NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.3

Go back to  'Introduction to Trigonometry'

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

Solution

Video Solution

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,

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

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

Solution

Video Solution
 

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,

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

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

Video Solution

Reasoning:

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

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:

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

Question 4

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

Solution

Video Solution

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:

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

Question 5

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

Solution

Video Solution

Reasoning:

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

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:

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

Question 6

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

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

Solution

Video Solution

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

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

Question 7

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

Solution

Video Solution

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 \({45^0}\) .

Download SOLVED Practice Questions of NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.3 for FREE
Ncert Class 10 Exercise 8.3
Ncert Solutions For Class 10 Maths Chapter 8 Exercise 8.3
  
Download SOLVED Practice Questions of NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.3 for FREE
Ncert Class 10 Exercise 8.3
Ncert Solutions For Class 10 Maths Chapter 8 Exercise 8.3
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