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) \(\begin{align}\text{cos}\,38{}^\circ \,\cos \,52{}^\circ  -\sin 38{}^\circ \ \sin 52{}^\circ  =0 \end{align}\)

 Video Solution
Introduction To Trigonometry
Ex 8.3 | Question 2

Text Solution

 

Reasoning:

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

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

Steps: 

(i)Taking

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

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

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

Hence,

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

(ii) Taking

\(\begin{align}\text{L.H.S }\!=\!\text{cos}\,38{}^\circ \,\cos \,52{}^\circ -\sin 38{}^\circ \ \sin 52{}^\circ\end{align}\)

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

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

Hence,

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

  
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