Question

Solve 67 and 68, I leave bad reviews if you leave the session.

Answer 1

67)

`\begin{gathered} sec\theta\text{ = }(1)/(cos\theta) \\ Cross\text{ multiply} \\ sec\theta\text{ }*\text{ cos}\theta \\ (1)/(cos\theta)\text{ }* cos\theta \\ =\text{ 1} \end{gathered}`

68)

Draw a right angle triangle:

`\begin{gathered} sin\theta\text{ = }(opposite)/(hypotenuse)\text{ = }(a)/(c) \\ cos\theta\text{ = }\frac{adjacent\text{ }}{Hypotenuse}\text{ = }(b)/(c) \\ tan\theta\text{ = }\frac{opposite}{adjacent\text{ }}\text{ = }(a)/(b) \end{gathered}`
`\begin{gathered} \\ \frac{sin\theta\begin{equation*}\end{equation*}}{cos\theta}\text{ = }(a)/(c)\text{ }/\text{ }(b)/(c) \\ =\text{ }(a)/(c)\text{ }*\text{ }(c)/(b) \\ =\text{ }(a)/(b) \\ Hence,tan\theta\text{= }(sin\theta)/(cos\theta) \end{gathered}`