Solution

Applying Pythagoras theorem for $\mathrm{△}ABC$, we obtain

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

$=(24cm{)}^2+(7cm{)}^2$

$=(576+49)c{m}^2$

$=625c{m}^2$

$\therefore AC=\sqrt{625}cm=25cm$

(i) $\sin A={\displaystyle \frac{\text{ Side opposite to }\mathrm{\angle }A}{\text{ Hypotenuse }}}={\displaystyle \frac{BC}{AC}}$

$={\displaystyle \frac{7}{25}}$

$\cos A=$

${\displaystyle \frac{\text{ Side adjacent to }\mathrm{\angle }A}{\text{ Hypotenuse }}}={\displaystyle \frac{AB}{AC}}={\displaystyle \frac{24}{25}}$

(ii)

$\begin{array}{rl}& \sin C={\displaystyle \frac{\text{ Side opposite to }\mathrm{\angle }C}{\text{ Hypotenuse }}}={\displaystyle \frac{AB}{AC}}\\ & ={\displaystyle \frac{24}{25}}\end{array}$

$\begin{array}{rl}& {\displaystyle \frac{\text{ Side adjacent to }\mathrm{\angle }C}{\text{ Hypotenuse }}}={\displaystyle \frac{BC}{AC}}\\ & \cos C=\\ & ={\displaystyle \frac{7}{25}}\end{array}$

Solution

Applying Pythagoras theorem for $\mathrm{△}PQR$, we obtain

$P{R}^2=P{Q}^2+Q{R}^2$

$(13cm{)}^2=(12cm{)}^2+Q{R}^2$

$169c{m}^2=144c{m}^2+Q{R}^2$

$25c{m}^2=Q{R}^2$

$QR=5cm$

$\begin{array}{rl}\tan P& ={\displaystyle \frac{\text{ Side opposite to }\mathrm{\angle }P}{\text{ Side adjacent to }\mathrm{\angle }P}}={\displaystyle \frac{QR}{PQ}}\\ & ={\displaystyle \frac{5}{12}}\end{array}$

$\begin{array}{rl}\cot R& ={\displaystyle \frac{\text{ Side adjacent to }\mathrm{\angle }R}{\text{ Side opposite to }\mathrm{\angle }R}}={\displaystyle \frac{QR}{PQ}}\\ & ={\displaystyle \frac{5}{12}}\end{array}$

$\tan P-\cot R={\displaystyle \frac{5}{12}}-{\displaystyle \frac{5}{12}}=0$

Solution

Let $\mathrm{△}ABC$ be a right-angled triangle, right-angled at point $B$.

Given that,

$\begin{array}{rl}& \sin A={\displaystyle \frac{3}{4}}\\ & {\displaystyle \frac{BC}{AC}}={\displaystyle \frac{3}{4}}\end{array}$

Let $BC$ be $3k$. Therefore, $AC$ will be $4k$, where $k$ is a positive integer.

Applying Pythagoras theorem in $\mathrm{△}ABC$, we obtain

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

$(4k{)}^2=A{B}^2+(3k{)}^2$

$16k^2-9k^2=A{B}^2$

$7k^2=A{B}^2$

$AB=\sqrt{7}k$

$\cos A={\displaystyle \frac{\text{ Side adjacent to }\mathrm{\angle }A}{\text{ Hypotenuse }}}$

$={\displaystyle \frac{AB}{AC}}={\displaystyle \frac{\sqrt{7k}}{4k}}={\displaystyle \frac{\sqrt{7}}{4}}$

$\tan A={\displaystyle \frac{\text{ Side opposite to }\mathrm{\angle }A}{\text{ Side adjacent to }\mathrm{\angle }A}}$

$={\displaystyle \frac{BC}{AB}}={\displaystyle \frac{3k}{\sqrt{7}k}}={\displaystyle \frac{3}{\sqrt{7}}}$

Solution

Consider a right-angled triangle, right-angled at $B$.

