If $ \tan (\mathbf{A}+\mathbf{B})=1 $ and $ \tan (\mathbf{A}-\mathbf{B})=\frac{1}{\sqrt3}, 0^{\circ} < A + B < 90^{\circ}, A > B, $ then find the values of $ \mathbf{A} $ and $ \mathbf{B} $.

Given:

\( \tan (A+B)=1 \) and \( \tan (A-B)=\frac{1}{\sqrt{3}}, 0^{\circ} < A+B < 90^{\circ}, A>B \)

To do:

We have to find the values of $A$ and $B$.

Solution:  

$\tan (A+B)=1$

$\tan (A+B)=\tan 45^{\circ}$          (Since $\tan 45^{\circ}=1$)       

$\Rightarrow  A+B=45^{\circ}$......(i)

$\tan (A-B)=\frac{1}{\sqrt3}$

$\tan (A-B)=\tan 30^{\circ}$             (Since $\tan 30^{\circ}=\frac{1}{\sqrt3}$)

$\Rightarrow A-B=30^{\circ}$

$\Rightarrow  A=30^{\circ}+B$........(ii)

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

$30^{\circ}+B+B=45^{\circ}$

$\Rightarrow  2B=15^{\circ}$

$\Rightarrow  B=\frac{15^{\circ}}{2}$

$\Rightarrow  B=7\frac{1}{2}^{\circ}$

$\Rightarrow  A=45^{\circ}-7\frac{1}{2}^{\circ}=37\frac{1}{2}^{\circ}$

The values of $A$ and $B$ are $37\frac{1}{2}^{\circ}$ and $7\frac{1}{2}^{\circ}$ respectively.  

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Updated on: 10-Oct-2022

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