Introduction to Trigonometry Test - 40

$$cosec(75^{0}+\theta)-\sec(15^{0}-\theta)-\tan(55^{0}+\theta)+\cot(35^{0}-\theta)$$
$$=\sec(90^{0}-75^{0}-\theta)-\sec(15^{0}-\theta)-\cot(90^{0}-55^{0}-\theta)+\cot(35^{0}-\theta)$$
$$=\sec(15^{0}-\theta)-\sec(15^{0}-\theta)-\cot(35^{0}+\theta)+\cot(35^{0}+\theta)$$
$$=0$$

$$sin\theta=\dfrac{12}{13}=\dfrac{P}{H}$$
$$P=12,\,\,H=13$$
$$\Rightarrow P^2+B^2=H^2\\B=\sqrt{H^2-P^2}=\sqrt{13^2-12^2}={5}$$

Hence
$$cos\theta=\dfrac{B}{H}=\dfrac{5}{13}$$
$$tan\theta=\dfrac{P}{B}=\dfrac{12}{5}$$

Substituting in the given equation, we get
$$\dfrac{2\dfrac{5}{13}+3\dfrac{12}{5}}{\dfrac{12}{13}+\dfrac{12\times12}{5\times13}}$$

$$=\dfrac{50+468}{144+60}$$

$$=\dfrac{518}{204}$$

$$=\dfrac{259}{102}$$

$$\sec { \theta  } +\tan { \theta  } =P\\ \Rightarrow { \left( \sec { \theta  } +\tan { \theta  }  \right)  }^{ 2 }={ P }^{ 2 }\\ \Rightarrow \sec ^{ 2 }{ \theta  } +\tan ^{ 2 }{ \theta  } +2\tan { \theta  } \sec { \theta  } ={ P }^{ 2 }$$
$$\displaystyle \Rightarrow 2\tan ^{ 2 }{ \theta  } +1+\frac { 2\sin { \theta  }  }{ \cos { \theta  }  } ={ P }^{ 2 }$$

$$\displaystyle \Rightarrow \frac { 2\sin ^{ 2 }{ \theta  } +\cos ^{ 2 }{ \theta  } +2\sin { \theta  }  }{ \cos ^{ 2 }{ \theta  }  } ={ P }^{ 2 }$$

Applying componendo and dividendo
$$\displaystyle \frac { 2\sin ^{ 2 }{ \theta  } +\cos ^{ 2 }{ \theta  } +2\sin { \theta  } -\cos ^{ 2 }{ \theta  }  }{ 2\sin ^{ 2 }{ \theta  } +\cos ^{ 2 }{ \theta  } +2\sin { \theta  } -\cos ^{ 2 }{ \theta  }  } =\frac { { P }^{ 2 }-1 }{ { P }^{ 2 }+1 } $$

$$\displaystyle \Rightarrow \frac { 2\sin { \theta  } \left( \sin { \theta  } +1 \right)  }{ 2\left( \sin { \theta  } +1 \right)  } =\frac { { P }^{ 2 }-1 }{ { P }^{ 2 }+1 } \Rightarrow \sin { \theta  } =\frac { { P }^{ 2 }-1 }{ { P }^{ 2 }+1 } $$

Given $$ \sin{x}+\sin^{2}{x}=1$$
$$\Rightarrow\sin{x}=1-\sin^{2}{x}$$
$$\Rightarrow \sin x=\cos^{2}{x}$$

Now,  $$\cos^{2}{x}+\cos^{4}{x}$$
             $$=\sin{x}+\sin^{2}{x}$$
             $$=1$$

$$\sin\theta.\sec^{7}\theta+\cos\theta.co\sec^{7}\theta$$
$$=\tan\theta.\sec^{6}\theta+\cot\theta.co\sec^{6}\theta$$.
$$=\tan\theta(1+\tan^{2}\theta)^{3}+\cot\theta(1+\cot^{2}\theta)^{3}$$
$$=\sqrt{\dfrac{a}{b}}(1+\dfrac{a}{b})^{3}+\sqrt{\dfrac{b}{a}}(1+\dfrac{b}{a})^{3}$$

$$\sin\theta +cosec\theta =2\\ \Rightarrow \sin\theta +\cfrac { 1 }{ \sin\theta  } =2$$
Squaring both sides
$${ \left( \sin\theta +\cfrac { 1 }{ \sin\theta  }  \right)  }^{ 2 }={ 2 }^{ 2 }\\ \Rightarrow { \sin }^{ 2 }\theta +\cfrac { 1 }{ { \sin }^{ 2 }\theta  } +2\sin\theta \cfrac { 1 }{ \sin\theta  } =4\\ \Rightarrow { \sin }^{ 2 }\theta +\cfrac { 1 }{ { \sin }^{ 2 }\theta  } =2$$
Again squaring bot sides
$${ \left( { \sin }^{ 2 }\theta +\cfrac { 1 }{ { \sin }^{ 2 }\theta  }  \right)  }^{ 2 }={ 2 }^{ 2 }\\ \Rightarrow { \sin }^{ 4 }\theta +\cfrac { 1 }{ { \sin }^{ 4 }\theta  } =2$$
Similarly
$${ \sin }^{ 8 }\theta +\cfrac { 1 }{ { \sin }^{ 8 }\theta  } =2$$
$$\Rightarrow { \sin }^{ 8 }\theta +{ cosec }^{ 8 }\theta =2$$

Given,
$$A+B=90^{ \circ  }$$
$$=>A=90^{ \circ  }-B$$
Given,
$$\tan A=\dfrac{3}{4}$$
$$=>\cot B=\dfrac{3}{4}$$

$$\quad (x-2)\sin^230^{\small\circ} + (x-3)\tan^260^{\small\circ} - x\cos^245^{\small\circ} = \displaystyle\frac{17}{4}$$

$$\quad (x-2)\left(\displaystyle\frac{1}{2}\right)^2 + (x-3)(\sqrt3)^2 - x\left(\displaystyle\frac{1}{\sqrt2}\right)^2 = \displaystyle\frac{17}{4}$$

$$\quad \Rightarrow \displaystyle\frac{x-2}{4} + 3(x-3) - \displaystyle\frac{x}{2} = \displaystyle\frac{17}{4}$$

$$\quad \Rightarrow (x-2) + 12x - 36 - 2x = 17$$

$$\quad \Rightarrow x + 12x - 2x = 17 + 36 + 2 \quad \Rightarrow 11x = 55$$

$$\quad \Rightarrow x = 5$$

$$\sqrt{(1-\cos^{2}\theta)\sec^{2}\theta}$$