$\cot A={\displaystyle \frac{\text{ Side adjacent to }\mathrm{\angle }A}{\text{ Side opposite to }\mathrm{\angle }A}}$

$={\displaystyle \frac{AB}{BC}}$

It is given that,

$\begin{array}{rl}\cot A& ={\displaystyle \frac{8}{15}}\\ {\displaystyle \frac{AB}{BC}}& ={\displaystyle \frac{8}{15}}\end{array}$

Let $AB$ be $8k$.Therefore, $BC$ will be $15k$, where $k$ is a positive integer.

Applying Pythagoras theorem in $\mathrm{△}ABC$, we obtain

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

$=(8k{)}^2+(15k{)}^2$

$=64k^2+225k^2$

$=289k^2$

$AC=17k$

$\begin{array}{rl}\sin A& ={\displaystyle \frac{\text{ Side opposite to }\mathrm{\angle }A}{\text{ Hypotenuse }}}={\displaystyle \frac{BC}{AC}}\\ & ={\displaystyle \frac{15k}{17k}}={\displaystyle \frac{15}{17}}\end{array}$

$\sec A={\displaystyle \frac{\text{ Hypotenuse }}{\text{ Side adjacent to }\mathrm{\angle }A}}$

$={\displaystyle \frac{AC}{AB}}={\displaystyle \frac{17}{8}}$

Solution

Consider a right-angle triangle $\mathrm{△}ABC$, right-angled at point $B$.

$\sec \theta ={\displaystyle \frac{\text{ Hypotenuse }}{\text{ Side adjacent to }\mathrm{\angle }\theta }}$

${\displaystyle \frac{13}{12}}={\displaystyle \frac{AC}{AB}}$

If $AC$ is $13k,AB$ will be $12k$, where $k$ is a positive integer.

Applying Pythagoras theorem in $\mathrm{△}ABC$, we obtain

$(AC{)}^2=(AB{)}^2+(BC{)}^2$

$(13k{)}^2=(12k{)}^2+(BC{)}^2$

$169k^2=144k^2+B{C}^2$

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

$BC=5k$

$\sin \theta ={\displaystyle \frac{\text{ Side opposite to }\mathrm{\angle }\theta }{\text{ Hypotenuse }}}={\displaystyle \frac{BC}{AC}}={\displaystyle \frac{5k}{13k}}={\displaystyle \frac{5}{13}}$

$\cos \theta ={\displaystyle \frac{\text{ Side adjacent to }\mathrm{\angle }\theta }{\text{ Hypotenuse }}}={\displaystyle \frac{AB}{AC}}={\displaystyle \frac{12k}{13k}}={\displaystyle \frac{12}{13}}$

$\tan \theta ={\displaystyle \frac{\text{ Side opposite to }\mathrm{\angle }\theta }{\text{ Side adjacent to }\mathrm{\angle }\theta }}={\displaystyle \frac{BC}{AB}}={\displaystyle \frac{5k}{12k}}={\displaystyle \frac{5}{12}}$

$\cot \theta ={\displaystyle \frac{\text{ Side adjacent to }\mathrm{\angle }\theta }{\text{ Side opposite to }\mathrm{\angle }\theta }}={\displaystyle \frac{AB}{BC}}={\displaystyle \frac{12k}{5k}}={\displaystyle \frac{12}{5}}$

$cosec\theta ={\displaystyle \frac{\text{ Hypotenuse }}{\text{ Side opposite to }\mathrm{\angle }\theta }}={\displaystyle \frac{AC}{BC}}={\displaystyle \frac{13k}{5k}}={\displaystyle \frac{13}{5}}$

Solution

Let us consider a triangle $ABC$ in which $CD\perp AB$.

It is given that

$\cos A=\cos B$ $\Rightarrow {\displaystyle \frac{AD}{AC}}={\displaystyle \frac{BD}{BC}}$

We have to prove $\mathrm{\angle }A=\mathrm{\angle }B$. To prove this, let us extend $AC$ to $P$ such that $BC=CP$.

From equation (1), we obtain

${\displaystyle \frac{AD}{BD}}={\displaystyle \frac{AC}{BC}}$

$\Rightarrow {\displaystyle \frac{AD}{BD}}={\displaystyle \frac{AC}{CP}}$

(By construction, we have $BC=CP$ )

By using the converse of B.P.T,

$CD\mid BP$

$\Rightarrow \mathrm{\angle }ACD=\mathrm{\angle }CPB$ (Corresponding angles) ..

