Trigonometry • Topic 1 of 3

Trigonometric Ratios and Identities

What is Trigonometry? The word trigonometry comes from Greek words meaning "triangle measuring." It is a branch of mathematics that studies the relationship between the side lengths and angles of triangles. Imagine you are standing near a tall mobile tower and looking up at its top. If you know your distance from the base of the tower and the angle at which you look up, trigonometry helps you find the height of the tower without physically climbing up to measure it!

Trigonometric Ratios. In a right-angled triangle, we name the sides relative to a specific acute angle, which we call $\theta$. The sides are:

  • Hypotenuse: the longest side, directly opposite the $90^{\circ}$ right angle.
  • Opposite side: the side directly facing our chosen angle $\theta$.
  • Adjacent side: the side that runs alongside angle $\theta$ and meets the right angle.

The six fundamental trigonometric ratios are simple fractions built by dividing one side by another:

RatioSymbolDefinitionReciprocal
Sine$\sin\theta$$\dfrac{\text{opposite}}{\text{hypotenuse}}$$\operatorname{cosec}\theta=\dfrac{1}{\sin\theta}$
Cosine$\cos\theta$$\dfrac{\text{adjacent}}{\text{hypotenuse}}$$\sec\theta=\dfrac{1}{\cos\theta}$
Tangent$\tan\theta$$\dfrac{\text{opposite}}{\text{adjacent}}$$\cot\theta=\dfrac{1}{\tan\theta}$
Cosecant$\operatorname{cosec}\theta$$\dfrac{\text{hypotenuse}}{\text{opposite}}$$\sin\theta$
Secant$\sec\theta$$\dfrac{\text{hypotenuse}}{\text{adjacent}}$$\cos\theta$
Cotangent$\cot\theta$$\dfrac{\text{adjacent}}{\text{opposite}}$$\tan\theta$

Quotient relations. Tangent and cotangent can be written purely in terms of sine and cosine, which makes most proofs easier:

  • $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$
  • $\cot\theta=\dfrac{\cos\theta}{\sin\theta}$

Reading a ratio as side lengths. A ratio such as $\sin\theta=\tfrac{5}{13}$ does not force the opposite side to equal $5$ and the hypotenuse to equal $13$. It only fixes their ratio. So we may take opposite $=5k$ and hypotenuse $=13k$ for some positive constant $k$, find the third side by the Pythagoras theorem, and the $k$ cancels in every ratio we compute. This single idea answers almost every "given one ratio, find the others" question.

Why ratios depend only on the angle. If you draw two right triangles with the same acute angle $\theta$, they are similar (AA similarity). Corresponding sides are in the same proportion, so $\tfrac{\text{opp}}{\text{hyp}}$ is identical in both. That is why $\sin\theta$, $\cos\theta$, $\tan\theta$ depend on the angle alone and not on the size of the triangle.

Range of values for an acute angle. Since the hypotenuse is the longest side, the opposite and adjacent sides are each shorter than it. Hence for $0^{\circ}<\theta<90^{\circ}$ we always have $0<\sin\theta<1$ and $0<\cos\theta<1$, while $\tan\theta$ can be any positive number. A "sine" or "cosine" answer bigger than $1$ is a sure sign of an arithmetic slip.

Trigonometric Identities. A trigonometric identity is an equation in these ratios that stays true for every permissible angle. The three Pythagorean identities are the backbone of the chapter. Take a right triangle with opposite $=a$, adjacent $=b$, hypotenuse $=c$, so $a^{2}+b^{2}=c^{2}$.

  • Divide by $c^{2}$: $\dfrac{a^{2}}{c^{2}}+\dfrac{b^{2}}{c^{2}}=1\;\Rightarrow\;\sin^{2}\theta+\cos^{2}\theta=1.$
  • Divide by $b^{2}$: $\dfrac{a^{2}}{b^{2}}+1=\dfrac{c^{2}}{b^{2}}\;\Rightarrow\;\tan^{2}\theta+1=\sec^{2}\theta.$
  • Divide by $a^{2}$: $1+\dfrac{b^{2}}{a^{2}}=\dfrac{c^{2}}{a^{2}}\;\Rightarrow\;1+\cot^{2}\theta=\operatorname{cosec}^{2}\theta.$

Each identity has handy rearrangements you should memorise: $\sin^{2}\theta=1-\cos^{2}\theta$, $\sec^{2}\theta-\tan^{2}\theta=1$, and $\operatorname{cosec}^{2}\theta-\cot^{2}\theta=1$. These rearranged forms are what you actually substitute while proving harder identities.

