# Ex.8.4 Q3 Introduction to Trigonometry Solution - NCERT Maths Class 10

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## Question

Evaluate

\begin{align} & \text{(i)}\,\,\,\,\frac{{{\sin }^{2}}{{63}^{{}^\circ }}+{{\sin }^{2}}{{27}^{{}^\circ }}}{{{\cos }^{2}}{{17}^{{}^\circ }}+{{\cos }^{2}}{{73}^{{}^\circ }}} \\ & \text{(ii)}\,\,\,\text{sin}{{25}^{{}^\circ }}\text{ cos}{{65}^{{}^\circ }}\text{ }+\text{ cos}{{25}^{{}^\circ }}\text{ sin}{{65}^{{}^\circ }} \end{align}

## Text Solution

#### Reasoning:

\begin{align}{\sin ^{2} A+\cos ^{2} A=1} \\ {\sin \left(90^{\circ}-\theta\right)=\cos \theta} \\ {\cos \left(90^{\circ}-\theta\right)=\sin \theta}\end{align}

#### Steps:

\begin{align}\text{ (i) }\frac{{{\sin }^{2}}{{63}^{{}^\circ }}+{{\sin }^{2}}{{27}^{{}^\circ }}}{{{\cos }^{2}}{{17}^{{}^\circ }}+{{\cos }^{2}}{{73}^{{}^\circ }}}\end{align}

\begin{align} & =\frac{{{\left[ \sin \left( {{90}^{{}^\circ }}-27 \right) \right]}^{2}}+{{\sin }^{2}}27}{{{\left[ \cos \left( {{90}^{{}^\circ }}-{{73}^{{}^\circ }} \right) \right]}^{2}}+{{\cos }^{2}}{{73}^{{}^\circ }}}\end{align}
\begin{align}& =\frac{{{[\cos 27]}^{2}}+{{\sin }^{2}}{{27}^{{}^\circ }}}{{{\left[ \sin {{73}^{{}^\circ }} \right]}^{2}}+{{\cos }^{2}}{{73}^{{}^\circ }}}\end{align} ( $$\rm sin(90^{\circ} - \theta)$$ $$=$$ $$\rm cos \;\theta$$ & $$\rm cos(90^{\circ} - \theta)$$ $$=$$ $$\rm sin \;\theta$$
)
\begin{align} =\frac{1}{1}\end{align}                    (By identity $$\sin ^{2} A+\cos ^{2} A=1$$)
\begin{align} =\;1 \end{align}

\begin{align}\text{(ii)}\,\,\,\sin {{25}^{{}^\circ }}\cos {{65}^{{}^\circ }}+\cos {{25}^{{}^\circ }}\sin {{65}^{{}^\circ }}\end{align}

\begin{align} &=\left( \sin {{25}^{{}^\circ }} \right)\left\{ \cos \left( {{90}^{{}^\circ }}-{{25}^{{}^\circ }} \right) \right\}\end{align}
\begin{align} +\cos {{25}^{{}^\circ }} \left\{ \sin \left( {{90}^{{}^\circ }}-{{25}^{{}^\circ }} \right) \right. \end{align}
\begin{align} \left( \sin \left( {{90}^{{}^\circ }}-\text{ }\!\!\theta\!\!\text{ } \right)=\cos \text{ }\!\!\theta\!\!\text{ }\;\;\And\text{ }\cos \left( {{90}^{{}^\circ }}-\theta \right)=\sin \theta \right) \end{align}
\begin{align} =\left( \sin {{25}^{{}^\circ }} \right)\left( \sin {{25}^{{}^\circ }} \right)+\cos {{25}^{{}^\circ }}\left( \cos {{25}^{{}^\circ }} \right) \end{align}
\begin{align} ={{\sin }^{2}}{{25}^{{}^\circ }}+{{\cos }^{2}}{{25}^{{}^\circ }} \end{align}
\begin{align} =1\qquad\qquad \qquad \end{align}  (By identity $$\sin ^{2} A+\cos ^{2} A=1$$
)

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