And, $\mathrm{\angle }BCD=\mathrm{\angle }CBP$ (Alternate interior angles)

By construction, we have $BC=CP$.

$\therefore \mathrm{\angle }CBP=\mathrm{\angle }CPB$ (Angle opposite to equal sides of a triangle) $\dots$ (5)

From equations (3), (4), and (5), we obtain

$\mathrm{\angle }ACD=\mathrm{\angle }BCD$.

In $\mathrm{△}CAD$ and $\mathrm{△}CBD$,

$\mathrm{\angle }ACD=\mathrm{\angle }BCD$ [Using equation (6)]

$\mathrm{\angle }CDA=\mathrm{\angle }CDB[.$ Both ${90}^{\circ }$ ]

Therefore, the remaining angles should be equal.

$\therefore \mathrm{\angle }CAD=\mathrm{\angle }CBD$

$\Rightarrow \mathrm{\angle }A=\mathrm{\angle }B$

Alternatively,

Let us consider a triangle $ABC$ in which $CD\perp AB$.

It is given that,

$\cos A=\cos B$

$\Rightarrow {\displaystyle \frac{AD}{AC}}={\displaystyle \frac{BD}{BC}}$

$\Rightarrow {\displaystyle \frac{AD}{BD}}={\displaystyle \frac{AC}{BC}}$

Let ${\displaystyle \frac{AD}{BD}}={\displaystyle \frac{AC}{BC}}=k$

$\Rightarrow AD=kBD$.

And, $AC=kBC$

Using Pythagoras theorem for triangles CAD and CBD, we obtain

$C{D}^2=A{C}^2-A{D}^2$..

And, $C{D}^2=B{C}^2-B{D}^2$.

From equations (3) and (4), we obtain

$A{C}^2-A{D}^2=B{C}^2-B{D}^2$

$\Rightarrow (kBC{)}^2-(kBD{)}^2=B{C}^2-B{D}^2$

$\Rightarrow k^2(B{C}^2-B{D}^2)=B{C}^2-B{D}^2$

$\Rightarrow k^2=1$

$\Rightarrow k=1$

Putting this value in equation (2), we obtain

$AC=BC$

$\Rightarrow \mathrm{\angle }A=\mathrm{\angle }B$ (Angles opposite to equal sides of a triangle)

Solution

Let us consider a right triangle $ABC$, right-angled at point $B$.

$\begin{array}{rl}\cot \theta & ={\displaystyle \frac{\text{ Side adjacent to }\mathrm{\angle }\theta }{\text{ Side opposite to }\mathrm{\angle }\theta }}={\displaystyle \frac{BC}{AB}}\\ & ={\displaystyle \frac{7}{8}}\end{array}$

If $BC$ is $7k$, then $AB$ will be $8k$, where $k$ is a positive integer.

Applying Pythagoras theorem in $\mathrm{△}ABC$, we obtain

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

$=(8k{)}^2+(7k{)}^2$

$=64k^2+49k^2$

$=113k^2$

$AC=\sqrt{113}k$

$\sin \theta ={\displaystyle \frac{\text{ Side opposite to }\mathrm{\angle }\theta }{\text{ Hypotenuse }}}={\displaystyle \frac{AB}{AC}}$

$={\displaystyle \frac{8k}{\sqrt{113}k}}={\displaystyle \frac{8}{\sqrt{113}}}$

$\cos \theta ={\displaystyle \frac{\text{ Side adjacent to }\mathrm{\angle }\theta }{\text{ Hypotenuse }}}={\displaystyle \frac{BC}{AC}}$

$={\displaystyle \frac{7k}{\sqrt{113}k}}={\displaystyle \frac{7}{\sqrt{113}}}$

(i) ${\displaystyle \frac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}}={\displaystyle \frac{(1-{\sin }^2\theta )}{(1-{\cos }^2\theta )}}$ $={\displaystyle \frac{1-({\displaystyle \frac{8}{\sqrt{113}}}{)}^2}{1-({\displaystyle \frac{7}{\sqrt{113}}}{)}^2}}={\displaystyle \frac{1-{\displaystyle \frac{64}{113}}}{1-{\displaystyle \frac{49}{113}}}}$

