consider the diagram above.

We all know that

$\sin&space;\theta&space;=&space;\frac{opposite}{hypotenuse}&space;\&space;\&space;\cos&space;\theta&space;=&space;\frac{adjacent}{hypotenuse}&space;\&space;\&space;\&space;and&space;\&space;\&space;\tan&space;\theta&space;=&space;\frac{opposite}{adjacent}$

we will now introduce

- cosec/csc (the reciprocal of sin)

- sec (the reciprocal of cos)

- cot (the reciprocal of tan)

This means $cosec&space;\&space;\theta&space;=&space;\frac{Hypotenuse}{Opposite}&space;\&space;\&space;sec&space;\&space;\theta&space;=&space;\frac{Hypotenuse}{Adjacent}&space;\&space;\&space;cot&space;\&space;\theta&space;=&space;\frac{Adjacent}{Opposite}$

from our diagram above, $\sin&space;\theta&space;=&space;\frac{b}{c}$ , $\csc&space;\theta&space;=&space;\frac{c}{b}$       note that csc is the short form of cosec

$\cos&space;\theta&space;=&space;\frac{a}{c},&space;\&space;\&space;\sec&space;\theta&space;=&space;\frac{c}{a}&space;\\&space;\tan&space;\theta&space;=&space;\frac{b}{a},&space;\&space;\&space;\cot&space;\theta&space;=&space;\frac{a}{b}$

from the above, we realized that sin and cosec are reciprocal of each other, same with cos and sec, also tan and cot.

PROOF 1: To show that sin and cosec are reciprocal of each other. i.e     $\sin&space;\theta&space;=&space;\frac{1}{\csc&space;\theta&space;}$

Solution

from our above triangle, $\&space;sin&space;\theta&space;=&space;\frac{b}{c}&space;\&space;\&space;while\&space;\csc&space;\theta&space;=&space;\frac{c}{b}$

$\sin&space;\theta&space;=&space;\frac{1}{\csc&space;\theta&space;}$

$\frac{b}{c}&space;=&space;\frac{1}{\frac{c}{b}}&space;\\&space;\frac{b}{c}&space;=&space;1&space;\div&space;\frac{c}{b}$

at the right side of the equation, if we change divide to X, then c/b will become b/c

$\frac{b}{c}&space;=&space;1&space;\times&space;\frac{b}{c}\\&space;\frac{b}{c}&space;=&space;\frac{b}{c}\\&space;\therefore&space;\sin&space;\theta&space;=&space;\frac{1}{\csc&space;\theta&space;}&space;\&space;as&space;\&space;required.$

Note: Once the expression on the right side of the equation is the same with that of the left, then the proof is correct.

the above proof can also be done for cos and sec, tan and cot.

also if   $\sin&space;\Theta&space;=&space;\frac{1}{\csc&space;\Theta&space;},$ then

$sin&space;^{2}\Theta&space;=&space;\frac{1}{\csc^{2}&space;\Theta&space;}$

$\cos^{2}&space;\Theta&space;=&space;\frac{1}{sec^{2}\Theta}&space;\\&space;\\&space;\tan^{2}&space;\Theta&space;=&space;\frac{1}{\cot^2&space;\Theta}$

Students can verify the last three proofs on their own.

PROOF 2: To show that $\tan&space;\Theta&space;=&space;\frac{\sin&space;\Theta&space;}{\cos&space;\Theta&space;}$

Solution

We will still make use of our triangle above. where   $\sin&space;\Theta&space;=&space;\frac{b}{c},&space;\cos&space;\Theta&space;=&space;\frac{a}{c},&space;\tan&space;\Theta&space;=&space;\frac{b}{a}$

$\tan&space;\Theta&space;=&space;\frac{\sin&space;\Theta&space;}{\cos&space;\Theta&space;}$

