Trigonometric Ratios

What Is Trigonometry?

The word trigonometry comes from Greek: tri (three), gon (sides), metron (measure) — literally 'measuring three-sided figures'. At Class 10 level, it studies the relationship between the angles and sides of a right-angled triangle.

Think of it this way: if you know one acute angle and one side of a right triangle, trigonometry lets you find every other side.

Key Point: All the trigonometric ratios in this chapter are defined for an acute angle inside a right-angled triangle.

[Board Important] The hypotenuse is always the side opposite the right angle — it never changes. But 'opposite' and 'adjacent' depend on which acute angle you are looking at.

The Three Sides Relative to an Angle

For an acute angle $\theta$ in a right triangle, the three sides are named relative to $\theta$:

• Hypotenuse — the side opposite the right angle (the longest side).
• Opposite (Perpendicular) — the side directly across from $\theta$.
• Adjacent (Base) — the remaining side, next to $\theta$ (not the hypotenuse).

Key Point: If you switch attention to the other acute angle, the 'opposite' and 'adjacent' sides swap. The hypotenuse stays the same.

[Board Important] Always identify the sides freshly for the angle in the question. A common error is using the opposite side of the wrong angle.

The Six Trigonometric Ratios

For an acute angle $\theta$:

$$\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}}, \quad \cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}, \quad \tan\theta = \frac{\text{Opposite}}{\text{Adjacent}}$$

$$\csc\theta = \frac{\text{Hyp}}{\text{Opp}}, \quad \sec\theta = \frac{\text{Hyp}}{\text{Adj}}, \quad \cot\theta = \frac{\text{Adj}}{\text{Opp}}$$

A classic mnemonic: 'Pandit Badri Prasad Har Har Bole' → P/H, B/H, P/B for sin, cos, tan (Perpendicular, Base, Hypotenuse). In English, SOH-CAH-TOA (Sin = Opp/Hyp, Cos = Adj/Hyp, Tan = Opp/Adj).

Key Point: sin, cos, tan are the three basic ratios; cosec, sec, cot are their reciprocals.

[JEE/NEET Tip] Memorise SOH-CAH-TOA cold — it is the single most-used fact in all of trigonometry.

Reciprocal and Quotient Relationships

The last three ratios are simply reciprocals of the first three: $$\csc\theta = \frac{1}{\sin\theta}, \quad \sec\theta = \frac{1}{\cos\theta}, \quad \cot\theta = \frac{1}{\tan\theta}$$

Also, two quotient relations connect them: $$\tan\theta = \frac{\sin\theta}{\cos\theta}, \quad \cot\theta = \frac{\cos\theta}{\sin\theta}$$

Key Point: Pair each ratio with its reciprocal: sin↔cosec, cos↔sec, tan↔cot. And $\tan\theta = \sin\theta / \cos\theta$.

[Board Important] These relations let you find all six ratios once you know any one of them (plus the triangle), using the Pythagoras theorem to get the third side.

Solved Examples

Example 1: Basic ratios

In a right triangle, the side opposite $\theta$ is 3 and the hypotenuse is 5. Find $\sin\theta$ and $\cos\theta$.

Solution:

1. $\sin\theta = \dfrac{\text{Opp}}{\text{Hyp}} = \dfrac{3}{5}$.
2. Adjacent $= \sqrt{5^2 - 3^2} = 4$, so $\cos\theta = \dfrac{4}{5}$.

Final Answer: $\sin\theta = \tfrac35$, $\cos\theta = \tfrac45$.

Takeaway: Use Pythagoras to get the missing side.

Example 2: All six ratios

If $\tan\theta = \dfrac{4}{3}$, find all six trigonometric ratios.

Solution:

1. Opposite $= 4$, Adjacent $= 3$, so Hyp $= \sqrt{16+9} = 5$.
2. $\sin\theta = \tfrac45$, $\cos\theta = \tfrac35$, $\tan\theta = \tfrac43$.
3. $\csc\theta = \tfrac54$, $\sec\theta = \tfrac53$, $\cot\theta = \tfrac34$.

Final Answer: As above.

