If sum of angle A and B is 45° and $t a n A + t a n B = 1.$ Find the value of $(t a n A)^{2} + (t a n B)^{2}$

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Solution

The correct option is B 1
Given $A + B = 45^{\circ}$ and $t a n A + t a n B = 1$
Using the formula, $(a + b)^{2} = a^{2} + b^{2} + 2 a b$ $(t a n A + t a n B)^{2} = (t a n A)^{2} + (t a n B)^{2} + 2 t a n A t a n B$
We know, $t a n (A + B) = (t a n A + t a n B) / (1 - t a n A t a n B)$
Since $A + B = 45^{\circ} \Rightarrow t a n (A + B) = t a n 45^{\circ} = 1$
Also $t a n A + t a n B = 1$
Therefore $1 = 1 / (1 - t a n A t a n B) \Rightarrow t a n A t a n B = 0$
Therefore $(t a n A + t a n B)^{2} = (t a n A)^{2} + (t a n B)^{2} + 2 t a n A t a n B$
$(1)^{2} = (t a n A)^{2} + (t a n B)^{2} + 0$
$\Rightarrow (t a n A)^{2} + (t a n B)^{2} = 1$

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