# NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.3

Go back to  'Introduction to Trigonometry'

## Chapter 8 Ex.8.3 Question 1

Evaluate:

(i)  $$\;\frac{\sin {{18}^{0}}}{\cos {{72}^{0}}}$$

(ii) $$\;\frac{\tan {{26}^{0}}}{\cot {{64}^{0}}}$$

(iii) $$\;\cos {{48}^{0}}-\sin {{42}^{0}}$$

(iv) $$\;\text{cosec}\,{{31}^{0}}-\sec \,{{59}^{0}}$$

#### Reasoning:

\begin{align} \sin \left(90^{\circ}-\theta\right) &=\cos \theta \\ \tan \left(90^{\circ}-\theta\right) &=\cot \theta \\ \sec \left(90^{\circ}-\theta\right) &=\text{cosec} \;\theta \end{align}

#### Solution:

\begin{align}\rm (i)\qquad \frac{\sin {{18}^{0}}}{\cos {{72}^{0}}}\end{align}

\begin{align}\text{Since,}&\\ &\sin \left( {{90}^{0}}-\theta \right)=\cos \theta \end{align}

Here $$\text{ }\!\!\theta\!\!\text{ }={{72}^{0}}$$

\begin{align} & ∴ =\frac{\sin ({{90}^{0}}-{{72}^{0}})}{\cos {{72}^{0}}} \\ & \,\,\,\,=\frac{\cos {{72}^{0}}}{\cos {{72}^{0}}} \\ & \,\,\,\,=1 \\ \end{align}

\begin{align}\text{(ii)} \qquad\frac{\tan {{26}^{0}}}{\cot {{64}^{0}}}\end{align}

\begin{align}\text{Since}&, \\ &\tan ({{90}^{0}}-\theta )=\cot \theta \end{align}

Here $$\theta ={{64}^{0}}$$

\begin{align} & ∴ =\frac{\tan ({{90}^{0}}-{{64}^{0}})}{\cot {{64}^{0}}} \\ & \,\,\,\,=\frac{\cot {{64}^{0}}}{\cot {{64}^{0}}} \\ & \,\,\,\,=1 \\ \end{align}

$$\text{(iii)}\qquad\cos {{48}^{0}}-\sin {{42}^{0}}$$

\begin{align}\text{Since}&, \\&\sin ({{90}^{0}}-\theta )=\cos \theta\end{align}

Here $$\text{ }\!\!\theta\!\!\text{ }={{48}^{0}}$$

\begin{align} ∴ \ &=\cos {{48}^{0}}-\sin \left( {{90}^{0}}-{{48}^{0}} \right) \\ & =\cos {{48}^{0}}-\cos {{48}^{0}} \\ & =0 \end{align}

$$\text{(iv) }\qquad\text{cosec}{{31}^{0}}-\sec {{59}^{0}}$$

\begin{align}\text{Since}&,\\ &\sec \,({{90}^{0}}-\theta )=\operatorname{cosec}\theta\end{align}

Here $$\theta ={{31}^{0}}$$

Therefore,

\begin{align} \,& \text{cosec}{{31}^{\circ}}-\sec {{59}^{\circ}}\\ & =\operatorname{cosec}{{31}^{\circ}}-\sec \left( {{90}^{\circ}}-{{31}^{\circ}} \right) \\ & =\operatorname{cosec}{{31}^{\circ}}-\operatorname{cosec}{{31}^{\circ}} \\ & =0 \end{align}

## Chapter 8 Ex.8.3 Question 2

Show that:

(i) \begin{align}\text{tan}\,48{}^\circ \,\text{tan}\,23{}^\circ \, \text{tan}\,42{}^\circ \ \text{tan}\,67{}^\circ =1\end{align}

(ii) \begin{align}\text{cos}\,38{}^\circ \,\cos \,52{}^\circ -\sin 38{}^\circ \ \sin 52{}^\circ =0 \end{align}

#### Reasoning:

$$\sin \left( {{90}^{\circ}}-\theta \right)=\cos \theta$$

$$\tan ({{90}^{\circ}}-\theta )=\cot \theta$$

#### Steps:

(i)Taking

$$\text{L.H.S }= \text{tan}\,48{}^\circ \,\text{tan}\,23{}^\circ \,\text{tan}\,42{}^\circ \ \text{tan}\,67{}^\circ$$

Since $$\tan \,({{90}^{\circ}}-\theta)=\cot \theta$$

\begin{align}\text{L.H.S }& =\tan({{90}^{\circ}}-{{42}^{\circ}}) \tan ({{90}^{\circ}}-{{67}^{\circ}}) \tan {{42}^{\circ}}\tan {{67}^{\circ}}\\ & =\cot {{42}^{\circ}}\cot {{67}^{\circ}}\tan {{42}^{\circ}}\tan {{67}^{\circ}} \\ & = (\cot {{42}^{\circ}}\tan {{42}^{\circ}}) (\cot {{67}^{\circ}}\tan {{67}^{\circ}})\\ & = \left( \frac{1}{\tan {{42}^{\circ}}}\tan {{42}^{\circ}} \right) \left( \frac{1}{\tan {{67}^{\circ}}}\tan {{67}^{\circ}} \right) \\ & =1\,\times\;1\\&=1 \\ & =\text{R.H.S}\end{align}

Hence,

$$\text{tan}\,48{}^\circ \,\text{tan}\,23{}^\circ \,\text{tan}\,42{}^\circ \ \text{tan}\,67{}^\circ =1$$

(ii) Taking

\begin{align}\text{L.H.S }\!=\!\text{cos}\,38{}^\circ \,\cos \,52{}^\circ -\sin 38{}^\circ \ \sin 52{}^\circ\end{align}

