Introduction to Trigonometry Test - 16

$co{t}^{2}A=\frac{9}{16}$

cot A = $\frac{3}{4}=\frac{Base}{Perpendicular}$

${\left(Hypotenuse\right)}^2={\left(Base\right)}^2+{\left(Perpendicular\right)}^2$

$H^2={\left(9\right)}^2+{\left(16\right)}^2=25$

$H=\sqrt{25}=5$

$\cos A = \frac{Base}{Hypotenuse}=\frac{3}{5}$

Therefore, option A is correct.

$a = (\mathrm{cosec} \theta + cot \theta ) and b = (\mathrm{cosec} \theta - cot \theta )$

$a \times b$$= (\mathrm{cosec} \theta + cot \theta ) (\mathrm{cosec} \theta - cot \theta )$

$={\mathrm{cosec}}^2\theta +cot\theta \mathrm{cosec}\theta - cot\theta \mathrm{cosec}\theta -co{t}^2\theta$

$={\mathrm{cosec}}^2\theta -co{t}^2\theta$

$=\frac{1}{{\sin }^2\theta }-\frac{1}{{\tan }^2\theta }$

$=\frac{1}{{\sin }^2\theta }-\frac{1}{{\displaystyle \frac{{\sin }^2\theta }{{\cos }^2\theta }}}$

$=\frac{1}{{\sin }^2\theta }-\frac{{\cos }^2\theta }{{\sin }^2\theta }$

$=\frac{1-{\cos }^2\theta }{{\sin }^2\theta }$

$=\frac{{\sin }^2\theta }{{\cos }^2\theta }=1$

Therefore, option B is correct.

$\sin A=\frac{1}{\sqrt{2}}$

squaring it we get,

${\sin }^{2}A=\frac{{\left(1\right)}^2}{{\left(\sqrt{2}\right)}^2}=\frac{1}{2}$

${\sin }^{2}A=\frac{{\left(Perpendicular\right)}^2}{{\left(Hypotenuse\right)}^2}=\frac{1}{2}$

${\left(Hypotenuse\right)}^2={\left(Base\right)}^2+{\left(Perpendicular\right)}^2$

${\left(Base\right)}^2={\left(Hypotenuse\right)}^2-{\left(Perpendicular\right)}^2$

$= 2 - 1 = {\left(1\right)}^2=1$ 

$\tan A= \frac{Perpendicular}{Base}=\frac{1}{1}$

$= 1$

Therefore, option A is correct.

$\cos (90° - \theta )$$. sec \theta . \tan (90° - \theta ). \mathrm{cosec} \theta = x$ 

By applying properties of trigonometry we get,

$\sin \theta ·\frac{1}{\cos \theta }·cot\theta ·\frac{1}{\sin \theta }=x$

$\Rightarrow \frac{cot\theta }{\cos \theta }=\frac{{\displaystyle \frac{1}{\tan \theta }}}{\cos \theta }$

$\Rightarrow \frac{{\displaystyle \frac{1}{{\displaystyle \frac{\sin \theta }{\cos \theta }}}}}{\cos \theta }=\frac{{\displaystyle \frac{\cos \theta }{\sin \theta }}}{\cos \theta }$

$\Rightarrow \frac{\cos \theta }{\cos \theta \times \sin \theta }=\frac{1}{\sin \theta }$

$=\mathrm{cosec} \theta$

Therefore, option C is correct.

$\sin P = \frac{Perpendicular}{Hypotenuse}=\frac{6}{10}$

squaring it we get,

${\sin }^{2}P=\frac{{\left(Perpendicular\right)}^2}{{\left(Hypotenuse\right)}^2}=\frac{{\left(6\right)}^2}{{\left(10\right)}^2}=\frac{36}{100}$

${\left(Hypotenuse\right)}^2={\left(Base\right)}^2+{\left(Perpendicular\right)}^2$

${\left(Base\right)}^2={\left(Hypotenuse\right)}^2-{\left(Perpendicular\right)}^2$

${\left(Base\right)}^2={\left(10\right)}^2-{\left(6\right)}^2=64$

${\cos }^{2}P=\frac{{\left(Base\right)}^2}{{\left(Hypotenuse\right)}^2}=\frac{64}{100}$

$=\frac{16}{25}$

Therefore, option B is correct.

