# NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.1

Go back to  'Introduction to Trigonometry'

## Question 1

In  $$\,\Delta ABC,$$ right-angled at $$\rm{B}$$, $$\rm AB = 24 \,\rm cm,$$ $$\rm BC = 7\rm\, cm$$, determine:

(i) $$\quad\text{sin}\,A,\text{cos}A$$

(ii) $$\quad \text{sin }C,\text{cos}\,C$$

#### What is the known?

Two sides of a right-angled triangle $$\Delta\!\!\text{ ABC}$$

#### What is the unknown?

Sine and cosine of angle $$A$$ and $$C$$.

#### Reasoning:

Applying Pythagoras theorem for  $$\Delta \text{ABC,}$$ we can find hypotenuse (side $$\rm AC$$). Once hypotenuse is known, we can find sine and cosine angle using trigonometric ratios.

#### Steps:

$$\Delta \text{ABC,}$$ we obtain.

\begin{align}\text{A}{{\text{C}}^{\text{2}}}\, &=\,\text{A}{{\text{B}}^{\text{2}}}\text{+B}{{\text{C}}^{\text{2}}} \\ &=\,{{\text{(24}\ \text{cm)}}^{\text{2}}}\,\text{+}\,{{\text{(7}\ \text{cm)}}^{\text{2}}} \\ &=\,\text{(576+49)}\ \text{c}{{\text{m}}^{\text{2}}} \\ &=\,\text{625}\ \text{c}{{\text{m}}^{\text{2}}} \end{align}

$$\therefore$$ Hypotenuse,

$$\text{AC}\,\text{=}\,\sqrt{\text{625}}\ \text{cm}\,\text{=}\,\text{25}\ \text{cm}$$

(i)

\begin{align} \sin \text{A} & =\frac{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{A}}{\text{hypotenuse}} \\ &=\frac{\text{BC}}{\text{AC}} \\ \sin \text{A} & =\frac{\text{7}\,\text{cm}}{\text{25}\,\text{cm}}=\frac{7}{25} \\ \sin \text{A}&=\frac{7}{25} \\ \cos \text{A} & =\frac{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{A}}{\text{hypotenuse}} \\ &=\frac{\text{AB}}{\text{AC}} \\ & =\frac{24\,\text{cm}}{25\,\text{cm}}=\frac{24}{25} \\ \cos \text{A} &=\frac{24}{25} \end{align}

(ii)

\begin{align} \text{sin}\,\text{C} & =\frac{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{C}}{\text{hypotenuse}} \\ & =\frac{\text{AB}}{\text{AC}} \\ \text{sin}\,\text{C} &=\frac{24\,\text{cm}}{\text{25}\,\text{cm}}\text{=}\frac{24}{\text{25}} \\ \text{sin}\,\text{C} &=\frac{\text{24}}{\text{25}} \\ \text{cos}\,\text{C} &=\frac{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{C}}{\text{hypotenuse}}\\ & =\frac{\text{BC}}{\text{AC}} \\ & =\frac{7\,\text{cm}}{\text{25}\,\text{cm}}\text{=}\frac{7}{\text{25}} \\ \text{cos}\,\text{C} &=\frac{7}{\text{25}} \end{align}

## Question 2

In the given figure, find $$\text{tan}\,\text{P }-\text{cot }\text{R}.$$

#### What is the known?

$$PQ = 12\,\rm{ cm}$$ and $$PR = 13\,\rm {cm}.$$

#### What is the unknown?

One side of right-angled triangle $$\Delta PQR$$

#### Reasoning:

Using Pythagoras theorem, we can find the length of the third side, then the required trignometric ratios.