$\frac{b}{a}&space;=&space;\frac{\frac{b}{c}}{\frac{a}{c}}&space;\\&space;\frac{b}{a}&space;=&space;\frac{b}{c}&space;\div&space;\frac{a}{c}&space;\\&space;\\&space;\frac{b}{a}&space;=&space;\frac{b}{c}&space;\times&space;\frac{c}{a}&space;\\&space;\\&space;at&space;\&space;the&space;\&space;left&space;\&space;side,&space;\&space;c&space;\&space;cancel&space;\&space;c&space;\\&space;\\&space;\frac{b}{a}&space;=&space;\frac{b}{1}&space;\times&space;\frac{1}{a}&space;\\&space;\\&space;\frac{b}{a}&space;=&space;\frac{b}{a}&space;\\&space;\\&space;\&space;since&space;\&space;left&space;\&space;side&space;\&space;of&space;\&space;the&space;\&space;equation&space;=&space;\&space;right&space;\&space;side&space;\\$

$\therefore&space;\tan&space;\Theta&space;=&space;\frac{\sin&space;\Theta&space;}{\cos&space;\Theta&space;}$

PROOF 3: Show that   $\tan&space;^2&space;\alpha&space;=&space;\frac{1}{\cot&space;^2\alpha&space;}$

Solution

NOTE: we can use β, α, γ, δ, π, ρ, σ, letters A..Z in place of θ

.$\tan&space;\alpha&space;=&space;\frac{b}{a},&space;\&space;\cot&space;\alpha&space;=&space;\frac{a}{b}$

$\tan&space;^2&space;\alpha&space;=&space;\frac{1}{\cot&space;^2\alpha&space;}$ , $\tan&space;^2&space;\alpha&space;\&space;means&space;\left&space;(&space;\tan&space;\alpha&space;\right&space;)^2&space;=&space;\left&space;(&space;\frac{b}{a}&space;\right&space;)^2,&space;\&space;while&space;\cot&space;^2\alpha&space;=&space;\left&space;(&space;\frac{a}{b}&space;\right&space;)^2$

$\left&space;(&space;\frac{b}{a}&space;\right&space;)^2&space;=&space;\frac{1}{\left&space;(&space;\frac{a}{b}\right&space;)^2}&space;\\&space;\\&space;\\&space;\left&space;(&space;\frac{b}{a}&space;\right&space;)^2&space;=&space;\left&space;(&space;\frac{b}{a}\right&space;)^2$

Since left side (LS) equals right side (RS) then $\tan&space;^2&space;\alpha&space;=&space;\frac{1}{\cot&space;^2\alpha&space;}$ .

Note: $\frac{1}{\left&space;(&space;\frac{a}{b}&space;\right&space;)^2}&space;=&space;\left&space;(&space;\frac{b}{a}&space;\right&space;)^2$ reciprocal of fractions can be obtained easily by removing the 1 and swapping the numerator and denominator.

** Notice that b initially at the denominator is now at the numerator and a is now at the denominator.

Exercise A

Show that:

(1)  $\cot&space;\Theta&space;=&space;\frac{\cos&space;\Theta&space;}{\sin&space;\Theta&space;}$

(2)   $\sec&space;^2&space;\Theta&space;=&space;\frac{1}{\cos&space;^2&space;\Theta&space;}$

(3)   $\tan^2&space;\Theta&space;=&space;\frac{\sin^2&space;\Theta&space;}{\cos^2&space;\Theta&space;}$

(4)   $\cot^2&space;\Theta&space;=&space;\frac{\cos^2&space;\Theta&space;}{\sin^2&space;\Theta&space;}$

PROOF 4: To show that Sin2θ + Cos2θ = 1

Solution

from our triangle at the top of this page, Sinθ = $\frac{b}{c}$  and Cosθ = $\frac{a}{c}$

Sin2θ + Cos2θ = 1

$\left&space;(&space;\frac{b}{c}&space;\right&space;)^{2}&space;+&space;\left&space;(&space;\frac{a}{c}&space;\right&space;)^{2}&space;=&space;1&space;\\&space;\frac{b^2}{c^2}&space;+&space;\frac{a^2}{c^2}&space;=&space;1&space;\\$

the lcm is C2

$\frac{b^2&space;+&space;a^2}{c^2}&space;=&space;1$, applying Pythagoras' theorem to our above right-angled triangle, we have C2 = a2 + b2

we will substitute, C2 for a2 + b2 in the above equation.

$\frac{c^2}{c^2}&space;=&space;1$

C2 will cancel C2 to give 1.

