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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