#### Steps:

Apply Pythagoras theorem for $$\Delta PQR$$ we obtain:

\begin{align} P{R^2}\, &= \,P{Q^2} + Q{R^2}\\ Q{R^2}\, &= \,P{R^2} - P{Q^2}\\ Q{R^2} &= {(13\rm cm)^2}\, - \,{(12\rm cm)^2}\\ Q{R^2} &= 169\,\rm c{m^2} - 144\,\rm c{m^2}\\ Q{R^2} &= 25\,\rm c{m^2}\\ QR\, &= \,5\,\rm cm \end{align}

\begin{align} \text{tan}\,\text{P} &= \frac{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{P}}{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{P}}\\ &= \frac{\text{QR}}{\text{PQ}} \\ & = \frac{\text{5}\,\text{cm}}{\text{12}\,\text{cm}} \\ \text{tan}\,\text{P} & = \frac{\text{5}}{\text{12}} \end{align}

\begin{align} \rm{cot}\,{R} &= \frac{ \text{side adjacent to}\ \angle {R}}{\text{side opposite to}\ \angle {R}} \\ & = \frac{{QR}}{{PQ}} \\ & =\frac{{5}\,\rm{cm}}{{12}\,\rm{cm}} \\ \rm {cot}\,{R} &= \frac{{5}}{{12}} \end{align}

\begin{align} \rm {\tan} \,{P}- {{cot}}\,{R} & = \frac{{5}}{{12}}-\frac{{5}}{{12}} \\ \rm{tan}\,P- {cot}\,{R} &= {0}\end{align}

## Question 3

If \begin{align}\text{sin A}=\frac{3}{4}\end{align} calculate $$\rm{cos\,A}$$ and $$\rm {tan\,A}.$$

#### What is the known?

Sine of $$\angle \text{A}$$ .

#### What is the unknown?

Cosine and tangent of $$\angle \text{A}$$

#### Reasoning:

Using sin A, we can find the ratio of the length of two sides of the right-angled triangle. Then by using Pythagoras theorem, the third side and required trigonometric ratios.

Steps:

Let $$\Delta \text{ABC}$$ be a right-angled triangle, right angled at point $$\rm{B}$$.

Given that:

\begin{align} \Rightarrow {\sin A}&={\frac{3}{4}} \\ {\frac{B C}{A C}}&={\frac{3}{4}}\end{align}

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

Applying Pythagoras theorem for $$\Delta \,\rm ABC,$$  we obtain:

\begin{align} A{C^2} &= A{B^2}\, + \,B{C^2}\\ A{B^2} &= A{C^2} - \,B{C^2}\\ A{B^{2\,}} &= {(4\,k)^2} - \,{(3\,k)^2}\\ A{B^2} &= 16{k^2} - 9\,{k^2}\\ A{B^2}\, &= \,7\,{k^2}\\ AB\, &= \sqrt {7\,} k \end{align}

\begin{align} \,\text{cosA}\,&=\frac{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{A}}{\text{hypotenuse}} \\ & =\frac{\text{AB}}{\text{AC}}\\ & =\frac{\sqrt{\text{7}}k}{\text{4}\,k} \\&=\frac{\sqrt{\text{7}}}{\text{4}} \end{align}

\begin{align} \text{tan}\,\text{A}&=\frac{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{A}}{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{A}} \\ & =\frac{\text{BC}}{\text{AB}}\,\,\,\text{=}\frac{\text{3}\,k}{\sqrt{\text{7}}\,k} \\ &=\frac{\text{3}}{\sqrt{\text{7}}} \end{align}

Thus,

\begin{align}\,\text{cos}\,\text{A=}\frac{\sqrt{\text{7}}}{\text{4}}\ \ \text{and}\ \text{tan}\ \text{A=}\frac{\text{3}}{\sqrt{\text{7}}}\end{align}

## Question 4

Given $$15 \,\rm{cot A} = 8,$$ find $$\rm{sin\, A}$$ and $$\rm{sec\,A.}$$

#### What is the known?

Cotangent of $$\angle A$$

#### What is the unknown?

Sine and Secant of $$\angle A$$ .

#### Reasoning:

Using cot A, we can find the ratio of the length of two sides of the right-angled triangle. Then by using Pythagoras theorem, the third side and required trigonometric ratios.

