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

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Question

Evaluate:

\(\begin{align}\rm{(i)}\;\frac{\sin {{18}^{0}}}{\cos {{72}^{0}}} \qquad (ii)\;\frac{\tan {{26}^{0}}}{\cot {{64}^{0}}}\end{align}\)

\(\begin{align}\rm(iii)\;\cos {{48}^{0}}-\sin {{42}^{0}}\qquad\rm(iv)\;\text{cosec}\,{{31}^{0}}-\sec \,{{59}^{0}}\end{align}\)

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 \(\theta ={{31}^{0}}\)

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