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.
Consider $\triangle ACB$, right-angled at $C$, in which $AB =29$ units, $BC =21$ units and $\angle ABC =\theta$ (see $Fig.$). Determine the values of
$(i)$ $\cos ^{2} \theta+\sin ^{2} \theta$
$(ii)$ $\cos ^{2} \theta-\sin ^{2} \theta$
In a right triangle $A B C$, right-angled at $B$. if $\tan A =1,$ then verify that $2 \sin A \cos A=1$
Evaluate:
$\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}$
Evaluate the following:
$\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}$
Prove that $\sec A(1-\sin A)(\sec A+\tan A)=1$