#### Steps:

Let us consider a right-angled $$\Delta \rm{ABC},$$ right angled at $$\rm{B.}$$

\begin{align}\text{cot}\,\text{A}\, \text{=}\frac{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{A}}{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{A}}\text{=}\frac{\text{AB}}{\text{BC}}\end{align}

It is given that

\begin{align}\text{cotA}\,\text{=}\,\frac{\text{8}}{\text{15}}\Rightarrow \frac{\text{AB}}{\text{BC}}\,\text{=}\,\frac{\text{8}}{\text{15}}\end{align}

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

Apply Pythagoras theorem in $$\text{ }\!\!\Delta\!\!\text{ }\,\text{ABC,}$$ we obtain.

\begin{align} A{C^2}\, &= \,A{B^2} + B{C^2}\\ A{C^2}\, &= \,{(8k)^2} + {(15k)^2}\\ A{C^2}\, &= \,64{k^2} + 225{k^2}\\ A{C^2}\, &= \,289{k^2}\\ \,\;\,AC\,\, &= \,17k \end{align}

\begin{align} {\rm{sin}}\,{\rm{A}} &= \frac{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\angle {\rm{A}}}}{{{\rm{hypotenuse}}}} \\ & = \frac{{BC}}{{AC}} = \frac{{15\,k}}{{17\,k}}\\ & = \frac{{{\rm{15}}}}{{{\rm{17}}}}\\ {\rm{sec}}\,{\rm{A}} & = \frac{{{\rm{hypotenuse}}}}{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\angle {\rm{A}}}}\\ & = \frac{{AC}}{{AB}} = \frac{{17\,k}}{{8\,k}}\\ &= \frac{{{\rm{17}}}}{{\rm{8}}} \end{align}

Thus,

\begin{align}\text{sin}\,\text{A}\,\text{=}\,\frac{\text{15}}{\text{17}}\ \ \text{and}\ \ \text{sec}\,\text{A}\,\text{=}\,\frac{\text{17}}{\text{8}}\end{align}

## Question 5

Given \begin{align}\text{sec}\,\theta =\frac{13}{12},\end{align} calculate all other trigonometric ratios.

#### What is the known?

$$\text{Secant}\ \rm{of} \,\theta$$

#### What is the unknown?

Other trigonometric ratios.

#### Reasoning:

Using $$\rm Sec\;\theta$$ , we can find the ratio of the length of two sides of the right-angled triangle. Then by using Pythagoras theorem, the third side and required trigonometric ratios.

#### Steps:

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

It is given that:

\begin{align}\text{sec}\ \text{ }\!\!\theta\!\!\text{ }\,&=\frac{\text{hypotenuse}}{\text{side}\ \text{adjacent}\ \text{to}\ \angle \theta }\\ &=\frac{\text{AC}}{\text{AB}} \\ & =\frac{\text{13}}{\text{12}}\end{align}

Let $$AC = 13\,k$$ and $$BC = 12\,k$$ where $$k$$ is a positive integer.

Apply Pythagoras theorem in \begin{align}\triangle \rm{ABC}\end{align} we obtain:

\begin{align} A{C^2} &= A{B^2} + B{C^2}\\ B{C^2} &= A{C^2} - A{B^2}\\ B{C^2} &= {(13\,k)^2} - {(12\,k)^2}\\ B{C^2} &= 169\,{k^2} - 144\,{k^2}\\ B{C^2} &= 25\,{k^2}\\ BC &= 5\,k\end{align}

