Introduction to Trigonometry Test - 33

Let $$\displaystyle \csc { \theta  } -\cot { \theta  } =q$$        ...(1)

$$\displaystyle \dfrac { \left( \csc { \theta  } -\cot { \theta  }  \right) \left( \csc { \theta  } +\cot { \theta  }  \right)  }{ \csc { \theta  } +\cot { \theta  }  } =q\\ \Rightarrow \dfrac { \csc ^{ 2 }{ \theta  } -\cot ^{ 2 }{ \theta  }  }{ \csc { \theta  } +\cot { \theta  }  } =q$$
$$\Rightarrow \csc { \theta  } +\cot { \theta  } =\dfrac { 1 }{ q } $$          ...(2)
Adding (1) & (2) we get,
$$2\csc { \theta  } =q+\dfrac { 1 }{ q } \\ \therefore \quad \csc { \theta  } =\dfrac { 1 }{ 2 } \left( q+\dfrac { 1 }{ q }  \right) $$

Ans: C

$$x =\cfrac { 2sin\alpha  }{ 1+sin\alpha +cos\alpha  } \\ =\cfrac { 2sin\alpha  }{ \left( 1+sin\alpha  \right) +\left( cos\alpha  \right)  } \times \cfrac { \left( 1+sin\alpha  \right) -\left( cos\alpha  \right)  }{ \left( 1+sin\alpha  \right) -\left( cos\alpha  \right)  } \\ =\cfrac { 2sin\alpha \left( 1+sin\alpha -cos\alpha  \right)  }{ { \left( 1+sin\alpha  \right)  }^{ 2 }-{ \left( cos\alpha  \right)  }^{ 2 } } \\ =\cfrac { 2sin\alpha \left( 1+sin\alpha -cos\alpha  \right)  }{ 1+{ sin }^{ 2 }\alpha +2sin\alpha -{ cos }^{ 2 }\alpha  } \\ =\cfrac { 2sin\alpha \left( 1+sin\alpha -cos\alpha  \right)  }{ 1+{ sin }^{ 2 }\alpha +2sin\alpha -1+{ sin }^{ 2 }\alpha  } \\ =\cfrac { 2sin\alpha \left( 1+sin\alpha -cos\alpha  \right)  }{ 2{ sin }^{ 2 }\alpha +2sin\alpha  } \\ =\cfrac { 2sin\alpha \left( 1+sin\alpha -cos\alpha  \right)  }{ 2{ sin }\alpha \left( 1+sin\alpha  \right)  } \\ =\cfrac { \left( 1+sin\alpha -cos\alpha  \right)  }{ \left( 1+sin\alpha  \right)  } $$

given $$ \sin\theta_1 + \sin\theta_2 + \sin\theta_3 = 3$$
we know maximum value of $$\sin\theta = 1$$  at $$\theta=\dfrac{\pi}{2}$$
so above equation only satisfy if $$ \sin\theta_1 = 1, \sin\theta_2 = 1, $$and $$ \sin\theta_3 = 1$$ and $$\theta_1=\theta_2 =\theta_3=\dfrac{\pi}{2}$$
$$\because \cos \dfrac{pi}{2}=0$$
$$\Rightarrow \cos\theta_1 = 0, \cos\theta_2 = 0, $$and $$ \cos\theta_3 = 0 $$
So, $$ \cos\theta_1 + \cos\theta_2 + \cos\theta_3 = 0$$
Hence, option $$'D'$$ is correct.

$$sinx+cosecx=2\\ sinx+\cfrac { 1 }{ sinx } =2$$
Squaring both sides
$${ sin }^{ 2 }x+\cfrac { 1 }{ { sin }^{ 2 }x } +2sinx.\cfrac { 1 }{ sinx } =4\\ { sin }^{ 2 }x+\cfrac { 1 }{ { sin }^{ 2 }x } =2$$
Again squaring both sides
$${ sin }^{ 4 }x+\cfrac { 1 }{ { sin }^{ 4 }x } +2{ sin }^{ 2 }x.\cfrac { 1 }{ { sin }^{ 2 }x } =4\\ { sin }^{ 4 }x+\cfrac { 1 }{ { sin }^{ 4 }x } =2$$
Similarly
$${ sin }^{ 8 }x+\cfrac { 1 }{ { sin }^{ 8 }x } =2$$
and
$${ sin }^{ n }x+\cfrac { 1 }{ { sin }^{ n }x } =2\\ { sin }^{ n }x+{ cosec }^{ n }x=2$$

$$\sin { \theta  } -\cos { \theta  } =1$$   ....(1)