$={\displaystyle \frac{{\displaystyle \frac{49}{113}}}{64}}={\displaystyle \frac{49}{64}}$

113

(ii) ${\cot }^2\theta =(\cot \theta {)}^2=({\displaystyle \frac{7}{8}}{)}^2={\displaystyle \frac{49}{64}}$

Solution

It is given that $3\cot A=4$

Or, $\cot A={\displaystyle \frac{4}{3}}$

Consider a right triangle $ABC$, right-angled at point $B$.

$\cot A={\displaystyle \frac{\text{ Side adjacent to }\mathrm{\angle }A}{\text{ Side opposite to }\mathrm{\angle }A}}$

${\displaystyle \frac{AB}{BC}}={\displaystyle \frac{4}{3}}$

If $AB$ is $4k$, then $BC$ will be $3k$, where $k$ is a positive integer.

In $\mathrm{△}ABC$,

$(AC{)}^2=(AB{)}^2+(BC{)}^2$

$=(4k{)}^2+(3k{)}^2$

$\begin{array}{rl}& =16k^2+9k^2\\ & =25k^2\\ & AC=5k\\ & \cos A={\displaystyle \frac{\text{ Side adjacent to }\mathrm{\angle }A}{\text{ Hypotenuse }}}={\displaystyle \frac{AB}{AC}}\\ & ={\displaystyle \frac{4k}{5k}}={\displaystyle \frac{4}{5}}\\ & \sin A={\displaystyle \frac{\text{ Side opposite to }\mathrm{\angle }A}{\text{ Hypotenuse }}}={\displaystyle \frac{BC}{AC}}\\ & ={\displaystyle \frac{3k}{5k}}={\displaystyle \frac{3}{5}}\\ & \tan A={\displaystyle \frac{\text{ Side opposite to }\mathrm{\angle }A}{\text{ Hypotenuse }}}={\displaystyle \frac{BC}{AB}}\\ & ={\displaystyle \frac{3k}{4k}}={\displaystyle \frac{3}{4}}\\ & {\displaystyle \frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}}={\displaystyle \frac{1-({\displaystyle \frac{3}{4}}{)}^2}{1+({\displaystyle \frac{3}{4}}{)}^2}}={\displaystyle \frac{1-{\displaystyle \frac{9}{16}}}{1+{\displaystyle \frac{9}{16}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{7}{16}}}{{\displaystyle \frac{25}{16}}}}={\displaystyle \frac{7}{25}}\\ & {\cos }^{2}A-{\sin }^{2}A=({\displaystyle \frac{4}{5}}{)}^2-({\displaystyle \frac{3}{5}}{)}^2\\ & ={\displaystyle \frac{16}{25}}-{\displaystyle \frac{9}{25}}={\displaystyle \frac{7}{25}}\\ & \therefore {\displaystyle \frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}}={\cos }^{2}A-{\sin }^{2}A\end{array}$

Solution

$\tan A={\displaystyle \frac{1}{\sqrt{3}}}$

${\displaystyle \frac{BC}{AB}}={\displaystyle \frac{1}{\sqrt{3}}}$

If $BC$ is $k$, then $AB$ will be $\sqrt{3}k$, where $k$ is a positive integer.

In $\mathrm{△}ABC$,

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

$=(\sqrt{3}k{)}^2+(k{)}^2$

$=3k^2+k^2=4k^2$

$\therefore AC=2k$

$\sin A={\displaystyle \frac{\text{ Side opposite to }\mathrm{\angle }A}{\text{ Hypotenuse }}}={\displaystyle \frac{BC}{AC}}={\displaystyle \frac{k}{2k}}={\displaystyle \frac{1}{2}}$

$\cos A={\displaystyle \frac{\text{ Side adjacent to }\mathrm{\angle }A}{\text{ Hypotenuse }}}={\displaystyle \frac{AB}{AC}}={\displaystyle \frac{\sqrt{3}k}{2k}}={\displaystyle \frac{\sqrt{3}}{2}}$

