Introduction to Trigonometry Test - 18

Let $AB=5k$ and $AC=13k$

Using Pythagorean theorem:

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

$\therefore BC=\sqrt{{\left(13k\right)}^2-{\left(5k\right)}^2}=\sqrt{169k^2-25k^2}$

$\Rightarrow BC=\sqrt{144k^2}=12k$

$\therefore \cos \theta =\frac{BC}{AC}=\frac{12k}{13k}=\frac{12}{13}$

Hence, the correct option is (D).

In $△ABC, \angle ABC=90°$

$\therefore \tan A=\frac{BC}{AB}$

Since $\tan A=1$ (Given) $\frac{BC}{AB}=1$

$\therefore BC=AB$

Let $AB=BC=k$, where $k$ is a positive number.

Now, $A{C}^2=A{B}^2+B{C}^2$

$\therefore AC=\sqrt{A{B}^2+B{C}^2}=\sqrt{k^2+k^2}$

$\therefore AC=k\sqrt{2}$

$\therefore \sin A=\frac{BC}{AC}=\frac{k}{k\sqrt{2}}=\frac{1}{\sqrt{2}}$

$\cos A=\frac{AB}{AC}=\frac{k}{k\sqrt{2}}=\frac{1}{\sqrt{2}}$

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

$\therefore 2 \sin A \cos A=1$

Hence, the correct option is (D).

$=\cos A sec A$

$=\cos A\times \frac{1}{\cos A}$

$=1\times 1$

$=1$

Therefore, $\cos A sec A=1$ is true.

Hence, the correct option is (B).

Given, $\tan \theta =\frac{20}{21}$

Dividing all terms of $\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }$ by $\cos \theta$,

$=\frac{1-\tan \theta }{1+\tan \theta }=\frac{1-{\displaystyle \frac{20}{21}}}{1+{\displaystyle \frac{20}{21}}}=\frac{21-20}{21+20}=\frac{1}{41}$

Hence, the correct option is (D).

Given, $5 \tan \alpha =4$

Dividing all terms of $\frac{5 \sin \alpha -3 \cos \alpha }{5 \sin \alpha +2 \cos \alpha }$ by $\cos \alpha$

$=\frac{5 \tan \alpha -3}{5 \tan \alpha +2}=\frac{4-3}{4+2}=\frac{1}{6}$

Hence, the correct option is (D).

Given, $\tan \theta =\frac{a}{b}$

Dividing all terms of $\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }$ by $\cos \theta$,

$=\frac{1+\tan \theta }{1-\tan \theta }$

$=\frac{1+{\displaystyle \frac{a}{b}}}{1-{\displaystyle \frac{a}{b}}}$

$=\frac{b+a}{b-a}$

Hence, the correct option is (B).

$Tan \theta =\frac{opposite side}{efficient side}=\frac{20}{21}$

Let $x$ be the hypotenuse by applying Pythagoras we get

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

$x^2={\left(20\right)}^2+{\left(21\right)}^2$

$x^2=400+441$

$x^2=841$

$\Rightarrow x=29$

$\sin \theta =\frac{AB}{AC}=\frac{20}{29}$

$\cos \theta =\frac{BC}{AC}=\frac{21}{29}$

Substitute $\sin \theta , \cos \theta$ in the equation we get

$=\frac{1-\sin \theta +\cos \theta }{1+\sin \theta +\cos \theta }$

$\Rightarrow \frac{1-{\displaystyle \frac{20}{29}}+{\displaystyle \frac{21}{29}}}{1+{\displaystyle \frac{20}{29}}+{\displaystyle \frac{21}{29}}}=\frac{{\displaystyle \frac{29-20+21}{29}}}{{\displaystyle \frac{29+20+21}{29}}}=\frac{30}{70}=\frac{3}{7}$

Hence, the correct option is (B).

Given, $4 \tan \theta =3$

Dividing all terms of $\frac{4 \sin \theta -\cos \theta }{4 \sin \theta +\cos \theta }$ by $\cos \theta$,

$=\frac{4 \tan \theta -1}{4 \tan \theta +1}=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2}$

Hence, the correct option is (D).