How to prove a harder identity. Pick the more complicated side (usually the LHS) and transform it until it equals the other side. The reliable toolkit is: (i) convert every ratio to $\sin$ and $\cos$; (ii) take an LCM to combine fractions; (iii) use $\sin^{2}\theta+\cos^{2}\theta=1$ to simplify; (iv) when a surd or a $1\pm\cos\theta$ term appears, rationalise by multiplying numerator and denominator by the conjugate. Never move terms across the equality sign as if solving an equation — an identity is proved by transforming one side, not by cross-multiplying both.

Common mistakes. Writing $\sin^{2}\theta$ to mean $\sin(\theta^{2})$ — it means $(\sin\theta)^{2}$. Forgetting that $\sin^{2}\theta+\cos^{2}\theta=1$ holds only when both ratios use the same angle. Treating $\tfrac{1}{\sin\theta+\cos\theta}$ as $\operatorname{cosec}\theta+\sec\theta$ — reciprocals do not distribute over a sum. And cancelling a side length instead of cancelling the common factor $k$.

Right triangle with sides named opposite, adjacent and hypotenuse relative to angle theta Naming the sides relative to angle θ θ C B A Adjacent (base) Opposite Hypotenuse The three primary trigonometric ratios as side fractions The three primary ratios sin θ opposite / hypotenuse cos θ adjacent / hypotenuse tan θ opposite / adjacent
Solved Examples
1
Worked Example
In right triangle $ABC$, right-angled at $B$, the side $AB = 3$ cm and $BC = 4$ cm. Find $\sin A$ and $\cos A$.
Solution
  1. Step 1: Find the hypotenuse $AC$ by the Pythagoras theorem: $AC^{2}=AB^{2}+BC^{2}=3^{2}+4^{2}=9+16=25$.
  2. Step 2: So $AC=\sqrt{25}=5$ cm.
  3. Step 3: For angle $A$, the opposite side is $BC=4$ and the adjacent side is $AB=3$, with hypotenuse $AC=5$.
  4. Step 4: $\sin A=\dfrac{\text{opp}}{\text{hyp}}=\dfrac{4}{5}$ and $\cos A=\dfrac{\text{adj}}{\text{hyp}}=\dfrac{3}{5}$.

Answer: $\sin A=\tfrac{4}{5},\ \cos A=\tfrac{3}{5}$.

2
Worked Example
If $\sin\theta=\dfrac{5}{13}$, find $\cos\theta$ and $\tan\theta$.
Solution
  1. Step 1: $\sin\theta=\dfrac{\text{opp}}{\text{hyp}}=\dfrac{5}{13}$, so take opposite $=5$ and hypotenuse $=13$.
  2. Step 2: Adjacent $=\sqrt{13^{2}-5^{2}}=\sqrt{169-25}=\sqrt{144}=12$.
  3. Step 3: $\cos\theta=\dfrac{\text{adj}}{\text{hyp}}=\dfrac{12}{13}$.
  4. Step 4: $\tan\theta=\dfrac{\text{opp}}{\text{adj}}=\dfrac{5}{12}$.

Answer: $\cos\theta=\tfrac{12}{13},\ \tan\theta=\tfrac{5}{12}$.

3
Worked Example
Given $\tan\theta=\dfrac{4}{3}$, find all the remaining five trigonometric ratios.
Solution
  1. Step 1: $\tan\theta=\dfrac{\text{opp}}{\text{adj}}=\dfrac{4}{3}$, so take opposite $=4$, adjacent $=3$.
  2. Step 2: Hypotenuse $=\sqrt{4^{2}+3^{2}}=\sqrt{16+9}=\sqrt{25}=5$.
  3. Step 3: $\sin\theta=\dfrac{4}{5}$ and $\cos\theta=\dfrac{3}{5}$.
  4. Step 4: The reciprocals are $\operatorname{cosec}\theta=\dfrac{5}{4}$, $\sec\theta=\dfrac{5}{3}$, $\cot\theta=\dfrac{3}{4}$.

Answer: $\sin\theta=\tfrac45,\cos\theta=\tfrac35,\operatorname{cosec}\theta=\tfrac54,\sec\theta=\tfrac53,\cot\theta=\tfrac34$.

