Ex.8.2 Q4 Introduction to Trigonometry Solution - NCERT Maths Class 10

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Question

State whether the following are true or false. Justify your answer.

(i) \(\begin{align}{\rm{sin}}\,\left( {A + B} \right) = {\rm{sin}}\,A + {\rm{sin}}\,B\,.\end{align}\)

(ii) The value of \(\sin\) \(\theta\) increases as \(\theta\) increases.

(iii) The value of \(\cos\) \(\theta\) increases as \(\theta\) increases.

(iv) \(\begin{align}{\rm{sin}}\,\theta = {\rm{cos}}\,\theta \end{align}\) for all values of \(\theta\).

(v) \(\cot A\) is not defined for \(A = 0°\).

 

Text Solution

Steps:

(i)  \(\begin{align}{\rm{sin}}\,\left( {A + B} \right) = {\rm{sin}}\,A + {\rm{sin}}\,B\,.\end{align}\)

For the purpose of verification, Let \(\begin{align}A = {30^\circ}\;{\rm{and}}\;B = {60^\circ}\end{align}\)

\[\begin{align}L.H.S\, & = \sin \,\left( {A + B} \right)\\ &= \sin \,\left( {{{30}^\circ} + {{60}^\circ}} \right)\\ &= \sin \,{90^\circ}\\ &= 1\end{align}\]

\[\begin{align}R.H.S\, & = \sin A + \sin B\\ &= \sin {30^\circ} + \sin {60^\circ}\\ &= \frac{1}{2} + \frac{{\sqrt 3 }}{2}\\ &= \frac{{1 + \sqrt 3 }}{2}\end{align}\]

Since, \(\begin{align}{\rm{sin}}\,\left( {A + B} \right) \ne {\rm{sin}}\,A + {\rm{sin}}\,B\,.\end{align}\)

Hence, the given statement is not true

(ii) The value of \(\begin{align}\sin \,\theta \end{align}\) increases from 0 to 1 as \(\theta\) increases from \(\begin{align}{0^\circ}\;{\rm{to}}\;{90^\circ}\end{align}\)

\[\begin{align}\sin {0^\circ} &= 0\\\sin {30^\circ} &= \frac{1}{2} = 0.5\\\sin {45^\circ} &= \frac{1}{{\sqrt 2 }} = 0.707\\\sin {60^\circ} &= \frac{{\sqrt 3 }}{2} = 0.866\\\sin {90^\circ} &= 1\end{align}\]

Hence, the given statement is true.

(iii) The value of \(\begin{align}\cos \theta \end{align}\) decreases from 1 to 0 as \(\theta\) increases from \(\begin{align}{0^\circ}\;{\rm{to}}\;{90^\circ}\end{align}\)

\[\begin{align}\cos {0^\circ} &= 1\\\cos {30^\circ} &= \frac{{\sqrt 3 }}{2} = 0.866\\\cos {45^\circ} &= \frac{1}{{\sqrt 2 }} = 0.707\\\cos {60^\circ} &= \frac{1}{2} = 0.5\\\cos {90^\circ} &= 0\end{align}\]

Hence, the given statement is false.

(iv)

This is true when \(\begin{align}\theta = {45^\circ}\end{align}\)

As \(\begin{align}\sin {45^\circ} = \frac{1}{{\sqrt 2 }}\;\;{\rm{and}}\;\;\cos {45^\circ} = \frac{1}{{\sqrt 2 }}\end{align}\)

It is not true for other values of \(\theta\)

As,

\[\begin{align}\sin {30^\circ} &= \frac{1}{2}\quad{\rm{and}}\quad\cos {30^\circ} = \frac{{\sqrt 3 }}{2}\\\sin {60^\circ} &= \frac{{\sqrt 3 }}{2}\quad{\rm{and}}\quad\cos {60^\circ} = \frac{1}{{\sqrt 2 }}\\\sin {90^\circ} &= 1\quad{\rm{and}}\quad\cos {90^\circ} = 0\end{align}\]

Hence, the given statement is false.

(v)

\[\begin{align}\cot A &= \frac{{\cos A}}{{\sin A}}\\∴\; \cot {0^\circ} &= \frac{{\cos {0^\circ}}}{{\sin {0^\circ}}}\,\, = \frac{1}{0} = \text{undefined}\end{align}\]

Hence the given statement is true.

  
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