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Exercise

$\frac{2-\tan3a}{1-\tan^23a}$

Step-by-step Solution

1

Rewrite $\frac{2-\tan\left(3a\right)}{1-\tan\left(3a\right)^2}$ in terms of sine and cosine functions

$\frac{2+\frac{-\sin\left(3a\right)}{\cos\left(3a\right)}}{1+\frac{-\sin\left(3a\right)^2}{\cos\left(3a\right)^2}}$
2

Combine $1+\frac{-\sin\left(3a\right)^2}{\cos\left(3a\right)^2}$ in a single fraction

$\frac{2+\frac{-\sin\left(3a\right)}{\cos\left(3a\right)}}{\frac{-\sin\left(3a\right)^2+\cos\left(3a\right)^2}{\cos\left(3a\right)^2}}$
3

Divide fractions $\frac{2+\frac{-\sin\left(3a\right)}{\cos\left(3a\right)}}{\frac{-\sin\left(3a\right)^2+\cos\left(3a\right)^2}{\cos\left(3a\right)^2}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$

$\frac{\left(2+\frac{-\sin\left(3a\right)}{\cos\left(3a\right)}\right)\cos\left(3a\right)^2}{-\sin\left(3a\right)^2+\cos\left(3a\right)^2}$
4

Applying the trigonometric identity: $\cos\left(\theta \right)^2-\sin\left(\theta \right)^2 = \cos\left(2\theta \right)$

$\frac{\left(2+\frac{-\sin\left(3a\right)}{\cos\left(3a\right)}\right)\cos\left(3a\right)^2}{\cos\left(6a\right)}$
Why is cos(2x) = cos(x)^2 - sin(x)^2 ?
5

Apply the trigonometric identity: $\frac{\sin\left(\theta \right)}{\cos\left(\theta \right)}$$=\tan\left(\theta \right)$, where $x=3a$

$\frac{\left(2-\tan\left(3a\right)\right)\cos\left(3a\right)^2}{\cos\left(6a\right)}$
Why is sin(x)/cos(x) = tan(x) ?

Final answer to the exercise

$\frac{\left(2-\tan\left(3a\right)\right)\cos\left(3a\right)^2}{\cos\left(6a\right)}$

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