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

Question.

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

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

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

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

(v) $\cot \mathrm{A}$ is not defined for $\mathrm{A}=0^{\circ}$.

Solution:

(i) False.

When A = 60°, B = 30°

LHS = sin (A + B) = sin (60° + 30°)

= sin 90° = 1

RHS = sin A + sin B

= sin 60° + sin 30°

$=\frac{\sqrt{3}}{2}+\frac{1}{2} \neq 1$

i.e., LHS $\neq$ RHS

(ii) True.

Note that $\sin 0^{\circ}=0, \quad \sin 30^{\circ}=\frac{\mathbf{1}}{\mathbf{2}}=0.5$,

$\sin 45^{\circ}=\frac{1}{\sqrt{2}}=0.7$ (approx.)

$\sin 60^{\circ}=\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}=0.87$ (approx.)

and $\quad \sin 90^{\circ}=1$

i.e., value of $\sin \theta$ increases as $\theta$ increases from $0^{\circ}$ to $90^{\circ}$.

(iii) False.

Note that $\cos 0^{\circ}=1$

$\cos 30^{\circ}=\frac{\sqrt{3}}{2}=0.87$ (approx.)

$\cos 45^{\circ}=\frac{1}{\sqrt{2}}=0.7$ (approx.)

$\cos 60^{\circ}=\frac{\mathbf{1}}{\mathbf{2}}=0.5$ and $\cos 90^{\circ}=0$

i.e., value of $\cos \theta$ decreases as $\theta$ increases from $0^{\circ}$ to $90^{\circ}$.

(iv) False, it is true for only $\theta=45^{\circ}$

(v) True, $\cot A=\frac{1}{\mathbf{0}}=$ not defined.