1 = 1.

therefore, Sin2θ + Cos2θ = 1.

From the above proof, if we divide both sides by Sin2θ, we have

$\frac{\sin&space;^2\theta&space;}{\sin&space;^2\theta&space;}&space;\&space;+&space;\frac{\cos^2\Theta&space;}{\sin&space;^2\theta&space;}&space;=&space;\frac{1}{\sin&space;^2\theta&space;}$

1 + cot2θ = cosec2θ.

Similarly, if we divide through by  Cos2θ, then

$\frac{\sin&space;^2\Theta&space;}{\cos&space;^2\Theta&space;}&space;\&space;+&space;\frac{\cos&space;^2\Theta&space;}{\cos&space;^2\Theta&space;}&space;=&space;\frac{1}{\cos&space;^2\Theta&space;}$

Tan2 θ + 1 = sec2 θ.

Take note of the above proofs, as we will use them later.

Sin2θ + Cos2θ = 1

1 + cot2θ = cosec2 θ.

Tan2 θ + 1 = sec2 θ.

EXAMPLES

1.   Given that sinθ = 4/5, find the value of cosθ and tanθ.

Solution

sinθ = 4/5, cosθ = ?

Sin2θ + Cos2θ = 1

$\left&space;(&space;\frac{4}{5}&space;\right&space;)^2&space;+&space;\cos&space;^{2}\Theta&space;=&space;1&space;\\&space;\frac{16}{25}&space;+&space;\cos&space;^{2}\Theta&space;=&space;1$

make Cos2θ the subject

$\cos&space;^{2}\Theta&space;=&space;1&space;-&space;\frac{16}{25}$

$\cos&space;^{2}\Theta&space;=&space;\frac{1}{1}&space;-&space;\frac{16}{25}$

L.C.M is 25

$\cos&space;^2\Theta&space;=&space;\frac{25&space;-&space;16}{25}&space;\\&space;\cos&space;^2\Theta&space;=&space;\frac{9}{25}&space;\\&space;\cos\Theta&space;=&space;\sqrt{\frac{9}{25}}&space;\\&space;\cos\Theta&space;=&space;\frac{3}{5}$

After getting cosθ, we can now get tanθ.

sinθ = 4/5, cosθ = 3/5, tanθ = ?

using $tan\Theta&space;=&space;\frac{sin\Theta}{cos\Theta&space;}&space;\\&space;tan\Theta&space;=&space;\frac{4/5}{3/5}&space;\\&space;tan\Theta&space;=&space;\frac{4}{5}&space;\div&space;\frac{3}{5}&space;\\&space;tan\Theta&space;=&space;\frac{4}{5}&space;\times&space;\frac{5}{3}$

5 will cancel 5

$tan\Theta&space;=&space;\frac{4}{1}&space;\times&space;\frac{1}{3}&space;\\&space;\therefore&space;tan&space;\Theta&space;=&space;\frac{4}{3}$

also we can get tanθ from Tan2 θ + 1 = sec2 θ.

sec2 θ = 1/Cos2θ

from our above calculation, cos2θ = 9/25

sec2 θ = $\frac{1}{9/25}$

therefore 25/9.

tan2 θ + 1 = sec2 θ

tan2 θ + 1 = 25/9

tan2 θ  = $\frac{25}{9}&space;-&space;1$

tan2 θ = $tan^2&space;\Theta&space;=&space;\frac{25}{9}&space;-&space;\frac{1}{1}&space;\&space;lcm&space;\&space;is&space;\&space;9&space;\\&space;\\&space;tan^2&space;\Theta&space;=&space;\frac{25&space;-&space;9}{9}&space;\\&space;\\&space;tan^2&space;\Theta&space;=&space;\frac{16}{9}&space;\\&space;\\&space;tan&space;\Theta&space;=&space;\sqrt{\frac{16}{9}}&space;\\&space;\\&space;tan&space;\Theta&space;=&space;\frac{4}{3}$

Our answer is still the same as before, so use any method you like.

try these on your own and check the answer in the next page.

1.    Given that cos α = 5/13, find the value of sin α, tan α and cosec α

2.    If tan B = 2/3, find the value of sin B, cos B and cosec B.