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

## Question

Evaluate

(i) $$\,\,\frac{{{\sin }^{2}}{{63}^{{}^\circ }}+{{\sin }^{2}}{{27}^{{}^\circ }}}{{{\cos }^{2}}{{17}^{{}^\circ }}+{{\cos }^{2}}{{73}^{{}^\circ }}}$$

(ii) $$\,\,\,\text{sin}{{25}^{^\circ }}\text{cos}{{65}^{^\circ }}+\text{cos}{{25}^{^\circ }}\text{ sin}{{65}^{^\circ }}$$

Video Solution
Introduction To Trigonometry
Ex 8.4 | Question 3

## 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:

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

(ii) $$\sin {{25}^{{}^\circ }}\cos {{65}^{{}^\circ }}+\cos {{25}^{{}^\circ }}\sin {{65}^{{}^\circ }}$$

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

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school