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Introduction to Trigonometry Test - 48

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Introduction to Trigonometry Test - 48
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  • Question 1
    1 / -0
    If  tan2θ=(1e2)\displaystyle \tan ^{2}\theta =\left ( 1-e^{2} \right ) then  secθ+tan3θcosecθ\displaystyle \sec \theta +\tan ^{3}\theta\, \text{cosec}\, \theta is equal to
    Solution
    secθ+tan3θcosecθ\displaystyle \sec \theta +\tan ^{3}\theta \,\text{cosec}\, \theta
    =1cosθ+sin2θcos3θ=cos2θ+sin2θcos3θ\displaystyle = \frac{1}{\cos \theta }+\frac{\sin ^{2}\theta }{\cos ^{3}\theta }=\frac{\cos ^{2}\theta+\sin ^{2}\theta}{\cos ^{3}\theta}
    =sec3θ\displaystyle =\sec ^{3}\theta
    Now tan2θ=(1e2)sec2θ=2e2\displaystyle \tan ^{2}\theta =\left ( 1-e^{2} \right )\Rightarrow \sec ^{2}\theta = 2-e^{2}
    sec3θ=(2e2)32\displaystyle \therefore \sec ^{3}\theta =\left ( 2-e^{2} \right )^{\tfrac32}
  • Question 2
    1 / -0
     If acos3θ+3acosθsin2θ=m\displaystyle a\cos ^{3}\theta +3a\cos \theta \sin ^{2}\theta =m and asin3θ+3acos2θsinθ=n\displaystyle a\sin ^{3}\theta +3a\cos ^{2}\theta \sin \theta=n then (m+n)23+(mn)23\displaystyle \left ( m+n \right )^{\tfrac23}+\left ( m-n \right )^{\tfrac23} is equal to
    Solution
    Given : acos3θ+3acosθsin2θ=m\displaystyle a\cos ^{3}\theta +3a\cos \theta \sin ^{2}\theta =m.....(1) 
                 asin3θ+3acos2θsinθ=n\displaystyle a\sin ^{3}\theta +3a\cos ^{2}\theta \sin \theta=n.....(2) 

    Adding equation (1) and (2) we get, 

    m+n=a(cos3θ+sin3θ )+3acosθsinθ(cosθ+sinθ )=a(cosθ+sinθ )3 \therefore \displaystyle m+n=a\left ( \cos ^{3}\theta +\sin ^{3}\theta  \right )+3a\cos \theta \sin \theta \left ( \cos \theta +\sin \theta  \right )=a\left ( \cos \theta +\sin \theta  \right )^{3}
    Similarly, we get
    mn=a(cosθsinθ )3\displaystyle m-n=a\left ( \cos \theta -\sin \theta  \right )^{3}

    =(m+n)23+(mn)23=\displaystyle \left ( m+n \right )^{\tfrac23}+\left ( m-n \right )^{\tfrac23}

    =a2/3[((cosθ+sinθ)3)2/3+((cosθsinθ)3)2/3]=\displaystyle a^{{2}/{3}}[((cos \theta + sin \theta)^3)^{2/3}+((cos \theta-sin\theta)^3)^{2/3}]

    =a2/3[(cosθ+sinθ)2+(cosθsinθ)2]=\displaystyle a^{2/3}[(cos\theta+sin\theta)^2+(cos\theta-sin\theta)^2]

    After solving the above two brackets we get,

    =a23[(2(cos2θ+sin2θ ))]=2a23=a^{\tfrac23}\left [ \left ( 2\left ( \cos ^{2}\theta +\sin ^{2}\theta  \right ) \right ) \right ]=2a^{\tfrac23}