\begin{align} \\{\sin \theta}&=\frac{\text { side opposite to } \angle \theta}{\text { hypotenuse }} \\ & =\frac{\mathrm{BC}}{\mathrm{AC}} \\ & =\frac{5}{13} \\ {\cos \theta}&=\frac{\text { side adjacent to } \angle \theta}{\text { hypotenuse }} \\ &=\frac{\mathrm{AB}}{\mathrm{AC}} \\ & =\frac{12}{13} \\ {\tan \theta}&=\frac{\text { side opposite to } \angle \theta}{\text { side adjacent to } \angle \theta} \\ &=\frac{\mathrm{BC}}{\mathrm{AB}} \\ & =\frac{5}{12} \\ {\cot \theta}&=\frac{\text { side adjacent to } \angle \theta}{\text { side opposite to } \angle \theta} \\ &=\frac{\mathrm{AB}}{\mathrm{BC}} \\ & =\frac{12}{5}\\ \text{cosec}\,\text{ }\!\!\theta\!\! &=\frac{\text{hypotenuse}}{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{ }\!\!\theta\!\!\text{ }} \\ & =\frac{\text{AC}}{\text{BC}} \\ & = \frac{\text{13}}{\text{5}}\end{align}

## Question 6

If $$\,\,\angle {A}$$ and $$\,\,\angle {B}$$ are acute angles such that $$\,\,\text{cos}\,{A}=\text{cos}\,{B},$$ then show that $$\angle {A}=\angle \,{B}\,{.}$$

#### What is the known?

$$\angle {A}$$ and $$\,\,\angle {B}$$ are acute angles and \begin{align}\text{cos}\, A = \text{cos}\, B\end{align}

#### What is the unknown?

To show that $$\angle A= \angle B$$

#### Reasoning:

Using $$\text{cos} \,A$$ and $$\text{cos}\, B,$$ we can find the ratio of the length of two sides of the right-angled triangle with respective angles. Then compare both the ratios.

#### Steps:

In the right-angled triangle $$ABC, \,$$$$\,\angle A$$  and  $$\angle B$$ are acute angles and $$\angle C$$ is right angle.

\begin{align} & \text{cos}\,{A}\,{=}\,\frac{\text{side adjacent to}\ \angle {A}}{\text{hypotenuse}}{=}\frac{{AC}}{{AB}} \\ & \text{cos}\,{B}\,{=}\,\frac{\text{side adjacent to}\,\angle {B}}{\text{hypotenuse}}{=}\,\frac{{BC}}{{AB}} \end{align}

Given that \begin{align} \;\text{cos} \,A&=\,\text{cos} B \end{align}

\begin{align} \text{Therefore, }\frac{{AC}}{{AB}}\,&=\,\frac{{BC}}{{AB}} \\ {AC}\,&=\,{BC}\end{align}

Hence, $$\angle A=\angle B\,,$$ (angles opposite to equal sides of triangle are equal)

Alternatively,

Reasoning:

Using $$\text{ cos} \,A$$ and $$\text{cos}\, B,$$ we can find the ratio of the length of two sides of the right-angled triangle with respective angles. Then by using Pythagoras theorem, relation between the sides

Let us consider a triangle $$ABC$$ in which \begin{align}{CO} \perp {AB}\end{align}

It is given that

\begin{align} \text{cos}\,{A}\,&=\, \text{cos}\,{B} \\ \frac{{AO}}{{AC}}&=\frac{{BO}}{{BC}} \\ \frac{{AO}}{{BO}}\,&=\,\frac{{AC}}{{BC}}\\ \\ \text{ Let }\; \frac{ {A O}}{{B O}}&=\frac{{A C}}{{B C}}={k} \\ { A O}&={k B O}\;\dots\rm{(i)} \\ {A C}&={k B C}\;\dots \rm{(ii)} \end{align}

By applying Pythagoras theorem in  $$\Delta {CAO}$$ and $$\Delta {CBO},$$ we get.

