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

## Question

(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 $$A = 0°$$.

Video Solution
Introduction To Trigonometry
Ex 8.2 | Question 4

## 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}}\;\;\cos {30^\circ} = \frac{{\sqrt 3 }}{2}\\\sin {60^\circ} &= \frac{{\sqrt 3 }}{2}\quad{\rm{and}} \;\;\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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