Introduction to Trigonometry Test - 12

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

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

${\left(\sqrt{2}\right)}^2={\left(1\right)}^2+{\left(Base\right)}^2$

${\left(Base\right)}^2={\left(\sqrt{2}\right)}^2-{\left(1\right)}^2$

$= 2 - 1 = 1$

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

= $\frac{1}{1}$ =$1$

Hence, the correct option is (A).

$\frac{Hypotenuse}{Side}$ adjacent to $\angle A=\frac{12}{5}$

$\frac{AC}{AB}=\frac{12}{5}$

Let $AC$ be $12k$, $AB$ will be $5k$, where $k$ is a positive integer.

Applying Pythagoras theorem in $△ABC$, we obtain

$A{C}^2=A{B}^2+B{C}^2$

${\left(12k\right)}^2={\left(5k\right)}^2+B{C}^2$

$144k^2=25k^2+B{C}^2$

$B{C}^2=119k^2$

$BC=10.9k$

It can be observed that for given two sides $AC=12k$ and $AB=5k$,

$BC$ should be such that,

$AC-AB<BC<AC+AB$

$12k-5k<BC<12k+5k$

$7k<BC<17k$

However, $BC=10.9k$. Clearly, such a triangle is possible and hence, such a value of $sec A$ is possible.

So, the given statement is true.

Hence, the correct option is (A).

cosec $\theta$ is the reciprocal of sin $\theta$ i.e., 

cosec $\theta = \frac{1}{\sin \theta } = \frac{Hypotenuse}{Perpendicular}$

So, sin $\theta$ will be $\frac{Perpendicular}{Hypotenuse}$

= $\frac{4}{5}$

∴ ${\sin }^2\theta =\frac{{\left(Perpendicular\right)}^2}{{\left(Hypotenuse\right)}^2}=\frac{{\left(4\right)}^2}{{\left(5\right)}^2}$

= $\frac{16}{25}$

Hence, the correct option is (D).

Given,

$15 cot A=8$

$cot A=\frac{8}{15}$

$\Rightarrow \tan A=\frac{15}{8} \left(\tan A=\frac{1}{cot A}\right)$

We know that,

$\tan \theta =\frac{opposite side}{adjacent side}$

Consider the attached figure, triangle ABC.

From Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$

$A{C}^2=8^2+{15}^2=64+225=289$

$AC=17$

$\cos A=\frac{adjacent side}{Hypotenuse}=\frac{AB}{AC}=\frac{8}{17}$

$sec A=\frac{1}{\cos A}=\frac{1}{\raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$17$}\right.}\frac{17}{8}$

$\sin A=\frac{opposite side}{Hypotenuse}=\frac{BC}{AC}=\frac{15}{17}$

Hence, the correct option is (A).

$se{c}^3\theta \times cot \theta$

$\Rightarrow \frac{1}{{\cos }^3\theta }\times \frac{1}{\tan \theta }$

Converting it into angle formula

$\Rightarrow {\left(\frac{1}{{\displaystyle \frac{Base}{Hypotenuse}}}\right)}^3\times \left(\frac{1}{{\displaystyle \frac{Perpendicular}{Base}}}\right)$

= $\frac{{\left(Hypotenuse\right)}^3}{{\left(Base\right)}^2}\times \frac{1}{Perpendicular}$

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

= $se{c}^2\theta \times \cos ec\theta$

Hence, the correct option is (A).

$\sin M =\frac{4}{4\sqrt{2}} = \frac{Perpendicular}{Hypotenuse}$

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

$MN = 4, MO = 4\sqrt{2}, NO = ?$

$N{O}^2=M{O}^2-M{N}^2$

= ${\left(4\sqrt{2}\right)}^2-{\left(4\right)}^2$

$N{O}^2= 32 - 16 = 16$

$NO=\sqrt{16} = 4$

$\cos M =\frac{Base}{Hypotenuse}=\frac{NO}{MO}$

$=\frac{4}{4\sqrt{2}}=\frac{1}{\sqrt{2}}$

Hence, the correct option is (A).

By identity, $Sec A = Cosec (90° - A)$

$Sec$ will be eliminated

$= A = 90° - B$

$=A+B=90°$

Hence, the correct option is (C).

Squaring $\frac{sec {45}^o}{cot {60}^o}=\sqrt{\frac{3x}{2}}$ we get,

$\frac{se{c}^2{45}^o}{co{t}^2{60}^o}={\left(\sqrt{\frac{3x}{2}}\right)}^2$

${\left(\frac{\sqrt{2}}{{\displaystyle \frac{1}{\sqrt{3}}}}\right)}^2={\left(\sqrt{\frac{3x}{2}}\right)}^2$

$\frac{2}{{\displaystyle \frac{1}{3}}}\times \frac{3x}{2}$

$\Rightarrow \frac{2}{1}\times \frac{3}{1}=\frac{3x}{2}$

$\Rightarrow x=\frac{2\times 2\times 3}{3}=4$

Hence, the correct option is (A).

