State whether the following are true or false. Justify your answer.
$\sin (A+B)=\sin A+\sin B$
State whether the following are true or false. Justify your answer.
$\sin (A+B)=\sin A+\sin B$
$\sin (A+B)=\sin A+\sin B$
Let $A=30^{\circ}$ and $B=60^{\circ}$
$\sin (A+B)=\sin \left(30^{\circ}+60^{\circ}\right)$
$=\sin 90^{\circ}$
$=1$
$\sin A+\sin B=\sin 30^{\circ}+\sin 60^{\circ}$
$=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}$
Clearly, $\sin (A+B) \neq \sin A+\sin B$
Hence, the given statement is false.
Similar Questions
If $\angle B$ and $\angle Q$ are acute angles such that $\sin B =\sin Q$, then prove that $\angle B =\angle Q$.
If $\angle B$ and $\angle Q$ are acute angles such that $\sin B =\sin Q$, then prove that $\angle B =\angle Q$.
$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=........$
$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=........$
Prove that
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta},$ using the identity
$\sec ^{2} \theta=1+\tan ^{2} \theta$
Prove that
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta},$ using the identity
$\sec ^{2} \theta=1+\tan ^{2} \theta$
Evaluate the following:
$\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}$
Evaluate the following:
$\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$