# Ex.8.3 Q1 Introduction to Trigonometry Solution - NCERT Maths Class 10

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## Question

Evaluate:

(i)  $$\;\frac{\sin {{18}^{0}}}{\cos {{72}^{0}}}$$

(ii) $$\;\frac{\tan {{26}^{0}}}{\cot {{64}^{0}}}$$

(iii) $$\;\cos {{48}^{0}}-\sin {{42}^{0}}$$

(iv) $$\;\text{cosec}\,{{31}^{0}}-\sec \,{{59}^{0}}$$

Video Solution
Introduction To Trigonometry
Ex 8.3 | Question 1

## Text Solution

#### Reasoning:

\begin{align} \sin \left(90^{\circ}-\theta\right) &=\cos \theta \\ \tan \left(90^{\circ}-\theta\right) &=\cot \theta \\ \sec \left(90^{\circ}-\theta\right) &=\text{cosec} \;\theta \end{align}

#### Solution:

\begin{align}\rm (i)\qquad \frac{\sin {{18}^{0}}}{\cos {{72}^{0}}}\end{align}

\begin{align}\text{Since,}&\\ &\sin \left( {{90}^{0}}-\theta \right)=\cos \theta \end{align}

Here $$\text{ }\!\!\theta\!\!\text{ }={{72}^{0}}$$

\begin{align} & ∴ =\frac{\sin ({{90}^{0}}-{{72}^{0}})}{\cos {{72}^{0}}} \\ & \,\,\,\,=\frac{\cos {{72}^{0}}}{\cos {{72}^{0}}} \\ & \,\,\,\,=1 \\ \end{align}

\begin{align}\text{(ii)} \qquad\frac{\tan {{26}^{0}}}{\cot {{64}^{0}}}\end{align}

\begin{align}\text{Since}&, \\ &\tan ({{90}^{0}}-\theta )=\cot \theta \end{align}

Here $$\theta ={{64}^{0}}$$

\begin{align} & ∴ =\frac{\tan ({{90}^{0}}-{{64}^{0}})}{\cot {{64}^{0}}} \\ & \,\,\,\,=\frac{\cot {{64}^{0}}}{\cot {{64}^{0}}} \\ & \,\,\,\,=1 \\ \end{align}

$$\text{(iii)}\qquad\cos {{48}^{0}}-\sin {{42}^{0}}$$

\begin{align}\text{Since}&, \\&\sin ({{90}^{0}}-\theta )=\cos \theta\end{align}

Here $$\text{ }\!\!\theta\!\!\text{ }={{48}^{0}}$$

\begin{align} ∴ \ &=\cos {{48}^{0}}-\sin \left( {{90}^{0}}-{{48}^{0}} \right) \\ & =\cos {{48}^{0}}-\cos {{48}^{0}} \\ & =0 \end{align}

$$\text{(iv) }\qquad\text{cosec}{{31}^{0}}-\sec {{59}^{0}}$$

\begin{align}\text{Since}&,\\ &\sec \,({{90}^{0}}-\theta )=\operatorname{cosec}\theta\end{align}

Here $$\theta ={{31}^{0}}$$

Therefore,

\begin{align} \,& \text{cosec}{{31}^{\circ}}-\sec {{59}^{\circ}}\\ & =\operatorname{cosec}{{31}^{\circ}}-\sec \left( {{90}^{\circ}}-{{31}^{\circ}} \right) \\ & =\operatorname{cosec}{{31}^{\circ}}-\operatorname{cosec}{{31}^{\circ}} \\ & =0 \end{align}

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