    Hence option C'C' is the answer.
  • Question 3
    1 / -0
    The value of 4(sin430+cos430)3(cos245+sin290)\displaystyle 4\left ( \sin ^{4}30^{\circ}+\cos ^{4}30^{\circ} \right )-3\left ( \cos ^{2}45^{\circ}+\sin ^{2}90^{\circ} \right )   is:
    Solution
    Given: 4(sin430°+cos430° )3(cos245°+sin290° )4\left( \sin ^{ 4 }{ 30° } +\cos ^{ 4 }{ 30° }  \right) -3\left( \cos ^{ 2 }{ 45° } +\sin ^{ 2 }{ 90° }  \right)

    =4((12 ) 4+(3 2 ) 4)3((12  ) 2+12)=4\left( { \left( \dfrac { 1 }{ 2 }  \right)  }^{ 4 }+{ \left( \dfrac { \sqrt { 3 }  }{ 2 }  \right)  }^{ 4 } \right) -3\left( { \left( \dfrac { 1 }{ \sqrt { 2 }  }  \right)  }^{ 2 }+{ 1 }^{ 2 } \right)

    =4(116+916 )3(12+1)=4\left( \dfrac { 1 }{ 16 } +\dfrac { 9 }{ 16 }  \right) -3\left( \dfrac { 1 }{ 2 } +1 \right)

    =(4×1016 )(3×32 )=\left( 4\times \dfrac { 10 }{ 16 }  \right) -\left( 3\times \dfrac { 3 }{ 2 }  \right)

    =5292=\dfrac { 5 }{ 2 } -\dfrac { 9 }{ 2 }

    =42=\dfrac { -4 }{ 2 }

    =2=-2

    Hence, the answer is 2.-2.
  • Question 4
    1 / -0
     sin285+sin25=\displaystyle  \sin^{2} 85^{\circ} + \sin ^{2}5^{\circ} = ______
    Solution
    Given, sin285o +sin25o  \sin ^{ 2 }{ { 85 }^{ o } }  + \sin ^{ 2 }{ { 5 }^{ o } } 

    We know that sin2Ao=cos2(90A)o  \sin ^{ 2 }{ { A }^{ o } } = \cos ^{ 2 }{ { (90-A) }^{ o } } 

     sin285o +sin25o =cos2(9085)o +sin25o =cos25o +sin25o=1 \Rightarrow  \sin ^{ 2 }{ { 85 }^{ o } }  + \sin ^{ 2 }{ { 5 }^{ o } }  = \cos ^{ 2 }{ {(90-85) }^{ o } }  + \sin ^{ 2 }{ { 5 }^{ o } }  = \cos ^{ 2 }{{ 5 }^{ o } }  + \sin ^{ 2 }{ { 5 }^{ o } } = 1

    (Since, cos2Ao +sin2Ao=1 \cos ^{ 2 }{ {A }^{ o } }  + \sin ^{ 2 }{ { A }^{ o } } = 1 )
  • Question 5
    1 / -0
    If sin4θ+cos4θ=12\displaystyle \sin ^{4}\theta +\cos ^{4}\theta =\dfrac{1}{2} then the value of  sinθcosθ \displaystyle \sin \theta \cos \theta   is  
    Solution
    We know that a4+b4=(a2+b2)22a2b2 { a }^{ 4 }+{ b }^{ 4 }=\quad \left( { a }^{ 2 }+b^{ 2 } \right) ^{ 2 }-2a^{ 2 }{ b }^{ 2 }

    So, sin4θ +cos40=(sin2θ +cos2θ  )22(sin2θ )2(cos2θ )2 \sin ^{ 4 }{ \theta  } +\cos ^{ 4 }{ 0 } =\quad \left( \sin ^{ 2 }{ \theta  } +\cos ^{ 2 }{ \theta  }  \right) ^{ 2 }-2(\sin ^{ 2 }{ \theta  } )^{ 2 }{ (\cos ^{ 2 }{ \theta  } ) }^{ 2 }

