Solution

Applying Pythagoras theorem for $\triangle A B C$, we obtain

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

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

$=(576+49) cm^{2}$

$=625 cm^{2}$

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

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

$=\dfrac{7}{25}$

$\cos A=$

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

(ii)

$ \begin{aligned} & \sin C=\dfrac{\text{ Side opposite to } \angle C}{\text{ Hypotenuse }}=\dfrac{AB}{AC} \\ & =\dfrac{24}{25} \end{aligned} $

$ \begin{aligned} & \dfrac{\text{ Side adjacent to } \angle C}{\text{ Hypotenuse }}=\dfrac{BC}{AC} \\ & \cos C= \\ & =\dfrac{7}{25} \end{aligned} $

Solution

Applying Pythagoras theorem for $\triangle PQR$, we obtain

$PR^{2}=PQ^{2}+QR^{2}$

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

$169 cm^{2}=144 cm^{2}+QR^{2}$

$25 cm^{2}=QR^{2}$

$QR=5 cm$

$ \begin{aligned} \tan P & =\dfrac{\text{ Side opposite to } \angle P}{\text{ Side adjacent to } \angle P}=\dfrac{QR}{PQ} \\ & =\dfrac{5}{12} \end{aligned} $

$ \begin{aligned} \cot R & =\dfrac{\text{ Side adjacent to } \angle R}{\text{ Side opposite to } \angle R}=\dfrac{QR}{PQ} \\ & =\dfrac{5}{12} \end{aligned} $

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

Solution

Let $\triangle A B C$ be a right-angled triangle, right-angled at point $B$.

Given that,

$ \begin{aligned} & \sin A=\dfrac{3}{4} \\ & \dfrac{BC}{AC}=\dfrac{3}{4} \end{aligned} $

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

Applying Pythagoras theorem in $\triangle A B C$, we obtain

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

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

$16 k^{2}-9 k^{2}=AB^{2}$

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

$AB=\sqrt{7} k$

$\cos A=\dfrac{\text{ Side adjacent to } \angle A}{\text{ Hypotenuse }}$

$=\dfrac{AB}{AC}=\dfrac{\sqrt{7 k}}{4 k}=\dfrac{\sqrt{7}}{4}$

$\tan A=\dfrac{\text{ Side opposite to } \angle A}{\text{ Side adjacent to } \angle A}$

$=\dfrac{BC}{AB}=\dfrac{3 k}{\sqrt{7} k}=\dfrac{3}{\sqrt{7}}$

Solution

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

$\cot A=\dfrac{\text{ Side adjacent to } \angle A}{\text{ Side opposite to } \angle A}$

$=\dfrac{AB}{BC}$

It is given that,

$ \begin{aligned} \cot A & =\dfrac{8}{15} \\ \dfrac{AB}{BC} & =\dfrac{8}{15} \end{aligned} $

Let $A B$ be $8 k$.Therefore, $B C$ will be $15 k$, where $k$ is a positive integer.

Applying Pythagoras theorem in $\triangle A B C$, we obtain

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

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

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

$=289 k^{2}$

$AC=17 k$

$ \begin{aligned} \sin A & =\dfrac{\text{ Side opposite to } \angle A}{\text{ Hypotenuse }}=\dfrac{BC}{AC} \\ & =\dfrac{15 k}{17 k}=\dfrac{15}{17} \end{aligned} $

$\sec A=\dfrac{\text{ Hypotenuse }}{\text{ Side adjacent to } \angle A}$

$ =\dfrac{AC}{AB}=\dfrac{17}{8} $

Solution

Consider a right-angle triangle $\triangle A B C$, right-angled at point $B$.

$\sec \theta=\dfrac{\text{ Hypotenuse }}{\text{ Side adjacent to } \angle \theta}$

$\dfrac{13}{12}=\dfrac{AC}{AB}$

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

Applying Pythagoras theorem in $\triangle ABC$, we obtain

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

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

$169 k^{2}=144 k^{2}+BC^{2}$

$25 k^{2}=BC^{2}$

$BC=5 k$

$\sin \theta=\dfrac{\text{ Side opposite to } \angle \theta}{\text{ Hypotenuse }}=\dfrac{B C}{A C}=\dfrac{5 k}{13 k}=\dfrac{5}{13}$

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

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

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

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

Solution

Let us consider a triangle $A B C$ in which $C D \perp A B$.

It is given that

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

We have to prove $\angle A=\angle B$. To prove this, let us extend $A C$ to $P$ such that $B C=C P$.

From equation (1), we obtain

$\dfrac{AD}{BD}=\dfrac{AC}{BC}$

$\Rightarrow \dfrac{AD}{BD}=\dfrac{AC}{CP}$

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

By using the converse of B.P.T,

$CD \mid BP$

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

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

By construction, we have $BC=CP$.

