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Question

If $$A+B=45^\circ$$, prove that $$(1+ \tan A)(1+ \tan B)=2$$ and hence deduce that $$\tan 22\dfrac{1}{2}^\circ=\sqrt{2}-1$$


Solution

$$\rightarrow A+B={ 45 }^{ 0 }$$
$$\rightarrow A={ 45 }^{ 0 }-B$$
$$\rightarrow \tan A=\tan\left( { 45 }^{ 0 }-B \right) =\dfrac { 1-\tan B }{ 1+\tan B } $$
$$\rightarrow \tan A+1=\dfrac { 1-\tan B }{ 1+\tan B } +1=\dfrac { 2 }{ 1+\tan B } $$
$$\rightarrow \left( 1+\tan A \right) \left( 1+\tan B \right) =2$$
Now, for $$A=B={ 22 }_{ 2 }^{ 1 }$$
$$\rightarrow { \left( 1+\tan A \right)  }^{ 2 }=2$$
$$\rightarrow \tan A=\tan\left( { 22 }_{ 2 }^{ 1 } \right) =\sqrt { 2 } -1$$

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