Given $$cosec \theta -\sin \theta =m$$ or  $$\displaystyle \frac{1}{\sin \theta }-\sin \theta =m$$
or  $$\displaystyle \frac{1-\sin ^{2}\theta }{\sin \theta }=m$$ or $$\displaystyle \frac{\cos ^{2}\theta }{\sin \theta }=m$$          (i)
Again   $$\sec \theta -\cos \theta =n$$ or $$\displaystyle \frac{1}{\cos \theta }-\cos \theta =n$$
or  $$\displaystyle \frac{1-\cos ^{2}\theta }{\cos \theta }=n$$ or

$$\displaystyle \frac{\sin ^{2}\theta }{\cos \theta }=n$$          (ii)
From Eq. (i),   $$\displaystyle \sin \theta =\frac{\cos ^{2}\theta }{m}$$          (iii)
Putting the value of $$\sin \theta $$ in Eq. (ii), we get
     $$\displaystyle \frac{\cos ^{4}}{m^{2}\cos \theta }=n$$ or $$\cos ^{3}=m^{2}n$$
$$\therefore

$$   $$\cos \theta =\left ( m^{2}n \right )^{1/3}$$ or $$\cos

^{2}\theta =\left ( m^{2}n \right )^{2/3}$$          (iv)
From Eq. (iii), $$\displaystyle \sin \theta =\frac{\cos ^{2}\theta

}{m}=\frac{\left ( m^{2}n \right

)^{2/3}}{m}=\frac{m^{4/3}n^{2/3}}{m}=m^{1/3}n^{2/3}=\left ( mn^{2}

\right )^{1/3}$$
$$\therefore $$   $$\sin ^{2}\theta =\left ( mn^{2} \right )^{2/3}$$          (v)
Adding Eqs. (iv) and (v), we get
     $$\left ( m^{2}n \right )^{2/3}+\left ( mn^{2} \right )^{2/3}=\cos ^{2}\theta +\sin ^{2}\theta $$
or   $$\left ( m^{2}n \right )^{2/3}+\left ( mn^{2} \right )^{2/3}=1$$

Given, $$\displaystyle \frac { x }{ a } \cos  \theta +\frac { y }{ b } \sin  \theta =1,\frac { x }{ a } \sin  \theta -\frac { y }{ b } \cos  \theta =1$$

Squaring and adding them
$$\displaystyle \Rightarrow \frac { { x }^{ 2 } }{ { a }^{ 2 } } \cos ^{ 2 }{ \theta  } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } \sin ^{ 2 }{ \theta  } +\frac { 2xy }{ ab } \sin { \theta  } \cos { \theta  }+ $$

$$\displaystyle \frac { { x }^{ 2 } }{ { a }^{ 2 } } \sin ^{ 2 }{ \theta  } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } \cos ^{ 2 }{ \theta  } -\frac { 2xy }{ ab } \sin { \theta  } \cos { \theta  } =1+1$$

$$\displaystyle \Rightarrow \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =2$$

$$\Rightarrow \displaystyle \dfrac { \tan ^{ 2 }{ { 20 }^{ 0 } } -\sin ^{ 2 }{ { 20 }^{ 0 } }  }{ \tan ^{ 2 }{ { 20 }^{ 0 }. } \sin ^{ 2 }{ { 20 }^{ 0 } }  } =\dfrac { \dfrac { \sin ^{ 2 }{ { 20 }^{ 0 } }  }{ \cos ^{ 2 }{ { 20 }^{ 0 } }  } -\sin ^{ 2 }{ { 20 }^{ 0 } }  }{ \dfrac { \sin ^{ 2 }{ { 20 }^{ 0 } }  }{ \cos ^{ 2 }{ { 20 }^{ 0 } }  } \sin ^{ 2 }{ { 20 }^{ 0 } }  } $$

$$\displaystyle =\dfrac { \sin ^{ 2 }{ { 20 }^{ 0 } } -\sin ^{ 2 }{ { 20 }^{ 0 }\cos ^{ 2 }{ { 20 }^{ 0 } }  }  }{ \sin ^{ 4 }{ { 20 }^{ 0 } }  } $$

$$\sec { \theta  } +\tan { \theta  } =p$$   ...(1)

$$1+\tan ^{ 2 }{ x } =\sec ^{ 2 }{ x } \\ \Rightarrow 1=\sec ^{ 2 }{ x } -\tan ^{ 2 }{ x } \\ \Rightarrow 1=\left( \sec { x } +\tan { x }  \right) \left( \sec { x } -\tan { x }  \right) $$