$\therefore \angle CBP=\angle CPB$ (Angle opposite to equal sides of a triangle) $\ldots$ (5)

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

$\angle ACD=\angle BCD$.

In $\triangle CAD$ and $\triangle CBD$,

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

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

Therefore, the remaining angles should be equal.

$\therefore \angle CAD=\angle CBD$

$\Rightarrow \angle A=\angle B$

Alternatively,

Let us consider a triangle $A B C$ in which $C D \perp A B$.

It is given that,

$\cos A=\cos B$

$\Rightarrow \dfrac{AD}{AC}=\dfrac{BD}{BC}$

$\Rightarrow \dfrac{AD}{BD}=\dfrac{AC}{BC}$

Let $\dfrac{AD}{BD}=\dfrac{AC}{BC}=k$

$\Rightarrow AD=k BD$.

And, $AC=k BC$

Using Pythagoras theorem for triangles CAD and CBD, we obtain

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

And, $CD^{2}=BC^{2}-BD^{2}$.

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

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

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

$\Rightarrow k^{2}(BC^{2}-BD^{2})=BC^{2}-BD^{2}$

$\Rightarrow k^{2}=1$

$\Rightarrow k=1$

Putting this value in equation (2), we obtain

$AC=BC$

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

Solution

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

$ \begin{aligned} \cot \theta & =\dfrac{\text{ Side adjacent to } \angle \theta}{\text{ Side opposite to } \angle \theta}=\dfrac{BC}{AB} \\ & =\dfrac{7}{8} \end{aligned} $

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

Applying Pythagoras theorem in $\triangle A B C$, we obtain

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

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

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

$=113 k^{2}$

$AC=\sqrt{113} k$

$\sin \theta=\dfrac{\text{ Side opposite to } \angle \theta}{\text{ Hypotenuse }}=\dfrac{AB}{AC}$

$=\dfrac{8 k}{\sqrt{113} k}=\dfrac{8}{\sqrt{113}}$

$\cos \theta=\dfrac{\text{ Side adjacent to } \angle \theta}{\text{ Hypotenuse }}=\dfrac{BC}{AC}$

$=\dfrac{7 k}{\sqrt{113} k}=\dfrac{7}{\sqrt{113}}$

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

$=\dfrac{\dfrac{49}{113}}{64}=\dfrac{49}{64}$

113

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

Solution

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

Or, $\cot A=\dfrac{4}{3}$

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

$\cot A=\dfrac{\text{ Side adjacent to } \angle A}{\text{ Side opposite to } \angle A}$

$\dfrac{AB}{BC}=\dfrac{4}{3}$

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

In $\triangle ABC$,

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

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

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

Solution

$\tan A=\dfrac{1}{\sqrt{3}}$

$\dfrac{BC}{AB}=\dfrac{1}{\sqrt{3}}$

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

In $\triangle ABC$,

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

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

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

$\therefore AC=2 k$

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

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

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

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

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

$ \begin{aligned} & =(\dfrac{1}{2})(\dfrac{1}{2})+(\dfrac{\sqrt{3}}{2})(\dfrac{\sqrt{3}}{2})=\dfrac{1}{4}+\dfrac{3}{4} \\ & =\dfrac{4}{4}=1 \end{aligned} $

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

Solution

Given that, $PR+QR=25$

$PQ=5$

Let PR be $x$.

Therefore, $QR=25-x$

Applying Pythagoras theorem in $\triangle PQR$, we obtain

$PR^{2}=PQ^{2}+QR^{2}$

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

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

$50 x=650$

$x=13$

Therefore, $PR=13 cm$

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

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

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

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

Solution

(i) Consider a $\triangle A B C$, right-angled at $B$.

$\tan A=\dfrac{\text{ Side opposite to } \angle A}{\text{ Side adjacent to } \angle A}$

$=\dfrac{12}{5}$

But $\dfrac{12}{5}>1$

$\therefore \tan A>1$

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

Hence, the given statement is false.

(ii)

$ \sec A=\dfrac{12}{5} $

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

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

Let $A C$ be $12 k, A B$ will be $5 k$, where $k$ is a positive integer.

Applying Pythagoras theorem in $\triangle ABC$, we obtain

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

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

$144 k^{2}=25 k^{2}+BC^{2}$

$BC^{2}=119 k^{2}$

$BC=10.9 k$

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

$B C$ should be such that,

$A C-A B<B C<A C+A B$

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

$7 k<BC<17 k$

However, $BC=10.9 k$. 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 $cosec A$. 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 $\angle A$.

Hence, the given statement is false.

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

We know that in a right-angled triangle,

$ \sin \theta=\dfrac{\text{ Side opposite to } \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