\begin{align} {A}{{{C}}^{{2}}} & ={A}{{{O}}^{{2}}}{+C}{{{O}}^{{2}}}[\text{from} \;\Delta\;{ CAO}] \\ {C}{{{O}}^{{2}}}&={A}{{{C}}^{{2}}}-{A}{{{O}}^{{2}}}\cdots \left(\rm {iii} \right) \\ {B}{{{C}}^{{2}}}&={B}{{{D}}^{{2}}}{+C}{{{O}}^{{2}}}[\text {from}\;\Delta\,{ CBO}] \\ {C}{{{O}}^{{2}}}&={B}{{{C}}^{{2}}}\,-{B}{{{O}}^{{2}}}\cdots \left( \rm{iv} \right) \\ \end{align}

From equation (iii) and equation (iv), we get

\begin{align} {A}{{{C}}^{{2}}}-{A}{{{O}}^{{2}}}&=\,{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}} \\ \ {{{(kBC)}}^{{2}}}-{{{(kBO)}}^{{2}}}&=\,{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}} \\ {{k}^{{2}}}{B}{{{C}}^{{2}}}-{{k}^{{2}}}{B}{{{O}}^{{2}}}\,&=\,{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}} \\ {{k}^{{2}}}{(B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}}{)}\,&=\,{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}} \\ {{k}^{{2}}}\,&=\,\frac{{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}}}{{B}{{{C}}^{{2}}}-{B}{{{O}}^{{2}}}} \\ &=1 \\k\,&=1 \end{align}

Putting this value in equation (ii) we obtain

${AC}\,{=}\,{BC}$

$${\angle A=\angle B}$$ (angles opposite to equal sides of triangle are equal.)

## Question 7

If $$\cot \theta = \frac{7}{8}$$, evaluate:

(i) \begin{align}\,\,{\rm{ }}\frac{{\left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}}\end{align},

(ii) \begin{align} {\rm{ co}}{{\rm{t}}^2}\theta \end{align}

### Solution

What is the known?

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

What is the unknown?

Value of \begin{align} \left( \rm{i} \right)\,\,{\rm{ }}\frac{{\left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}} \end{align}, and   \begin{align} \left( \rm{ii} \right){\rm{ co}}{{\rm{t}}^2}\theta \end{align}

Reasoning:

Using $$\cot \theta = \frac{7}{8}$$, we can find the ratio of the length of two sides of the right-angled triangle. Then by using Pythagoras theorem, the third side and required trigonometric ratios.

Steps:

Let $${\rm{\Delta }}\,{{ABC}}$$, in which angle $$B$$ is right angle.

\begin{align} \cot \theta &= \frac{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\theta }}{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\theta }}\\ &= \frac{{AB}}{{BC}}\\&= \frac{7}{8} \end{align}

Let $$AB = 7k$$and $$BC = 8k$$, where $$k$$ is a positive integer.

By applying Pythagoras theorem in we get.

\begin{align}{\rm A}{C^2}\, & = {\rm A}{{\rm B}^2} + {\rm B}{C^2}\\ &= {(7k)^2} + {(8k)^2}\\ &= 49{k^2} + 64{k^2}\\ &= 113{k^2}\\ AC\, &= \,\sqrt {113k{\,^2}} \\ &= \,\sqrt {113} k\end{align}

Therefore,

\begin{align}{\rm{sin }}\theta &= \frac{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\theta }}{{{\rm{hypotenuse}}}} = \frac{{BC}}{{AC}}\\ &= \frac{{8k}}{{\sqrt {113} k}} = \frac{8}{{\sqrt {113} }}\\\cos \theta &= \frac{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\theta }}{{{\rm{hypotenuse}}}} = \frac{{AB}}{{AC}} \\ & = \frac{{7k}}{{\sqrt {113} k}} = \frac{7}{{\sqrt {113} }}\end{align}

\begin{align}\left(\rm{i} \right)\,\,{\rm{ }}\frac{{\left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}} \end{align}

\begin{align}&\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[ \because {\left( {a + b} \right)\left( {a - b} \right) = \left( {{a^2} - {b^2}} \right)} \right]\\ &= \frac{{1 - {{\left( {\frac{8}{{\sqrt {113} }}} \right)}^2}}}{{1 - {{\left( {\frac{7}{{\sqrt {113} }}} \right)}^2}}}\\ &= \frac{{1 - \frac{{64}}{{113}}}}{{1 - \frac{{49}}{{113}}}}\\ &= \frac{{\frac{{49}}{{113}}}}{{\frac{{64}}{{113}}}}\\ &= \frac{{49}}{{64}}\end{align}

