I have memorized this chart and know that a= Sine, b= Tangent, c= Secant, and d= cosine.
However, while a and d make sense intuitively, I do not understand how Secant is c and b is tangent. Can someone explain? 
[Math] Express the length a, b, c, and d in the figure in terms of the trigonometric ratios of θ.
algebra-precalculus
Related Solutions
I have marked 5 points P,Q,R,S,T in your diagram.
Consider the triangle formed by points P, S and T. Do you agree that
- Length of PS = c
- Length of ST = b
- Length of PT = 1
It is a right-angle triangle, so apply Pythagoras theorem,
square of the hypotenuse is equal to the sum of the squares of the other two sides
We get $b^2+1^2=c^2$
We know that $\sin \theta=\Large \frac{\text{opposite side}}{\text{hypotenuse}}$and $\cos\theta = \Large \frac{\text{adjacent}}{\text{hypotenuse}}$(Now, refer triangle PST and write down write $\sin\theta,\cos\theta$)
Now, simplify, substitue to express b and c only in terms of trig ratio, that is b=some tri ratio($\theta$) and c=some trig ratio($\theta$)
Similarly, for triangle PRQ Do you agree that
- Length of PQ = d
- Length of RQ = a
- Length of PR = 1
Recall that $\tan(\alpha)=\displaystyle\frac{\text{opposite}}{\text{adjacent}}=\dfrac{2}{1}$
By the Pythagorean theorem, the hypotenuse is $\sqrt{2^2+1^2}=\sqrt{5}$. Thus $$\sin(\alpha)=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{2}{\sqrt{5}}$$ and $$\cos(\alpha)=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac{1}{\sqrt{5}}$$
Best Answer
According to the written formula,
$\tan \theta = \frac {\text{opposite side}}{\text{adjacent side}} = \frac b{\text{horizontal radius}} = \frac b1 = b$
The secant can be tackled similarly.