Since, $$\tan ({{90}^{circ}}-\theta)=\cos \theta$$

\begin{align}\text{L.H.S }& = \cos {{38}^{\circ}}\cos {{52}^{\circ}} -\sin \,({{90}^{\circ}}-{{57}^{\circ}}) \sin \,({{90}^{\circ}}-{{38}^{\circ}})\\ & =\cos {{38}^{\circ}}\cos {{52}^{\circ}}-\cos {{52}^{\circ}}\cos {{38}^{\circ}} \\ & =0 \\ & =\text{R.H.S} \\ \end{align}

Hence,

$$\cos {{38}^{\circ}}\cos {{52}^{\circ}}-\sin {{38}^{\circ}}\sin {{52}^{\circ}}=0$$

## Chapter 8 Ex.8.3 Question 3

If $${\rm{tan}}2A = {\rm{cot}}\left( {A-18^\circ } \right)$$ , where $$2A$$ is an acute angle, find the value of $$A.$$

### Solution

#### Steps:

Given that:

$${\rm{tan}}2{\rm{A}} = {\rm{cot}}\left( {{\rm{A}}-18^\circ } \right)$$…....(i)

But $$\tan 2{\rm{A}} = \cot \,({90^0} - 2{\rm{A}})$$

By substituting this in equation (i) we get:

\begin{align} \cot \,\left( {{{90}^0} - 2A} \right) &= \cot \,\left( {A - {{18}^0}} \right)\\ {90^0} - 2A &= A - {18^0}\\ 3A &= {108^0}\\ A &= \frac{{{{108}^0}}}{3} = {36^0}\\ A &= {36^0} \end{align}

## Chapter 8 Ex.8.3 Question 4

If $$\tan A = \cot B\,,$$ prove that $$A + B = {90^0}.$$

#### Reasoning:

$$\tan ({90^0} - \theta ) = \cot \theta$$

#### Steps:

Given that:

\begin{align}\,\tan A = \cot B \dots \rm (i)\end{align}

We know that,

$$\tan A = \cot \left( {{{90}^0} - A} \right)$$

By substituting this in equation (i) we get:

\begin{align} \cot \,({90^0} - A) &= \cot B\\ {90^0} - A &= B\\ A + B &= {90^0} \end{align}

## Chapter 8 Ex.8.3 Question 5

If $$\,{\rm{sec}}4A = {\rm{cosec}}\left( {A-20^\circ } \right),$$where $$4A$$ is an acute angle, find the value of $$A$$.

#### Steps:

Given that: $$\text{sec} \;4\;A= \text{cosec} (A -20^{\circ})$$….(i)

Since, $$\sec A = {\rm{cosec}}\,\left( {{{90}^0} - A} \right)$$

By using property in equation (i) we get:

\begin{align} {\rm{cosec}}\left( {{\rm{9}}{{\rm{0}}^\circ} - 4A} \right) &= {\rm{cosec}}\left( {A - {{20}^\circ}} \right)\\ {90^\circ} - 4A &= A - {20^\circ}\\ 5A &= {110^\circ}\\ A &= \frac{{{{110}^\circ}}}{5}\\ &= {22^\circ}\\ A &= {22^\circ} \end{align}

## Chapter 8 Ex.8.3 Question 6

If $$A$$, $$B$$ and $$C$$ are interior angles of a triangle $$ABC$$, then show that

\begin{align} \sin \,\left( {\frac{{B + C}}{2}} \right) = \cos \frac{A}{2}\end{align}

#### Reasoning:

$$\sin \left( {{{90}^0} - \theta } \right) = \cos \theta$$

#### Steps:

We know that for $$\Delta \,ABC,$$

\begin{align} \angle \,A + \angle \,B + \angle \,C &= {180^0}\\ \angle \,B + \angle \,C &= {180^0} - \angle \,A \end{align}

On dividing both sides by $$2$$, we get:

\begin{align} \frac{{\angle \,B + \angle \,C}}{2} &= \frac{{{{180}^0} - \angle \,A}}{2}\\ \frac{{\angle \,B + \angle \,C}}{2} &= {90^0} - \frac{{\angle \,A}}{2} \end{align}

Applying sine angles on both the sides:

\begin{align}\sin \,\left( {\frac{{B + C}}{2}} \right) = \sin \,\left( {{{90}^0} - \frac{A}{2}} \right)\end{align}

Since

\begin{align} \sin \,({90^0} - \theta ) &= \cos \theta \\ ∴ \sin \,\left( {\frac{{B + C}}{2}} \right) &= \cos \,\left( {\frac{A}{2}} \right) \end{align}

## Chapter 8 Ex.8.3 Question 7

Express $$\sin {67^0} + \cos {75^0}$$ in terms of trigonometric ratios of angles between $${0^0}$$ and $${45^0}$$ .

#### Reasoning:

$$\cos \left( {{{90}^0} - \theta } \right) = \sin \theta$$

#### Steps:

Given that: $$\sin {67^0} + \cos {75^0}$$ ….(i)

Since $$\cos \left( {{{90}^0} - \theta } \right) = \sin \theta$$

By using property in equation (i) we get:

\begin{align} &=\! \sin \left( {{{90}^\circ} - {{23}^\circ}} \right) \!+\! \cos \left( {{{90}^\circ} \!-\! {{15}^\circ}} \right)\\ &= \cos {23^\circ} + \sin {15^\circ} \end{align}

Hence, the expression $$\cos {23^0} + \sin {15^0}$$ has trigonometric ratios of angles between $${0^0}$$ and $${45^0}$$ .