$co{t}^{2}A=\frac{{\left(Base\right)}^2}{{\left(Perpendicular\right)}^2}$

${\tan }^{2}A=\frac{{\left(Perpendicular\right)}^2}{{\left(Base\right)}^2}$

$co{t}^{2}A\times {\tan }^{2}A=\left(\frac{{\left(Base\right)}^2}{{\left(Perpendicular\right)}^2}\times \frac{{\left(Perpendicular\right)}^2}{{\left(Base\right)}^2}\right)$

= 1

Therefore, option B is correct.

$\cos 55°. \cos ec 35° - 1$

$\Rightarrow \cos \left({90}^o-{35}^o\right)$ $· \cos ec 35° - 1$

$\Rightarrow \sin {35}^o·\frac{1}{\sin {35}^o}-1$

$\Rightarrow 1 - 1 = 0$

Therefore, option D is correct.

$\cos (A + B) = \cos A \cos B - \sin A \sin B$

$\cos 120°=\cos (90°+30°)$

$\cos (90° + 30°) = \cos 90° \cos 30° - \sin 90° \sin 30°$

Putting values we get,

$=0\times \frac{\sqrt{3}}{2}-1\times \frac{1}{2}=0-\frac{1}{2}$

$=\frac{-1}{2}$

Therefore, option B is correct.

$\cos 0° . \sin 25° . sec 85° . \tan 45° . \mathrm{cosec} 5° . \cos 65° . \sin 90°$

$= \cos 0° . \sin 25° . sec 85° . \tan 45° .$$\mathrm{cosec}\left({90}^o-{85}^o\right)·\cos \left({90}^o-{25}^o\right)·\sin \left({90}^o-{90}^o\right)$

$= \cos 0° . \sin 25° . sec 85° . \tan 45° ·sec {85}^o·\sin {25}^o·\cos 0^o$

$=1·\sin {25}^o·\sin {25}^o·sec {85}^o·sec {85}^o·1·1$

$={\sin }^2{25}^o·se{c}^2 {85}^o$

Therefore, option B is correct.

$$$\left[\left(\frac{se{c}^2\theta -{\tan }^2\theta }{1+co{t}^2\theta }\right)-{\sin }^2\theta \right]$

$\Rightarrow \left[\left(\frac{{\displaystyle \frac{1}{{\cos }^2\theta }}-{\displaystyle \frac{{\sin }^2\theta }{{\cos }^2\theta }}}{1+{\displaystyle \frac{1}{{\tan }^2\theta }}}\right)-{\sin }^2\theta \right]$

$\Rightarrow \left[\left(\frac{{\displaystyle \frac{1-{\sin }^2\theta }{{\cos }^2\theta }}}{1+{\displaystyle \frac{{\cos }^2\theta }{{\sin }^2\theta }}}\right)-{\sin }^2\theta \right]$

$\Rightarrow \left[\left(\frac{{\displaystyle \frac{{\cos }^2\theta }{{\cos }^2\theta }}}{{\displaystyle \frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta }}}\right)-{\sin }^2\theta \right]$

$\Rightarrow \left[\left(\frac{1}{{\displaystyle \frac{1}{{\sin }^2\theta }}}\right)-{\sin }^2\theta \right]$

$\Rightarrow {\sin }^2\theta -{\sin }^2\theta$

$=0$

Therefore, option A is correct.

