If $ \sin (A+B)=1 $ and $ \cos (A-B)=1,0^{\circ} < A + B \leq 90^{\circ}, A \geq B $ find A and B.

Given:

\( \sin (A+B)=1 \) and \( \cos (A-B)=1,0^{\circ} < A + B \leq 90^{\circ}, A \geq B \)

To do:

We have to find A and B.

Solution:  

$\sin (A+B)=1$

$\sin (A+B)=\sin 90^{\circ}$          (Since $\sin 90^{\circ}=1$)       

$\Rightarrow  A+B=90^{\circ}$......(i)

$\cos (A-B)=1$

$\cos (A-B)=\cos 0^{\circ}$             (Since $\cos 0^{\circ}=1$)

$\Rightarrow A-B=0^{\circ}$

$\Rightarrow  A=B$........(ii)

Substituting (ii) in (i), we get,

$B+B=90^{\circ}$

$\Rightarrow  2B=90^{\circ}$

$\Rightarrow  B=\frac{90^{\circ}}{2}$

$\Rightarrow  B=45^{\circ}$

$\Rightarrow  A=45^{\circ}$

The values of both $A$ and $B$ is $45^{\circ}$ each.

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Updated on: 10-Oct-2022

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