$$\left( \rm{ii} \right){\rm{ co}}{{\rm{t}}^2}\theta$$

\begin{align}{\rm{co}}{{\rm{t}}^2}\theta &= {\left( {\frac{7}{8}} \right)^2}\\ &= \frac{{49}}{{64}}\end{align}

## Question 8

If $$3 \;cot A = 4$$, check whether \begin{align} \frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A \end{align} or not.

### Solution

What is known?

Cotangent of angle $$A$$

What is unknown?

Whether \begin{align} \frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A \end{align}

Reasoning:

Using $$3\cot {{A}} = 4$$, we can find the ratio of the length of two sides of the right-angled triangle. Then by using Pythagoras theorem, the third side and required trigonometric ratios.

Steps:

\begin{align}{\rm{3}}\,{\rm{cot}}\,{{A}}\,\, &= \,\,{{4}}\\{\rm{cot}}\,{{A}}\,\, &= \,\,\frac{{\rm{4}}}{{\rm{3}}}\end{align}

Let , in which angle $$B$$ is right angle.

\begin{align} \cot A & = \frac{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\angle A}}{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\angle A}} \\ & = \frac{{AB}}{{BC}} = \frac{4}{3} \end{align}

Let $$AB = 4k\;{\rm{ }}{\rm{and }}\;BC = 3k$$ where $$k$$ is a positive integer.

By applying Pythagoras theorem in $$\Delta ABC$$ we get.

\begin{align}{A}{C^2}\, & = {A}{{B}^2} + {B}{C^2}\\ &= {(4k)^2} + {(3k)^2}\\ &= 16{k^2} + 9{k^2}\\ &= 25{k^2}\\{{AC}}\,& = \,\sqrt {{{25k}}{\,^{{2}}}} \\&={{5k}}\end{align}

Therefore,

\begin{align} {\rm{tan}}\,A &= \frac{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\angle A}}{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\angle A}}\\ & = \frac{{BC}}{{AB}} \\ & = \frac{{3k}}{{4k}} \\ &= \frac{3}{4} \end{align}

\begin{align}{\text{sin}} A &= \frac{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\angle {{A}}}}{{{\rm{hypotenuse}}}}{\rm{ = }}\frac{{{{BC}}}}{{{{AC}}}}\\&=\frac{{{\rm{3k}}}}{{{\rm{5k}}}}= \frac{{\rm{3}}}{{\rm{5}}} \end{align}

\begin{align} \cos A & = \frac{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\angle A}}{{{\rm{hypotenuse}}}} \\ & = \frac{{AB}}{{AC}} \\ & = \frac{{4k}}{{5k}} \\ & = \frac{4}{5} \end{align}

\begin{align}L.H.S &= \frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}\\ &= \frac{{1 - {{\left( {\frac{3}{4}} \right)}^2}}}{{1 + {{\left( {\frac{3}{4}} \right)}^2}}}\\ &= \frac{{1 - \frac{9}{{16}}}}{{1 + \frac{9}{{16}}}}\\& = \frac{{16 - 9}}{{16 + 9}}\\ &= \frac{7}{{25}}\end{align}

\begin{align}R.H.S\,\, &= {\cos ^2}A - {\sin ^2}A\\&= {\left( {\frac{4}{5}} \right)^2} - {\left( {\frac{3}{5}} \right)^2}\\ &= \frac{{16}}{{25}} - \frac{9}{{25}}\\&= \frac{{16 - 9}}{{25}}\\& = \frac{7}{{25}}\end{align}

Therefore, \begin{align}\frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A \end{align}