$\sin C={\displaystyle \frac{\text{ Side opposite to }\mathrm{\angle }C}{\text{ Hypotenuse }}}={\displaystyle \frac{AB}{AC}}={\displaystyle \frac{\sqrt{3}k}{2k}}={\displaystyle \frac{\sqrt{3}}{2}}$

$\cos C={\displaystyle \frac{\text{ Side adjacent to }\mathrm{\angle }C}{\text{ Hypotenuse }}}={\displaystyle \frac{BC}{AC}}={\displaystyle \frac{k}{2k}}={\displaystyle \frac{1}{2}}$

(i) $\sin A\cos C+\cos A\sin C$

$\begin{array}{rl}& =({\displaystyle \frac{1}{2}})({\displaystyle \frac{1}{2}})+({\displaystyle \frac{\sqrt{3}}{2}})({\displaystyle \frac{\sqrt{3}}{2}})={\displaystyle \frac{1}{4}}+{\displaystyle \frac{3}{4}}\\ & ={\displaystyle \frac{4}{4}}=1\end{array}$

(ii) $\cos A\cos C-\sin A\sin C$ $=({\displaystyle \frac{\sqrt{3}}{2}})({\displaystyle \frac{1}{2}})-({\displaystyle \frac{1}{2}})({\displaystyle \frac{\sqrt{3}}{2}})={\displaystyle \frac{\sqrt{3}}{4}}-{\displaystyle \frac{\sqrt{3}}{4}}=0$

Solution

Given that, $PR+QR=25$

$PQ=5$

Let PR be $x$.

Therefore, $QR=25-x$

Applying Pythagoras theorem in $\mathrm{△}PQR$, we obtain

$P{R}^2=P{Q}^2+Q{R}^2$

$x^2=(5{)}^2+(25-x{)}^2$

$x^2=25+625+x^2-50x$

$50x=650$

$x=13$

Therefore, $PR=13cm$

$QR=(25-13)cm=12cm$

$\sin P={\displaystyle \frac{\text{ Side opposite to }\mathrm{\angle }P}{\text{ Hypotenuse }}}={\displaystyle \frac{QR}{PR}}={\displaystyle \frac{12}{13}}$

$\cos P={\displaystyle \frac{\text{ Side adjacent to }\mathrm{\angle }P}{\text{ Hypotenuse }}}={\displaystyle \frac{PQ}{PR}}={\displaystyle \frac{5}{13}}$

$\tan P={\displaystyle \frac{\text{ Side opposite to }\mathrm{\angle }P}{\text{ Side adjacent to }\mathrm{\angle }P}}={\displaystyle \frac{QR}{PQ}}={\displaystyle \frac{12}{5}}$

Solution

(i) Consider a $\mathrm{△}ABC$, right-angled at $B$.

$\tan A={\displaystyle \frac{\text{ Side opposite to }\mathrm{\angle }A}{\text{ Side adjacent to }\mathrm{\angle }A}}$

$={\displaystyle \frac{12}{5}}$

But ${\displaystyle \frac{12}{5}}>1$

$\therefore \tan A>1$

So, $\tan A<1$ is not always true.

Hence, the given statement is false.

(ii)

$\sec A={\displaystyle \frac{12}{5}}$

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

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

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

Applying Pythagoras theorem in $\mathrm{△}ABC$, we obtain

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

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

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

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

$BC=10.9k$

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 value of sec $A$ is possible.

Hence, the given statement is true.

(iii) Abbreviation used for cosecant of angle $A$ is $cosecA$. And $\cos A$ is the abbreviation used for cosine of angle $A$.

Hence, the given statement is false.

(iv) $\cot A$ is not the product of $\cot$ and $A$. It is the cotangent of $\mathrm{\angle }A$.

Hence, the given statement is false.

(v) $\sin \theta ={\displaystyle \frac{4}{3}}$

We know that in a right-angled triangle,

$\sin \theta ={\displaystyle \frac{\text{ Side opposite to }\mathrm{\angle }\theta }{\text{ Hypotenuse }}}$

In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of $\sin \theta$ is not possible.

Hence, the given statement is false