Ex.8.3 Q2 Introduction to Trigonometry Solution - NCERT Maths Class 10

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Question

Show that:

(i) \(\begin{align}\text{tan}\,48{}^\circ \,\text{tan}\,23{}^\circ \,\text{tan}\,42{}^\circ \ \text{tan}\,67{}^\circ =1\end{align}\)
(ii) \(\begin{align}\text{cos}\,38{}^\circ \,\cos \,52{}^\circ -\sin 38{}^\circ \ \sin 52{}^\circ=0 \end{align}\)

Text Solution

 

Reasoning:

\(\sin \left( {{90}^{0}}-\theta \right)=\cos \theta \)

\(\tan ({{90}^{0}}-\theta )=\cot \theta \)

Steps: 

(i)Taking L.H.S =\(\text{tan}\,48{}^\circ \,\text{tan}\,23{}^\circ \,\text{tan}\,42{}^\circ \ \text{tan}\,67{}^\circ \)

Since \(\tan \,({{90}^{0}}-\theta)=\cot \theta\)

\[\begin{align} & =\tan \,({{90}^{0}}-{{42}^{0}})\tan ({{90}^{0}}-{{67}^{0}})\tan {{42}^{0}}\tan {{67}^{0}} \\ & =\cot {{42}^{0}}\cot {{67}^{0}}\tan {{42}^{0}}\tan {{67}^{0}} \\ & =(\cot {{42}^{0}}\tan {{42}^{0}})\,(\cot {{67}^{0}}\tan {{67}^{0}}) \\ & =\left( \frac{1}{\tan {{42}^{0}}}\tan {{42}^{0}} \right)\,\,\left( \frac{1}{\tan {{67}^{0}}}\tan {{67}^{0}} \right) \\ & =1\,{\rm x }\;1\\&=1 \\ & =\text{R.H.S} \\ \end{align}\]

Hence\(\text{tan}\,48{}^\circ \,\text{tan}\,23{}^\circ \,\text{tan}\,42{}^\circ \ \text{tan}\,67{}^\circ =1\)

(ii) Taking L.H.S =\(\text{cos}\,38{}^\circ \,\cos \,52{}^\circ -\sin 38{}^\circ \ \sin 52{}^\circ \)

Since, \(\tan ({{90}^{0}}-\theta)=\cos \theta\)

\[\begin{align} & =\cos {{38}^{0}}\cos {{52}^{0}}-\sin \,({{90}^{0}}-{{57}^{0}})\ \sin \,({{90}^{0}}-{{38}^{0}}) \\ & =\cos {{38}^{0}}\cos {{52}^{0}}-\cos {{52}^{0}}\cos {{38}^{0}} \\ & =0 \\ & =\text{R.H.S} \\ \end{align}\]

Hence, \(\cos {{38}^{0}}\cos {{52}^{0}}-\sin {{38}^{0}}\sin {{52}^{0}}=0\)

  
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