Question

$\left(i\right)\text{If}\sin (A+B)=1\text{and}\cos (A-B)=1,$ find $A$ and $B$.
$\left(ii\right)\text{If}\tan (2A+B)=\sqrt{3}\text{and}\cot (3A-B)=\sqrt{3}$, find $A$ and $B$.

• A. $\left(i\right)A=B={45}^{\circ }$$\left(ii\right)A={24}^{\circ }andB={18}^{\circ }$
• B. $\left(i\right)A=B={45}^{\circ }$$\left(ii\right)A={18}^{\circ }andB={24}^{\circ }$
• C. $\left(i\right)A=B={90}^{\circ }$$\left(ii\right)A={18}^{\circ }andB={24}^{\circ }$
• D. $\left(i\right)A=B={45}^{\circ }$$\left(ii\right)A={36}^{\circ }andB={24}^{\circ }$

Answer 1

The correct option is B $\left(i\right)A=B={45}^{\circ }$

$\left(ii\right)A={18}^{\circ }andB={24}^{\circ }$

$sin(A+B)=1$
Hence
$A+B={90}^0$
And
$cos(A-B)=1$
Hence
$A-B=0^0$
$A=B$
Substituting we get
$2A={90}^0$
$A={45}^0=B$
Similarly
$tan(2A+B)=\sqrt{3}$
Hence
$2A+B={60}^0$
And
$cot(3A-B)=\sqrt{3}$
Hence
$3A-B={30}^0$
Hence
$5A={90}^0$
$A={18}^0$
Therefore
$B={60}^0-{36}^0$
$B={24}^0$