Takeaway: From one ratio, build the triangle, then read off all six.

Example 3: Reciprocal

If $\sin\theta = \dfrac{7}{25}$, find $\csc\theta$.

Solution:

1. $\csc\theta = \dfrac{1}{\sin\theta} = \dfrac{25}{7}$.

Final Answer: $\csc\theta = \tfrac{25}{7}$.

Takeaway: cosec is just the reciprocal of sin.

Example 4: Find cos from sin

If $\sin\theta = \dfrac{12}{13}$, find $\cos\theta$ and $\tan\theta$.

Solution:

1. Adjacent $= \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = 5$.
2. $\cos\theta = \tfrac{5}{13}$, $\tan\theta = \tfrac{12}{5}$.

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

Takeaway: $(5, 12, 13)$ is a Pythagorean triple.

Example 5: Quotient relation

If $\sin\theta = \dfrac35$ and $\cos\theta = \dfrac45$, verify $\tan\theta$.

Solution:

1. $\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{4/5} = \dfrac34$.

Final Answer: $\tan\theta = \tfrac34$.

Takeaway: $\tan\theta = \sin\theta / \cos\theta$ always holds.

Example 6: cot from a triangle

In a right triangle, adjacent to $\theta$ is 8 and opposite is 15. Find $\cot\theta$ and $\sec\theta$.

Solution:

1. $\cot\theta = \dfrac{\text{Adj}}{\text{Opp}} = \dfrac{8}{15}$.
2. Hyp $= \sqrt{64 + 225} = 17$, so $\sec\theta = \dfrac{17}{8}$.

Final Answer: $\cot\theta = \tfrac{8}{15}$, $\sec\theta = \tfrac{17}{8}$.

Takeaway: $(8, 15, 17)$ is a Pythagorean triple.

Example 7: Value of an expression

If $\tan\theta = \dfrac34$, find $\dfrac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta}$.

Solution:

1. With Opp 3, Adj 4: $\sin\theta = \tfrac35$, $\cos\theta = \tfrac45$.
2. $\dfrac{3/5 + 4/5}{3/5 - 4/5} = \dfrac{7/5}{-1/5} = -7$.

Final Answer: $-7$.

Takeaway: Substitute the actual ratio values, then simplify.

Example 8: Both acute angles

In right $\triangle ABC$ (right angle at $B$), $AB = 24$, $BC = 7$. Find $\sin A$ and $\sin C$.

Solution:

1. $AC = \sqrt{24^2 + 7^2} = 25$.
2. $\sin A = \dfrac{BC}{AC} = \dfrac{7}{25}$ (opposite to $A$ is $BC$).
3. $\sin C = \dfrac{AB}{AC} = \dfrac{24}{25}$ (opposite to $C$ is $AB$).

Final Answer: $\sin A = \tfrac{7}{25}$, $\sin C = \tfrac{24}{25}$.

Takeaway: 'Opposite' changes with the angle you choose.

Example 9: Show a ratio cannot exceed 1

Can $\sin\theta = \dfrac54$ for an acute angle?

Solution:

1. $\sin\theta = \dfrac{\text{Opp}}{\text{Hyp}}$, and the hypotenuse is the longest side, so Opp < Hyp.
2. Hence $\sin\theta < 1$ always. $\tfrac54 > 1$ is impossible.

Final Answer: No — $\sin\theta$ of an acute angle is always less than 1.

Takeaway: Both $\sin\theta$ and $\cos\theta$ lie between 0 and 1 for acute angles.

Example 10: Mixed ratio expression

If $\cos\theta = \dfrac{15}{17}$, find $\dfrac{1 - \tan^2\theta}{1 + \tan^2\theta}$.

Solution:

1. Opp $= \sqrt{17^2 - 15^2} = 8$, so $\tan\theta = \tfrac{8}{15}$.
2. $\tan^2\theta = \tfrac{64}{225}$.
3. $\dfrac{1 - 64/225}{1 + 64/225} = \dfrac{161/225}{289/225} = \dfrac{161}{289}$.

Final Answer: $\dfrac{161}{289}$.

Takeaway: Build the triangle, get tan, then substitute.