    But sin2θ +cos2θ =1 \sin ^{ 2 }{ \theta  } +\cos ^{ 2 }{ \theta  } = 1

    So, sin4θ +cos4θ=122(sinθcosθ)2 \sin ^{ 4 }{ \theta  } +\cos ^{ 4 }{ \theta }= 1^{2} -2(\sin \theta \cos \theta) ^ {2}
     12=12(sinθcosθ)2 \Rightarrow  \dfrac {1}{2} = 1 - 2(\sin \theta \cos \theta) ^ {2}
     2(sinθcosθ)2=12 \Rightarrow  2(\sin \theta \cos \theta) ^ {2}= \dfrac {1}{2}
     (sinθcosθ)2=14 \Rightarrow  (\sin \theta \cos \theta) ^ {2}= \dfrac {1}{4}
     sinθcosθ=±12 \Rightarrow  \sin \theta \cos \theta = \pm \dfrac {1}{2}
  • Question 6
    1 / -0
    If  3tanθ=3sinθ\displaystyle  \sqrt{3}\tan \theta =3\sin \theta then the value of  sin2θcos2θ\displaystyle  \sin ^{2}\theta -\cos ^{2}\theta is
    Solution
    Given 3tanθ=3sinθ\sqrt{3}\tan\theta=3\sin \theta

    3sinθcosθ=3sinθ\sqrt 3\dfrac{\sin \theta}{\cos \theta}=3\sin \theta

    3sinθcosθ3sinθ=0\sqrt 3\dfrac{\sin \theta}{\cos \theta}-3\sin \theta=0

    3sinθ(1cosθ3)=0\sqrt3\sin \theta \big(\dfrac{1}{\cos \theta}-\sqrt3\big)=0

    Now either sinθ=0\sin \theta=0 or cosθ=13\cos \theta=\dfrac{1}{\sqrt3}

    But sinθ0\sin \theta \neq 0

    Hence cosθ=13 \cos \theta =\dfrac{1}{\sqrt3} 

    Now u\sin g sin2θ+cos2θ=1\sin ^2\theta+\cos ^2\theta=1,we will get sinθ=23\sin \theta=\sqrt{\dfrac{2}{3}}

    Hence sin2θcos2θ=2313=13\sin ^2\theta-\cos ^2\theta=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}
  • Question 7
    1 / -0
    If a=cosθ+bsinθ=c\displaystyle a=\cos \theta +b\sin \theta =c, then the value of (asinθbcosθ )\displaystyle \left ( a\sin \theta -b\cos \theta  \right ) is
    Solution
    acosθ +bsinθ =ca\cos { \theta  } +b\sin { \theta  } =c

    Squaring both sides, we get
    (acosθ +bsinθ  ) 2=c2\Rightarrow { \left( a\cos { \theta  } +b\sin { \theta  }  \right)  }^{ 2 }={ c }^{ 2 }

    a2cos2θ +b2sin2θ +2absinθ cosθ =c2\Rightarrow { a }^{ 2 }\cos ^{ 2 }{ \theta  } +{ b }^{ 2 }\sin ^{ 2 }{ \theta  } +2ab\sin { \theta  } \cos { \theta  } ={ c }^{ 2 }

    a2(1sin2θ  )+b2(1cos2θ  )+2absinθ cosθ =c2\Rightarrow { a }^{ 2 }\left( 1-\sin ^{ 2 }{ \theta  }  \right) +{ b }^{ 2 }\left( 1-\cos ^{ 2 }{ \theta  }  \right) +2ab\sin { \theta  } \cos { \theta  } ={ c }^{ 2 }

    a2a2sin2θ +b2b2cos2θ +2absinθ cosθ =c2\Rightarrow { a }^{ 2 }-{ a }^{ 2 }\sin ^{ 2 }{ \theta  } +{ b }^{ 2 }-{ b }^{ 2 }\cos ^{ 2 }{ \theta  } +2ab\sin { \theta  } \cos { \theta  } ={ c }^{ 2 }

