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

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## Question

Evaluate the following:

(i) $$\;\;\; \sin {60^\circ}\cos {30^\circ} + \sin {30^\circ}\cos {60^\circ}$$

(ii) $$\;\;\; \;2{\tan ^2}\,{45^\circ} + {\cos ^2}\,{30^\circ} - {\sin ^2}\,{60^\circ}$$

(iii) $$\;\;\; \dfrac{{\cos {{45}^\circ}}}{{\sec {{30}^\circ} + {\rm{cosec}}\,{{30}^\circ}}}$$

(iv) $$\;\;\; \dfrac{{\sin {{30}^\circ} + \tan {{45}^\circ} - {\rm{cosec}}\,{{60}^\circ}}}{{\sec {{30}^\circ} + \cos {{60}^\circ} - \cot {{45}^\circ}}}$$

(v) $$\;\;\;\dfrac{{5{{\cos }^2}\,{{60}^\circ} + 4{{\sec }^2}\,{{30}^\circ} - {{\tan }^2}\,{{45}^\circ}}}{{{{\sec }^2}\,{{30}^\circ} + {{\cos }^2}\,{{30}^\circ}}}$$

Video Solution
Introduction To Trigonometry
Ex 8.2 | Question 1

## 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}\\ &= \left( {\frac{{\sqrt 3 }}{2}} \right)\left( {\frac{{\sqrt 3 }}{2}} \right) + \left( {\frac{1}{2}} \right)\left( {\frac{1}{2}} \right)\\ &= \frac{3}{4} + \frac{1}{4} = \frac{{3 + 1}}{4}\\ &= \frac{4}{4} = 1\end{align}

(ii)

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