Ex.8.2 Q1 Introduction to Trigonometry Solution - NCERT Maths Class 10

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Question

Evaluate the following:

\(\begin{align}&({\rm{i}}) \;\;\; \sin {60^\circ}\cos {30^\circ} + \sin {30^\circ}\cos {60^\circ} \qquad \rm{(ii)} \;\;\; \;2{\tan ^2}\,{45^\circ} + {\cos ^2}\,{30^\circ} - {\sin ^2}\,{60^\circ}\\\\&\rm (iii) \; \frac{{\cos {{45}^\circ}}}{{\sec {{30}^\circ} + {\rm{cosec}}\,{{30}^\circ}}} \qquad\qquad\qquad\quad {\rm{(iv}}) \;\;\; \frac{{\sin {{30}^\circ} + \tan {{45}^\circ} - {\rm{cosec}}\,{{60}^\circ}}}{{\sec {{30}^\circ} + \cos {{60}^\circ} - \cot {{45}^\circ}}}\\\\&\rm{(v)} \;\;\;\frac{{5{{\cos }^2}\,{{60}^\circ} + 4{{\sec }^2}\,{{30}^\circ} - {{\tan }^2}\,{{45}^\circ}}}{{{{\sec }^2}\,{{30}^\circ} + {{\cos }^2}\,{{30}^\circ}}}\end{align}\)

Text Solution

Reasoning:

We know that,

Exact Values of Trigonometric Functions
Angle (\(\theta\)) sin (\(\theta\)) cos (\(\theta\)) tan (\(\theta\))
Degrees Radians
\(0^{\circ}\) \(0\) \(0\) \(1\) \(0\)

\(30^{\circ}\)

\(\begin{align}\frac{\pi }{6}\end{align}\) \(\begin{align}\frac{1}{2}\end{align}\) \(\begin{align}\frac{{\sqrt 3 }}{2}\end{align}\) \(\begin{align}\frac{1}{{\sqrt 3 }}\end{align}\)

\(45^{\circ}\)

\(\begin{align}\frac{\pi }{4}\end{align}\) \(\begin{align}\frac{1}{{\sqrt 2 }}\end{align}\) \(\begin{align}\frac{1}{{\sqrt 2 }}\end{align}\) \(1\)

\(60^{\circ}\)

\(\begin{align}\frac{\pi }{3}\end{align}\) \(\begin{align}\frac{{\sqrt 3 }}{2}\end{align}\) \(\begin{align}\frac{1}{2}\end{align}\) \(\begin{align}\sqrt 3 \end{align}\)

\(90^{\circ}\)

\(\begin{align}\frac{\pi }{2}\end{align}\) \(1\) \(0\) Not Defined

Steps:

(i)

\[\begin{align}&\sin {60^\circ}\cos {30^\circ} + \sin {30^\circ}\cos {60^\circ}\\ &\qquad= \left( {\frac{{\sqrt 3 }}{2}} \right)\;\left( {\frac{{\sqrt 3 }}{2}} \right)\; + \;\left( {\frac{1}{2}} \right)\;\left( {\frac{1}{2}} \right)\;\\ &\qquad= \frac{3}{4}\; + \;\frac{1}{4} = \frac{{3 + 1}}{4}\\ &\qquad= \frac{4}{4} = 1\end{align}\]

(ii)

\[\begin{align}&2{\tan ^2}\,{45^\circ } + {\cos ^2}\,{30^\circ } - {\sin ^2}\,{60^\circ }\\& \quad= 2{\left( {\tan {{45}^\circ }} \right)^2} + {\left( {\cos \,{{30}^\circ }} \right)^2} - {\left( {\sin \,{{60}^\circ }} \right)^2}\\& \quad= 2{(1)^2} + {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} - {\left( {\frac{{\sqrt 3 }}{2}} \right)^2}\\ &\quad= 2 + \frac{3}{4} - \frac{3}{4}\\ &\quad= 2\end{align}\]