4
Worked Example
If $15\cot A = 8$, find $\sin A$ and $\sec A$.
Solution
  1. Step 1: $\cot A=\dfrac{8}{15}=\dfrac{\text{adj}}{\text{opp}}$, so take adjacent $=8$, opposite $=15$.
  2. Step 2: Hypotenuse $=\sqrt{8^{2}+15^{2}}=\sqrt{64+225}=\sqrt{289}=17$.
  3. Step 3: $\sin A=\dfrac{\text{opp}}{\text{hyp}}=\dfrac{15}{17}$.
  4. Step 4: $\sec A=\dfrac{\text{hyp}}{\text{adj}}=\dfrac{17}{8}$.

Answer: $\sin A=\tfrac{15}{17},\ \sec A=\tfrac{17}{8}$.

5
Worked Example
If $\cos\theta=\dfrac{7}{25}$, evaluate $\dfrac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}$.
Solution
  1. Step 1: $\cos\theta=\dfrac{7}{25}$ gives adjacent $=7$, hypotenuse $=25$.
  2. Step 2: Opposite $=\sqrt{25^{2}-7^{2}}=\sqrt{625-49}=\sqrt{576}=24$, so $\sin\theta=\dfrac{24}{25}$.
  3. Step 3: Numerator $=\dfrac{24}{25}-\dfrac{7}{25}=\dfrac{17}{25}$; denominator $=\dfrac{24}{25}+\dfrac{7}{25}=\dfrac{31}{25}$.
  4. Step 4: The quotient $=\dfrac{17/25}{31/25}=\dfrac{17}{31}$.

Answer: $\dfrac{17}{31}$.

6
Worked Example
Simplify $(1+\tan^{2}\theta)\cos^{2}\theta$.
Solution
  1. Step 1: Use the identity $1+\tan^{2}\theta=\sec^{2}\theta$.
  2. Step 2: The expression becomes $\sec^{2}\theta\cdot\cos^{2}\theta$.
  3. Step 3: Since $\sec\theta=\dfrac{1}{\cos\theta}$, we have $\sec^{2}\theta=\dfrac{1}{\cos^{2}\theta}$.
  4. Step 4: So $\dfrac{1}{\cos^{2}\theta}\cdot\cos^{2}\theta=1$.

Answer: $1$.

7
Worked Example
Prove that $\dfrac{1}{1+\sin\theta}+\dfrac{1}{1-\sin\theta}=2\sec^{2}\theta$.
Solution
  1. Step 1: Take the LHS and add the fractions over the common denominator $(1+\sin\theta)(1-\sin\theta)$.
  2. Step 2: LHS $=\dfrac{(1-\sin\theta)+(1+\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}=\dfrac{2}{1-\sin^{2}\theta}$.
  3. Step 3: Use $1-\sin^{2}\theta=\cos^{2}\theta$: LHS $=\dfrac{2}{\cos^{2}\theta}$.
  4. Step 4: Since $\dfrac{1}{\cos^{2}\theta}=\sec^{2}\theta$, LHS $=2\sec^{2}\theta=$ RHS.

Answer: Identity proved.

8
Worked Example
Prove that $\dfrac{\sin\theta}{1+\cos\theta}+\dfrac{1+\cos\theta}{\sin\theta}=2\operatorname{cosec}\theta$.
Solution
  1. Step 1: Add the two fractions of the LHS over the common denominator $\sin\theta(1+\cos\theta)$.
  2. Step 2: Numerator $=\sin^{2}\theta+(1+\cos\theta)^{2}=\sin^{2}\theta+1+2\cos\theta+\cos^{2}\theta$.
  3. Step 3: Use $\sin^{2}\theta+\cos^{2}\theta=1$, so numerator $=1+1+2\cos\theta=2+2\cos\theta=2(1+\cos\theta)$.
  4. Step 4: LHS $=\dfrac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)}=\dfrac{2}{\sin\theta}=2\operatorname{cosec}\theta=$ RHS.

Answer: Identity proved.

9
Worked Example
Prove that $\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$ (for acute $A$).
Solution
  1. Step 1: Rationalise inside the root by multiplying numerator and denominator by $(1+\sin A)$.
  2. Step 2: $\dfrac{1+\sin A}{1-\sin A}\cdot\dfrac{1+\sin A}{1+\sin A}=\dfrac{(1+\sin A)^{2}}{1-\sin^{2}A}=\dfrac{(1+\sin A)^{2}}{\cos^{2}A}$.
  3. Step 3: Take the square root (all quantities positive for acute $A$): $\dfrac{1+\sin A}{\cos A}$.
  4. Step 4: Split: $\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}=\sec A+\tan A=$ RHS.

Answer: Identity proved.

