Quadrants:

sin θ is positive in the first and second quadrants and negative in the third and fourth quadrants.

cos θ is positive in the first and fourth quadrants and negative in the second and third quadrants.

tan θ is positive in the first and third quadrants and negative in the second and fourth quadrants.

Key terms:

TOA, CAH, SOH for right-angled triangles:

$ \tan \theta = $ $ \phantom{.} {Opp \over Adj} $, $ \cos \theta = $ $ \phantom{.} {Adj \over Hyp} $, $ \sin \theta = $ $ \phantom{.} {Opp \over Hyp} $

cot θ, sec θ, cosec θ:

(Hint: Look at the third letter of each term)

$ \cot \theta = $ $ \phantom{.} {1 \over \tan \theta} $, $ \sec \theta = $ $ \phantom{.} {1 \over \cos \theta} $, $ \text{cosec } \theta = $ $ \phantom{.} {1 \over \sin \theta} $

1. $\theta$ is an obtuse angle and $\sin \theta = k$, where $k$ is a constant. Express $\cot \theta$ in terms of $k$.

(from A Maths 360 2nd edition Ex 10.1)

Answer: $ \cot \theta = - {\sqrt{1 - k^2} \over k} $

Solutions

\begin{align} \sin \theta & = k \\ {Opp \over Hyp} & = {k \over 1} \end{align}

\begin{align} \text{By Pytha} & \text{goras theorem,} \\ \sqrt{(1)^2 - (k)^2} & = \sqrt{1 - k^2} \\ \\ \cot \theta & = {1 \over \tan \theta} \\ & = {1 \over {- {k \over \sqrt{1 - k^2} } } } \\ & = -{ \sqrt{1 - k^2} \over k } \end{align}

2. Given that $\tan \theta = -{4 \over 3}$ and $ \cos \theta > 0$, find the value of $\sec \theta$ without using a calculator.

Answer: $ \sec \theta = {5 \over 3} $

Solutions

\begin{align} \tan \theta & < 0 \phantom{000000} [\text{2nd or 4th quadrant}] \\ \\ \cos \theta & > 0 \phantom{000000} [\text{1st or 4th quadrant}] \\ \\ \therefore \theta \text{ lies in } & \text{the 4th quadrant} \\ \\ \tan \theta & = -{4 \over 3} \\ {Opp \over Adj} & = {-4 \over 3} \end{align}

\begin{align} \text{By Pytha} & \text{goras theorem,} \\ \sqrt{3^2 + 4^2} & = 5 \\ \\ \sec \theta & = {1 \over \cos \theta} \\ & = {1 \over {3 \over 5}} \\ & = {5 \over 3} \end{align}