You should already know that:
\(tan\theta = \frac{sin\theta}{cos\theta}\)
Plugging that in we get:
\(cos\theta + sin\theta (\frac{sin\theta}{cos\theta}) =\\\\cos\theta +\frac{sin^2\theta}{cos\theta}\)
Multiply the cos theta by cos theta on the numerator and denominator so we can add them :)
\((cos\theta \times \frac{cos\theta}{cos\theta})+\frac{sin^2\theta}{cos\theta} = \\\\\frac{cos^2\theta}{cos\theta} + \frac{sin^2\theta}{cos\theta}=\\\\\frac{cos^2\theta + sin^2\theta}{cos\theta}\)
You should also know the identity that cos^2 theta + sin^2 theta = 1
\(\frac{cos^2\theta + sin^2\theta}{cos\theta} = \\\\\frac{1}{cos\theta}\)
Therefore,
\(cos\theta + sin\theta tan\theta = \boxed{\frac{1}{cos\theta}}\)