(iii)

\[\begin{align}
\frac{{\cos \,{{45}^0}}}{{\sec \,{{30}^0} + {\rm{cosec}}\,{{30}^0}}} &= \frac{{\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\left( {\frac{2}{{\sqrt 3 }}} \right) + \left( {\frac{2}{1}} \right)}}\\
 &= \frac{{\frac{1}{{\sqrt 2 }}}}{{\frac{{2 + 2\sqrt 3 }}{{\sqrt 3 }}}}\\
 &= \frac{{1 \times \sqrt 3 }}{{\sqrt 2  \times \left( {2 + 2\sqrt 3 } \right)}}\\
 &= \frac{{\sqrt 3 }}{{2\sqrt 2 \left( {\sqrt 3  + 1} \right)}}
\end{align}\]

Multiplying numerator and denominator by \(\sqrt 2 \left( {\sqrt 3  - 1} \right)\) , we get

\[\begin{align}
 &= \frac{{\sqrt 3 }}{{2\sqrt 2 \;\left( {\sqrt 3  + 1} \right)}} \times \frac{{\sqrt 2 (\sqrt 3  - 1)}}{{\sqrt 2 (\sqrt 3  - 1)}}\\
 &= \frac{{3\sqrt 2  - \sqrt 6 }}{{4\;\left( {3 - 1} \right)}}\\
 &= \frac{{3\sqrt 2  - \sqrt 6 }}{8}
\end{align}\]

(iv)

\[\begin{align}
\frac{{\sin \,{{30}^0} + \tan \,{{45}^0} - {\rm{cosec}}\,{{60}^0}}}{{\sec \,{{30}^0} + \cos \,{{60}^0} + \cot \,{{45}^0}}} &= \frac{{\frac{1}{2} + 1 - \frac{2}{{\sqrt 3 }}}}{{\frac{2}{{\sqrt 3 }} + \frac{1}{2} + 1}}\\
 &= \frac{{\frac{3}{2} - \frac{2}{{\sqrt 3 }}}}{{\frac{2}{{\sqrt 3 }} + \frac{3}{2}}}\\
 &= \frac{{\frac{{3\sqrt 3  - 4}}{{2\sqrt 3 }}}}{{\frac{{4 + 3\sqrt 3 }}{{2\sqrt 3 }}}}\\
 &= \frac{{3\sqrt 3  - 4}}{{3\sqrt 3  + 4}}
\end{align}\]

Multiplying numerator and denominator by \(\left( {3\sqrt 3  - 4} \right)\) ,we get

\[\begin{align}
 &= \frac{{(3\sqrt 3  - 4)\;(3\sqrt 3  - 4)}}{{(3\sqrt 3  + 4)\;(3\sqrt 3  - 4)}} & \\
 &= \frac{{27 + 16 - 24\sqrt 3 }}{{27 - 16}}\\
 &= \frac{{43 - 24\sqrt 3 }}{{11}}
\end{align}\]

(v)

\[\begin{align}
\frac{{5{{\cos }^2}{{60}^0} + 4{{\sec }^2}{{30}^0} - {{\tan }^2}{{45}^0}}}{{{{\sin }^2}{{30}^0} + {{\cos }^2}{{30}^0}}} &= \frac{{5 \times {{\left( {\frac{1}{2}} \right)}^2} + 4 \times {{\left( {\frac{2}{{\sqrt 3 }}} \right)}^2} - {{( - 1)}^2}}}{{{{\left( {\frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}\\
 &= \frac{{\left( {\frac{5}{4} + \frac{{16}}{3} - 1} \right)}}{{\left( {\frac{1}{4} + \frac{3}{4}} \right)}}\\
 &= \frac{{\left( {\frac{{15 + 64 - 12}}{{12}}} \right)}}{{\left( {\frac{{3 + 1}}{4}} \right)}}\\
& = \frac{{\left( {\frac{{67}}{{12}}} \right)}}{{\left( {\frac{4}{4}} \right)}}\\
 &= \frac{{67}}{{12}}
\end{align}\]

  
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