    a2+b2[a2sin2θ +b2cos2θ 2absinθ cosθ  ]=c2\Rightarrow { a }^{ 2 }+{ b }^{ 2 }-\left[ { a }^{ 2 }\sin ^{ 2 }{ \theta  } +{ b }^{ 2 }\cos ^{ 2 }{ \theta  } -2ab\sin { \theta  } \cos { \theta  }  \right] ={ c }^{ 2 }

    a2sin2θ +b2cos2θ 2absinθ cosθ =a2+b2c2\Rightarrow { a }^{ 2 }\sin ^{ 2 }{ \theta  } +{ b }^{ 2 }\cos ^{ 2 }{ \theta  } -2ab\sin { \theta  } \cos { \theta  } ={ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }

    (acosθ bsinθ  ) 2=a2+b2c2\Rightarrow { \left( a\cos { \theta  } -b\sin { \theta  }  \right)  }^{ 2 }={ { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }

    (acosθ bsinθ  ) =±a2+b2c2\Rightarrow { \left( a\cos { \theta  } -b\sin { \theta  }  \right)  }=\pm \sqrt { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }

    Hence, the answer is ±a2+b2c2.\pm \sqrt { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } } .
  • Question 8
    1 / -0
    If (tanx+secx)(tany+secy)(tanz+secz)=(secxtanx)(secytanz)=K\displaystyle \left ( \tan x+\sec x \right )\left ( \tan y+\sec y \right )\left ( \tan z+\sec z \right )=\left ( \sec x-\tan x \right )\left ( \sec y-\tan z \right )=K, then the value of KK is
    Solution
    (tanx+secx)2(tany+secy)2(tanz+secz)2\displaystyle \left ( \tan x+\sec x \right )^{2}\left ( \tan y+\sec y \right )^{2}\left ( \tan z+\sec z \right )^{2}
    =(sec2xtan2x)(sec2ytan2y)(sec2ztan2z)\displaystyle =\left ( \sec ^{2}x-\tan ^{2}x \right )\left ( \sec ^{2}y-\tan ^{2}y \right )\left ( \sec ^{2}z-\tan ^{2}z \right )
    =1×1×1=1=k2\displaystyle = 1\times 1\times 1=1=k^{2}
    K=±1\displaystyle \therefore K=\pm 1
  • Question 9
    1 / -0
    If sinΘ+cosec Θ=2\displaystyle \sin \Theta +\text{cosec }\Theta =2 find the value of cotΘ+cosΘ \displaystyle \cot \Theta +\cos \Theta  
    Solution
    Given, sinΘ +cosec Θ =2 \sin \Theta  + \text{cosec } \Theta  = 2

    This is possible only when Θ=90o \Theta = 90^o as sin900=cosec 90o=1 \sin 90^0 = \text{cosec } 90^o = 1

    So, cotΘ+cosΘ=cot90o+cos900= 0+0=0 \cot \Theta + \cos \Theta = \cot 90^o + \cos 90^ 0 =  0 + 0 = 0
  • Question 10
    1 / -0
    If cosecθcotθ=x\displaystyle \text{cosec}\, \theta -\cot \theta =x then cosecθ+cotθ=\displaystyle \text{cosec}\, \theta +\cot \theta = _____
    Solution
    cosec  θcotθ =xcosec\;\theta -\cot { \theta  } =x
    Now, cosec2θcot2θ =1cosec^{ 2 }\theta -\cot ^{ 2 }{ \theta  } =1
    (cosec  θcotθ  )(cosec  θ+cotθ  )=1\Rightarrow \left( cosec\;\theta -\cot { \theta  }  \right) \left( cosec\;\theta +\cot { \theta  }  \right) =1
    x(cosec  θ+cotθ  )=1\Rightarrow x\left( cosec\;\theta +\cot { \theta  }  \right) =1
    cosec  θ+cotθ =1x\Rightarrow cosec\;\theta +\cot { \theta  } =\dfrac { 1 }{ x }
    Hence, the answer is 1x.\dfrac { 1 }{ x } .
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