Question

$\frac{cos{22}^{\circ }-sin{22}^{\circ }}{cos{22}^{\circ }+sin{22}^{\circ }}$ equal to tanA, $0^{\circ }$ < A < ${90}^{\circ }$ . Find the value of A.

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Answer 1

It will be easy to guess the method for those who have seen the expansion of tan (45+A) or tan (45-A).
Otherwise, there is a good chance that we will be tempted to multiply both numerator and denominator with cos22° ± sin 22°. If you do that you will see it does not simplify the expression considerably.
At the same time we have to keep in mind that, after simplifying we have to get it of the form tanA. This can be an inspiration for considering the expansion of angles like tan (A+B) or tan (A-B). Still we don't have any tan in the expression given. But once we know that tan (A+B) or tan (A-B) can be used, it is easy to get tan in the expression. We can divide by $cos{22}^{\circ }$ , both numerator and denominator.

$\Rightarrow$ $\frac{cos{22}^{\circ }-sin{22}^{\circ }}{cos{22}^{\circ }+sin{22}^{\circ }}=\frac{1-tan{22}^{\circ }}{1+tan{22}^{\circ }}$

We can write it as $\frac{tan45-tan22}{1+tan45tan22}$
Which is tan 45 -22 = tan 23

$\Rightarrow$ A = 23.
Intuition behind last step: We have already guessed that we will be using tan (A$\pm$B). For the

numerator to be of that form, instead of 1 it should be tan of some angle. Once we think this way it is obvious that we have replace tan 22 by tan 45 tan 22.
Key steps/concepts: (1) Prior experience with expanding tan (45 $\pm$ A)
(2) tan (A - B) = $\frac{tanA-tanB}{1+tanAtanB}$
(3) Guessing that we will be using tan (A $\pm$ B)

(4) Dividing by cos ${22}^{\circ }$

(5) Replacing 1 with tan ${45}^{\circ }$