10
Worked Example
If $\sin\theta+\cos\theta=\sqrt{2}\cos\theta$, show that $\tan\theta=\sqrt{2}-1$.
Solution
  1. Step 1: Divide every term by $\cos\theta$ (valid since $\cos\theta\ne0$).
  2. Step 2: $\dfrac{\sin\theta}{\cos\theta}+1=\sqrt{2}$, i.e. $\tan\theta+1=\sqrt{2}$.
  3. Step 3: Rearrange: $\tan\theta=\sqrt{2}-1$.

Answer: $\tan\theta=\sqrt{2}-1$.

11
Worked Example
If $\sec\theta+\tan\theta=p$, express $\sin\theta$ in terms of $p$.
Solution
  1. Step 1: Use the identity $\sec^{2}\theta-\tan^{2}\theta=1$, i.e. $(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1$.
  2. Step 2: Since $\sec\theta+\tan\theta=p$, we get $\sec\theta-\tan\theta=\dfrac{1}{p}$.
  3. Step 3: Subtract the two equations: $2\tan\theta=p-\dfrac{1}{p}=\dfrac{p^{2}-1}{p}$; add them: $2\sec\theta=\dfrac{p^{2}+1}{p}$.
  4. Step 4: Then $\sin\theta=\dfrac{\tan\theta}{\sec\theta}=\dfrac{(p^{2}-1)/2p}{(p^{2}+1)/2p}=\dfrac{p^{2}-1}{p^{2}+1}$.

Answer: $\sin\theta=\dfrac{p^{2}-1}{p^{2}+1}$.

12
Worked Example
Prove that $(\sin\theta+\operatorname{cosec}\theta)^{2}+(\cos\theta+\sec\theta)^{2}=7+\tan^{2}\theta+\cot^{2}\theta$.
Solution
  1. Step 1: Expand both squares: $\sin^{2}\theta+2\sin\theta\operatorname{cosec}\theta+\operatorname{cosec}^{2}\theta+\cos^{2}\theta+2\cos\theta\sec\theta+\sec^{2}\theta$.
  2. Step 2: Each cross term is $2$ because $\sin\theta\operatorname{cosec}\theta=1$ and $\cos\theta\sec\theta=1$; also $\sin^{2}\theta+\cos^{2}\theta=1$.
  3. Step 3: So far: $1+2+2+\operatorname{cosec}^{2}\theta+\sec^{2}\theta=5+\operatorname{cosec}^{2}\theta+\sec^{2}\theta$.
  4. Step 4: Use $\operatorname{cosec}^{2}\theta=1+\cot^{2}\theta$ and $\sec^{2}\theta=1+\tan^{2}\theta$: $5+(1+\cot^{2}\theta)+(1+\tan^{2}\theta)=7+\tan^{2}\theta+\cot^{2}\theta=$ RHS.

Answer: Identity proved.

Key Points

  • Trigonometry links the values of interior acute angles to the ratio of side lengths in right triangles.
  • The three primary ratios are sin (Opp/Hyp), cos (Adj/Hyp), and tan (Opp/Adj).
  • cosec, sec, and cot are the direct multiplicative reciprocals of sin, cos, and tan.
  • The square identity \(\sin^{2}\theta\) + \(\cos^{2}\theta\) = 1 is derived directly from the Pythagoras theorem.
  • Tangent can also be expressed as a quotient: \(\tan \theta\) = \(\sin \theta\) / \(\cos \theta\).
  • When one ratio is given, set the two named sides as multiples of $k$, find the third side by Pythagoras, and the $k$ cancels in every ratio.
  • For an acute angle, $0<\sin\theta<1$ and $0<\cos\theta<1$ — a sine or cosine above $1$ signals an error.
  • Prove identities by transforming ONE side (usually convert to $\sin$ and $\cos$, take an LCM, then rationalise); never cross-multiply both sides.
  • Useful rearrangements: $\sin^{2}\theta=1-\cos^{2}\theta$, $\sec^{2}\theta-\tan^{2}\theta=1$, $\operatorname{cosec}^{2}\theta-\cot^{2}\theta=1$.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.$\sin\theta=$
Explanation: Sine $=$ opposite over hypotenuse.
Q2.$\sin^2\theta+\cos^2\theta=$
Explanation: Fundamental identity.
Q3.$\tan\theta=$
Explanation: $\tan=\sin/\cos$.
Q4.$1+\tan^2\theta=$
Explanation: Pythagorean identity.
Practice more MCQs → 📝 Assignment · 35 marks