$\frac{\sin {75}^o}{\cos {15}^o}÷\frac{\cos ec {60}^o}{sec {30}^o}-1$

$=\frac{\sin \left(90°-15°\right)}{\cos 15°}÷\frac{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}$}\right.}}{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}$}\right.}}-1$

$=\frac{\cos 15°}{\cos 15°}÷1-1$

$=1÷1-1$

$=0$

Hence, the correct option is (D).

Let us assume an $△ABC$ in which $\angle B = 90°$ and $\angle C = \theta$

Given:

$cot \theta = \frac{BC}{AB} = \frac{7}{8}$

Let $BC = 7k$ and $AB = 8k$, where $k$ is a positive real number

According to Pythagoras theorem in $△ABC$ we get.

$A{C}^2 = A{B}^2+B{C}^2$

$A{C}^2 = {\left(8k\right)}^2+{\left(7k\right)}^2$

$A{C}^2 = 64k^2+49k^2$

$AC = \sqrt{113} k$

According to the sine and cos function ratios, it is written as:

$\sin \theta = \frac{AB}{AC}=$ $\frac{Opposite Side}{Hypotenuse} = \frac{8k}{\sqrt{113}k} = \frac{8}{\sqrt{113}}$ and

$\cos \theta =$ $\frac{Adjacent Side}{Hypotenuse}= \frac{BC}{AC} = \frac{7k}{\sqrt{113}k} = \frac{7}{\sqrt{113}}$

Now apply the values of sin function and cos function:

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

$=\frac{1-{\left({\displaystyle \frac{8}{\sqrt{113}}}\right)}^2}{1-{\left({\displaystyle \frac{7}{\sqrt{113}}}\right)}^2}=\frac{49}{64}$

Hence, the correct option is (C).

$m = 1 + {\tan }^2\theta$ ...(1)

$n = 1 + co{t}^2\theta$ ...(2)

$\frac{m}{n}=\frac{1+{\tan }^2\theta }{1+co{t}^2\theta }$

= $\frac{1+{\tan }^2\theta }{{\displaystyle \frac{1}{1}}+{\displaystyle \frac{1}{{\tan }^2\theta }}}$

= $\frac{1+{\tan }^2\theta }{{\displaystyle \frac{{\tan }^2\theta +1}{{\tan }^2\theta }}}$

= $\frac{\left(1+{\tan }^2\theta \right)\left({\tan }^2\theta \right)}{\left(1+{\tan }^2\theta \right)}$

= $\frac{m}{n}={\tan }^2\theta$

from (1)

${\tan }^2\theta =m-1$

∴ $\frac{m}{n}=m-1$

Hence, the correct option is (B).

Let us take ${45}^o$

square the equation and solve it 

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

${\sin }^2\left(x\right)+{\cos }^2\left(x\right)+2 \sin \left(x\right) \cos \left(x\right)$

${\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2+2\times \left(\frac{1}{\sqrt{2}}\right)\times \left(\frac{1}{\sqrt{2}}\right)$

$\frac{1}{2}+\frac{1}{2}+2\times \frac{1}{2}$

$\frac{1+1+2}{2}=\frac{4}{2}$ = $2$

Hence, the correct option is (A).

From the basic formula we can get the value of the following:

$\sin {45}^o = \frac{1}{\sqrt{2}}$

$\cos {90}^o = 0$

$\tan {60}^o = \sqrt{3}$

$\cos {75}^o = \frac{\left(\sqrt{6}-\sqrt{2}\right)}{4}$

$cot {30}^o =\sqrt{3}$

$\therefore \sin {45}^o·\cos {90}^o·\tan {60}^o·\cos {75}^o·cot {30}^o$

= $\frac{1}{\sqrt{2}}\times 0\times \sqrt{3}\times \frac{\left(\sqrt{6}-\sqrt{2}\right)}{4}\times \sqrt{3}$

$=0$

Hence, the correct option is (A).

$\sin {30}^o$ = $\frac{1}{2}$

$2x = \sin 30°$

$\Rightarrow 2x=\frac{1}{2}$

$\Rightarrow x=\frac{1}{4}$

put the value of $x$ in $x+y=\cos ec {45}^o$

$\frac{1}{4}+y=\sqrt{2}$

$y =\sqrt{2}-\frac{1}{4}$

Hence, the correct option is (B).

$\cos A=\frac{1}{\sqrt{5}}=\frac{Base}{Hypotenuse}$

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

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

= ${\left(\sqrt{5}\right)}^2 - {\left(1\right)}^2$

$= 5 - 1 = 4$

$Perpendicular=\sqrt{4}=2$

$\sin A=\frac{Perpendicular}{Hypotenuse}=\frac{2}{\sqrt{5}}$

Hence, the correct option is (B).