We have, tanθ = (x sinϕ)/ (1−xcosϕ)

⇒ (1−xcos ϕ) / (x sin ϕ) = 1/ tanθ 

⇒ (1/ xsin ϕ) −cotϕ=cotθ

⇒ 1/ xsin ϕ= =cot θ+cot ϕ and 

tan ϕ = y sinθ / (1−y cosθ) 

⇒ (1−y cosθ)/ y sinθ = 1/ tan ϕ

⇒ (1/y sin θ) – cot θ = cotϕ

⇒ (1/ y sin θ) =cot ϕ+cot θ

⇒ (1/y sin θ) = (1/ x sin ϕ) 

⇒ x/y = sin θ/ sin ϕ

Therefore, option C is correct.

$\sin P =\frac{6}{3\sqrt{5}}$ , $\cos P = \frac{3}{3\sqrt{5}}$

squaring both of them we get,

${\sin }^{2}P = \frac{36}{45}, {\cos }^{2}P = \frac{9}{45}$

${\sin }^{2}P=\frac{{\left(Perpendicular\right)}^2}{{\left(Hypotenuse\right)}^2}=\frac{36}{45}$

${\cos }^{2}P=\frac{{\left(Base\right)}^2}{{\left(Hypotenuse\right)}^2}=\frac{9}{45}$

∴ ${\left(Hypotenuse\right)}^2=45$

${\left(Base\right)}^2=9$

${\left(Perpendicular\right)}^2=36$

$se{c}^{2}P=\frac{{\left(Hypotenuse\right)}^2}{{\left(Base\right)}^2}=\frac{45}{9}$

${\mathrm{cosec}}^{2}P=\frac{{\left(Hypotenuse\right)}^2}{{\left(Perpendicular\right)}^2}=\frac{45}{36}$

$se{c}^{2}P·{\mathrm{cosec}}^{2}P=\frac{45}{9}·\frac{45}{36}=\frac{25}{4}$

$sec P·\mathrm{cosec} P=\sqrt{\frac{25}{4}}=\frac{5}{2}$ 

Therefore, option B is correct.

$\sin {60}^o·\sin {30}^o+\cos {60}^o·\cos {30}^o-\frac{\sqrt{3}}{2}$

Putting the values we get:

$\Rightarrow \frac{\sqrt{3}}{2}·\frac{1}{2}+\frac{1}{2}·\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}$

$\Rightarrow \frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{2}$

$\Rightarrow \frac{\sqrt{3}+\sqrt{3}-2\sqrt{3}}{4}= \frac{2\sqrt{3}-2\sqrt{3}}{4}$

$\Rightarrow \frac{0}{4}=0$

Therefore, option D is correct.

$\sin \theta =\frac{1}{sec\theta }$

$sec\theta =\frac{1}{\cos \theta }$

∴ $\sin \theta =\frac{1}{{\displaystyle \frac{1}{\cos \theta }}}$

$\sin \theta =\cos \theta$

and as we know that the value of $\sin \theta and \cos \theta$ is equal at ${45}^o$.

i.e., $\frac{1}{\sqrt{2}}$

So, $\sin \theta =\frac{1}{sec\theta }={45}^o$

Therefore, option C is correct.

$\mathrm{cosec} \theta =$$\frac{1}{\sin \theta }$

$\Rightarrow cot \theta =$$\frac{1}{\tan \theta }$

$\Rightarrow \tan \theta =$$\frac{\sin \theta }{\cos \theta }$

$\Rightarrow \mathrm{cosec} \theta · cot \theta =$$\frac{1}{\sin \theta }·\frac{1}{\tan \theta }$

$\Rightarrow \frac{1}{\sin \theta }·\frac{1}{{\displaystyle \frac{\sin \theta }{\cos \theta }}}$

$\Rightarrow \frac{1}{\sin \theta }·\frac{\cos \theta }{\sin \theta }$

$\Rightarrow \frac{\cos \theta }{{\sin }^2\theta }$

Therefore, option C is correct.