Given, $cot \theta =\frac{7}{8}$

Now, $\frac{\left(1+\sin \theta \right)\left(1-\sin \theta \right)}{\left(1+\cos \theta \right)\left(1-\cos \theta \right)}$

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

$={\left(\frac{7}{8}\right)}^2$

$=\frac{49}{64}$

Hence, the correct option is (C).

We have

$\sin A+{\sin }^{2}A=1$

$\Rightarrow \sin A=1-{\sin }^{2}A$

$\Rightarrow \sin A={\cos }^{2}A$...(i)

Squaring both sides

$\Rightarrow {\sin }^{2}A={\cos }^{4}A$...(ii)

From equation (i) and (ii), we have

${\cos }^{2}A+{\cos }^{4}A=\sin A+{\sin }^{2}A=1$

Hence, the correct option is (A).

$\cos ec\left(75°+\theta \right)-sec\left(15°-\theta \right)-\tan \left(55°+\theta \right)+cot\left(35°-\theta \right)$

$=\cos ec\left(75°+\theta \right)-\cos ec\left[90°-\left(15°-\theta \right)\right]-\tan \left(55°+\theta \right)+\tan \left[90°-\left(35°-\theta \right)\right]$

$=\cos ec\left(75°+\theta \right)-\cos ec\left(75°+\theta \right)-\tan \left(55°+\theta \right)+\tan \left(55°+\theta \right)$

$=0$

Hence, the correct option is (C).

$\tan 90°=1=\frac{AB}{BC}$

$\Rightarrow AB=BC$

Let $AB=BC=R$

Applying Pythagoras theorem

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

$\Rightarrow AC=\sqrt{A{B}^2+B{C}^2}$

$\therefore AC=\sqrt{{\left(k\right)}^2+{\left(k\right)}^2}=\sqrt{k^2+k^2}$

$\Rightarrow AC=\sqrt{2k^2}=\sqrt{2}k$

$\therefore \cos C \sin A-\cos A \sin C=\frac{k}{\sqrt{2}k}\times \frac{k}{\sqrt{2}k}-\frac{k}{\sqrt{2}k}$$\times \frac{k}{\sqrt{2}k}$

$=\frac{1}{2}-\frac{1}{2}=0$

Hence, the correct option is (A).

From trigonometric identity

$1+{\tan }^{2}A=se{c}^{2}A$

$\Rightarrow se{c}^{2}A-1={\tan }^{2}A$

$\Rightarrow \frac{1}{{\cos }^{2}A}-1={\tan }^{2}A$

$\Rightarrow {\left(\frac{5}{4}\right)}^2-1={\tan }^{2}A$

$\Rightarrow \frac{9}{16}={\tan }^{2}A$

$\Rightarrow \tan A=\frac{3}{4}$

Hence, the correct option is (B).

We have

Given, $\sin A = \frac{a}{b}$

As, $\sin A=\frac{Perpendicular}{Hypotenus}$

Let the complete ratio be $x$

Perpendicular $=ax$

Hypotenuse $=bx$

Now,

$Bas{e}^2+Perpendicula{r}^2=Hypotenu{s}^2$

$Bas{e}^2+{\left(ax\right)}^2={\left(bx\right)}^2$

$Bas{e}^2={\left(bx\right)}^2-{\left(ax\right)}^2$

$Base=x\sqrt{b^2-a^2}$

$\cos A=\frac{Base}{Hypotenus}=\frac{x\sqrt{b^2-a^2}}{bx}=\frac{\sqrt{b^2-a^2}}{b}$

Hence, the correct option is (B).

Given,

$\sin A=\frac{12}{13}$

$\cos A=\sqrt{1-{\sin }^{2}A}$

$=\sqrt{1-{\left(\frac{12}{13}\right)}^2}$

$=\sqrt{1-\frac{144}{169}}$

$=\sqrt{\frac{169-144}{169}}$

$=\sqrt{\frac{25}{169}}$

$\sqrt{{\left(\frac{5}{13}\right)}^2}$

$\cos A=\frac{5}{13}$

And,

$\tan A=\frac{\sin A}{\cos A}$

$=\frac{{\displaystyle \raisebox{1ex}{$12$}\!\left/ \!\raisebox{-1ex}{$13$}\right.}}{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$13$}\right.}}$

$\therefore \tan A=\frac{12}{5}$

Hence, the correct option is (A).