## Question 9

In the $$\Delta {ABC}$$ right-angled at $${B}$$, if \begin{align}\text{tan}\,A=\frac{\text{1}}{\sqrt{\text{3}}}\end{align} find the value of:

(i) $$\quad \text{sin }A \text{ cos }{C }+\text{ }\text{cos }A \;\text{sin }{C }$$

(ii) $$\quad \text{cos }{A}\;\text{cos }{C }-\text{ }\text{sin }{A }\text{ sin }{C }$$

#### Reasoning:

Using $$\tan {{A}} = \frac{1}{{\sqrt 3 }}$$, we can find the ratio of the length of two sides of the right-angled triangle. Then by using Pythagoras theorem, the third side and required trigonometric ratios.

#### Steps:

(i) Let \begin{align}\triangle {ABC}\end{align} be a right-angled triangle \begin{align}\text{tan}\,A=\frac{\text{1}}{\sqrt{\text{3}}}\end{align}

\begin{align}\text{tan}\,A\,& =\frac{\text{side}\ \text{opposite}\ \text{to}\ \angle A}{\text{side}\ \text{adjacent}\ \text{to}\ \angle A} \\ & =\frac{BC}{AB}=\frac{1}{\sqrt{3}}\end{align}

Let  $${BC} = {k}$$ and $${AB} = \sqrt{3\,}{k}$$  where $$k$$ is a positive real number.

By applying Pythagoras theorem for $$\text{ }\!\!\Delta\!\!\text{ }\,ABC$$

\begin{align} {AC}^{2} &={AB}^{2}+{BC}^{2} \\ &=(\sqrt{3} {k})+({k})^{2} \\ &=3 {k}^{2}+{k}^{2} \\ &=4 {k}^{2} \\ {AC} &=\sqrt{4 {k}^{2}} \\ &=2 {k} \end{align}

Therefore,

\begin{align}\sin A&=\frac{\text { side opposite to } \angle A}{\text { hypotenuse }} \\ &=\frac{B C}{A C}=\frac{1}{2} \\ \cos A&=\frac{\text { side adjacent to } \angle A}{\text { hypotenuse }} \\&=\frac{A B}{A C}=\frac{\sqrt{3}}{2} \\ \sin C&=\frac{\text { side opposite to } \angle C}{\text { hypotenuse }} \\ &=\frac{A B}{A C}=\frac{\sqrt{3}}{2} \\ \cos C&=\frac{\text { side adjacent to } \angle C}{\text { hypotenuse }} \\ & =\frac{B C}{A C}=\frac{1}{2}\end{align}

$$\text{(i) sin }A\text{ cos }C+\text{cos }A\text{ sin }C$$

(By substituting the values of the trigonometric functions in the above equation.)

\begin{align} & \text{ sin }A\text{ cos }C+\text{cos }A\text{ sin }C \\ &= \left( \frac{1}{2} \right)\! \left( \frac{1}{2} \right)\!+\!\left( \frac{\sqrt{3}}{2} \right)\! \left( \frac{\sqrt{3}}{2} \right) \\ & =\frac{1}{4}+\frac{3}{4} \\ & =\frac{1+3}{4} \\ & =\frac{4}{4} \\ & =1 \end{align}

$$\text{(ii)}\cos \,A\,\cos \,C-\sin \,A\,\sin \,C$$

By substituting the values of the trigonometric functions in the above equation.

(ii)

\begin{align} & \cos \,A\,\cos \,C-\sin \,A\,\sin \,C \\ &= \left( \frac{\sqrt{3}}{2} \right)\! \left( \frac{1}{2} \right) \!-\!\left( \frac{1}{2} \right)\! \left( \frac{\sqrt{3}}{2} \right) \\ & =\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4} \\ & =0 \end{align}

## Question 10

In $$\,\,\Delta PQR,$$ right-angled at $$\rm{Q,}$$ $$PR+QR=25 \text{cm}$$ and $${PQ}=5 \text{cm}.$$ Determine the values of $$\text{sin} \,P, \text{cos} \,P$$ and $$\text{tan}\, P.$$

