tan { { (90 }^{ o } } -{ 70 }^{ o })\tan { { 70 }^{ o } } $$
    $$=\cos { \left( { 40 }^{ o }-\theta  \right)  } -\cos { \left( { 40 }^{ o }-\theta  \right)  } +\cot { { 89 }^{ o } } \tan { { 89 }^{ o } } \cot { { 80 }^{ o } } \tan { { 80 }^{ o } } \cot { { 70 }^{ o } } \tan { { 70 }^{ o } } $$
    $$=0+1$$        $$\because \cot {\theta} \tan {\theta}=1$$]
    $$=1$$ 
    Hence, $$\sin { \left( { 50 }^{ o }+\theta  \right)  } -\cos { \left( { 40 }^{ o }-\theta  \right)  } +\tan {1}^{o} \tan {10}^{o} \tan {20}^{o} \tan {70}^{o} \tan {80}^{o} \tan {89}^{o}=1$$
  • Question 8
    1 / -0
    The value of $$\displaystyle \cos ^{4}\theta +\sin ^{4}\theta +2\cos ^{2}\theta \sin ^{2}\theta $$ when $$\displaystyle \theta=60^{\circ} $$ is
    Solution
    $$\displaystyle \cos ^{4}\theta +\sin ^{4}\theta +2\cos ^{2}\theta \sin ^{2}\theta $$
    = $$\displaystyle (\cos ^{2}\theta +\sin ^{2}\theta)^2$$
    = $$\displaystyle (\cos ^{2} 60^0 +\sin ^{2} 60^0)^2$$
    = $$\displaystyle (\frac{1}{4} +\frac{3}{4})^2$$
    = $$\displaystyle 1^2$$
    = $$1$$
  • Question 9
    1 / -0
    Evaluate: $$\cfrac { \cos ^{ 2 }{ { 20 }^{ o } } +\cos ^{ 2 }{ { 70 }^{ o } }  }{ \sec ^{ 2 }{ { 50 }^{ o } } -\cot ^{ 2 }{ { 40 }^{ o } }  } +2\text{cosec} ^{ 2 }{ { 58 }^{ o } } -2\cot { { 58 }^{ o } } \tan { { 32 }^{ o } } -4\tan { { 13 }^{ o } } \tan { { 37 }^{ o } } \tan { { 45 }^{ o } } \tan { { 53 }^{ o } } \tan { { 77 }^{ o } } $$
    Solution
    $$\cfrac { \cos ^{ 2 }{ { 20 }^{ o } } +\cos ^{ 2 }{ { 70 }^{ o } }  }{ \sec ^{ 2 }{ { 50 }^{ o } } +\cot ^{ 2 }{ { 40 }^{ o } }  } +2\csc ^{ 2 }{ { 58 }^{ o } } -2\cot { { 58 }^{ o } } \tan { { 32 }^{ o } } -4\tan { { 13 }^{ o } } \tan { { 37 }^{ o } } \tan { { 45 }^{ o } } \tan { { 53 }^{ o } } \tan { { 77 }^{ o } } $$
    $$=\cfrac { \cos ^{ 2 }{ { ({ 90 }^{ o }-70 }^{ o }) } +\cos ^{ 2 }{ { 70 }^{ o } }  }{ \sec ^{ 2 }{ { ({ 90 }^{ o }-40 }^{ o } } )+\cot ^{ 2 }{ { 40 }^{ o } }  }$$
     $$+2\csc ^{ 2 }{ { 58 }^{ o } } -2\cot { { 58 }^{ o } } \tan { ({ 90 }^{ o }-58^{ o }) } -4(\tan { { 13 }^{ o } } \tan { { 77 }^{ o } } )(\tan { { 37 }^{ o } } \tan { { 53 }^{ o } } )\tan { { 45 }^{ o } } $$
    $$=\cfrac { \sin ^{ 2 }{ { 70 }^{ o } } +\cos ^{ 2 }{ { 70 }^{ o } }  }{ co\sec ^{ 2 }{ { 40 }^{ o } } -\cot ^{ 2 }{ { 40 }^{ o } }  } +2(\csc ^{ 2 }{ { 58 }^{ o } } -\cot ^{ 2 }{ { 58 }^{ o } }) -4[\tan { ({ 90 }^{ o }-77^{ o }) } \tan { { 77 }^{ o } } ][\tan { ({ 90 }^{ o }-53^{ o }) } \tan { { 53 }^{ o } } ][1]$$
    $$=\cfrac{1}{1}+2-4$$
    $$=-1$$
  • Question 10
    1 / -0
    Evaluate:
    $$\cfrac { \sec ^{ 2 }{ { 54 }^{ o } } -\cot ^{ 2 }{ { 36 }^{ o } }  }{ co\sec ^{ 2 }{ { 57 }^{ o } } -\tan ^{ 2 }{ { 33 }^{ o } }  } +2\sin ^{ 2 }{ { 38 }^{ o } } \sec ^{ 2 }{ { 52 }^{ o } } -\sin ^{ 2 }{ { 45 }^{ o } } +\cfrac { 2 }{ \sqrt { 3 }  } \tan { { 17 }^{ o } } \tan { { 60 }^{ o } } \tan { { 73 }^{ o } } $$
    Solution
    $$\cfrac { \sec ^{ 2 }{ { 54 }^{ o } } -\cot ^{ 2 }{ { 36 }^{ o } }  }{ co\sec ^{ 2 }{ { 57 }^{ o } } -\tan ^{ 2 }{ { 33 }^{ o } }  } +2\sin ^{ 2 }{ { 38 }^{ o } } \sec ^{ 2 }{ { 52 }^{ o } } -\sin ^{ 2 }{ { 45 }^{ o } } +\cfrac { 2 }{ \sqrt { 3 }  } \tan { { 17 }^{ o } } \tan { { 60 }^{ o } } \tan { { 73 }^{ o } } $$

