NCERT Solutions
Class 10 - Mathematics - Chapter 8: INTRODUCTION TO TRIGONOMETRY - Exercise 8.2

Question 4

We check each statement using trigonometric identities and basic behaviour of \(\sin\theta\) and \(\cos\theta\).

(i) \(\sin(A+B)=\sin A+\sin B\)

False. The correct identity is:

\(\sin(A+B)=\sin A\cos B+\cos A\sin B\).

Justification by example: Take \(A=30^\circ\), \(B=30^\circ\).

LHS: \(\sin(30^\circ+30^\circ)=\sin 60^\circ=\dfrac{\sqrt{3}}{2}\).

RHS: \(\sin 30^\circ+\sin 30^\circ=\dfrac{1}{2}+\dfrac{1}{2}=1\).

Since \(\dfrac{\sqrt{3}}{2} \ne 1\), the statement is false.

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

True (for \(0^\circ \le \theta \le 90^\circ\)).

In the first quadrant, as \(\theta\) increases, the opposite side (compared to hypotenuse) effectively increases, so \(\sin\theta\) increases.

Quick table check: \(\sin 0^\circ=0\), \(\sin 30^\circ=\dfrac{1}{2}\), \(\sin 45^\circ=\dfrac{\sqrt{2}}{2}\), \(\sin 60^\circ=\dfrac{\sqrt{3}}{2}\), \(\sin 90^\circ=1\). These values clearly increase.

Student Note: This “increasing” behaviour is commonly discussed for \(0^\circ\) to \(90^\circ\). Outside this range, \(\sin\theta\) does not keep increasing forever.

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

False (for \(0^\circ \le \theta \le 90^\circ\)).

In the first quadrant, \(\cos\theta\) actually decreases as \(\theta\) increases.

Quick table check: \(\cos 0^\circ=1\), \(\cos 30^\circ=\dfrac{\sqrt{3}}{2}\), \(\cos 45^\circ=\dfrac{\sqrt{2}}{2}\), \(\cos 60^\circ=\dfrac{1}{2}\), \(\cos 90^\circ=0\). These values go down.

(iv) \(\sin\theta=\cos\theta\) for all values of \(\theta\)

False. \(\sin\theta\) equals \(\cos\theta\) only for some specific angles, not for all.

Justification by example: At \(\theta=45^\circ\), \(\sin 45^\circ=\cos 45^\circ=\dfrac{\sqrt{2}}{2}\) (so it is true here).

But at \(\theta=30^\circ\): \(\sin 30^\circ=\dfrac{1}{2}\) and \(\cos 30^\circ=\dfrac{\sqrt{3}}{2}\), which are not equal. Hence it is not true for all \(\theta\).

(v) “\(\cot A\) is not defined for \(A=0^\circ\)”

True. We know:

\(\cot A = \dfrac{\cos A}{\sin A}\).

At \(A=0^\circ\), \(\sin 0^\circ = 0\) and \(\cos 0^\circ = 1\).

So, \(\cot 0^\circ = \dfrac{1}{0}\), which is not defined.

Final Answers: (i) False, (ii) True, (iii) False, (iv) False, (v) True.