#### Reasoning:

Using Pythagoras theorem, we can find the length of the all three sides. Then the required trignometric ratios

#### Steps:

Given, $$\text{ }\!\!\Delta ABC$$ be a right angle at $${Q.}$$

\begin{align} & {PQ=5}\,\text{cm} \\ & {PR+QR=25} \\ \end{align}

Let $${PR}= x \, \text{cm}$$

Therefore,

\begin{align} QR\, &=25\,\text{cm}- PR \\ &=25\,\text{cm}-x\,\text{cm} \end{align}

By applying Pythagoras theorem for \begin{align}\Delta {PQR}\end{align} we obtain.

\begin{align} P{{R}^{2}}&=P{{Q}^{2}}+Q{{R}^{2}} \\ {{x}^{2}}&={{(5)}^{2}}+{{(25-x)}^{2}} \\ {{x}^{2}}&=25+625-50x+{{x}^{2}} \\ 50x&=650 \\ x&=\frac{650}{50} \\ &=130\ \rm{ m} \end{align}

Therefore,

\begin{align} {PR} &=13 \rm{cm} \\ {QR} &=(25-13) \rm{cm} \\ &=12 \rm{cm} \end{align}

By substituting the values obtained above in the trigonometric functions below.

\begin{align}\sin {P}&=\frac{\text { side opposite to } \angle {P}}{\text { hypotenuse }}\\ &=\frac{{QR}}{{PR}}=\frac{12}{13} \\\\ \cos {P} & =\frac{\text { side adjacent to } \angle {P}}{\text { hypotenuse }} \\ & =\frac{{PQ}}{{PR}}=\frac{5}{13} \\\\ \tan {P} & =\frac{\text { side opposite to } \angle {P}}{\text { side adjacent to } \angle {P}} \\ & =\frac{{QR}}{{PQ}}=\frac{12}{5}\end{align}

## Question 11

(i) The value of $$\text{tan}\,A$$ is always less than $$1.$$

(ii) \begin{align}\text{sec}A=\frac{\text{12}}{\text{5}}\end{align} for some value of $$\angle A$$.

(iii) $$\text{cos}\,A$$ is the abbreviation used for the cosecant of $${\angle A}$$

(iv) $$\text{cot}\, A$$ is the product of $$\rm{cot}$$ and $${A.}$$

(v) \begin{align}\sin \,\theta =\frac{4}{3},\end{align} for some \begin{align}\angle \theta\end{align}

#### Steps:

(i) False, because sides of a right-angled triangle may have any length. So $$\text{tan}\,A$$ may have any value.

(ii) \begin{align}\text{sec}\,{A=}\,\frac{\text{hypotenuse}}{\text{side}\ \text{adjacent}\ \text{to}\ \angle {A}}\end{align}

As hypotenuse is largest side, the ratio on RHS will be greater than 1. Hence \begin{align}\text{sec }\text{A}>1.\end{align} Thus, the given statement is true.

(iii) Abbreviation used for cosecant of $$\angle A$$ is $$\text{cosec}\,A$$ and $$\text{cos}\,A$$ is the abbreviation used for cosine of $$\rm \angle A$$. Hence the given statement is false.

(iv) $$\text{cot}\,A$$ is not the product of $$\rm{cot}$$ and $${A.}$$ It is the cotangent of $$\rm \angle A$$ . Hence, the given statement is false.

(v)\begin{align}\text{Sin}\,\theta =\frac{4}{3}\end{align}

We know that in a right-angled triangle,

\begin{align}\text{Sin}\,\theta =\frac{\text{side}\ \text{adjacent}\ \text{to}\ \angle \theta }{\text{hypotenuse}}\end{align}

In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Also, the value of Sine should be less than $$1.$$ Therefore, such value of $$\rm{Sin\, \theta}$$ is not possible. Hence the given statement is false.