    $$\Rightarrow \cfrac { \sec ^{ 2 }{ { (90 }^{ o }- } { 36 }^{ o })-\cot ^{ 2 }{ { 36 }^{ o } }  }{ co\sec ^{ 2 }{ { (90 }^{ o }- } { 33 }^{ o })-\tan ^{ 2 }{ { 33 }^{ o } }  } +2\sin ^{ 2 }{ { 38 }^{ o } } \sec ^{ 2 }{ { (90 }^{ o }- } { 38 }^{ o })-\sin ^{ 2 }{ { 45 }^{ o } } +\cfrac { 2 }{ \sqrt { 3 }  } \tan { \left( { 90 }^{ o }-{ 73 }^{ o } \right)  } \tan { { 73 }^{ o } } \tan { { 60 }^{ o } } $$

    $$\Rightarrow \cfrac { co\sec ^{ 2 }{ { 36 }^{ o } } -\cot ^{ 2 }{ { 36 }^{ o } }  }{ \sec ^{ 2 }{ { 33 }^{ o } } -\tan ^{ 2 }{ { 33 }^{ o } }  } +2\sin ^{ 2 }{ { 38 }^{ o } } co\sec ^{ 2 }{ { 38 }^{ o } } -{ \left( \cfrac { 1 }{ \sqrt { 2 }  }  \right)  }^{ 2 }+\cfrac { 2 }{ \sqrt { 3 }  } \cot { { 73 }^{ o } } \tan { { 73 }^{ o } } \times \sqrt { 3 } $$

    $$\Rightarrow \cfrac { 1 }{ 1 } +2\sin ^{ 2 }{ { 38 }^{ o } } \times \cfrac { 1 }{ \sin ^{ 2 }{ { 38 }^{ o } }  } -\cfrac { 1 }{ 2 } +\cfrac { 2 }{ \sqrt { 3 }  } \times \cfrac { 1 }{ \tan { { 73 }^{ o } }  } \times \tan { { 73 }^{ o } } \times \sqrt { 3 } $$

    $$\quad [\because \quad co\sec ^{ 2 }{ 0 } -\cot ^{ 2 }{ 0 } =1,\sec ^{ 2 }{ 0 } -\tan ^{ 2 }{ 0 } =1]$$

    $$\Rightarrow 1+2-\cfrac{1}{2}+2=5-\cfrac{1}{2}$$

    $$\Rightarrow \cfrac{9}{2}$$
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