Tan Theta Formula (original) (raw)

Last Updated : 19 Feb, 2026

Tangent is a simple trigonometry function; it is used for the representation of the angle of a right triangle as a ratio of the lengths of the opposite side and the adjacent side.

It can also be represented as a ratio of the sine of the angle to the cosine of the angle.

Right-angled Triangle

**Main definition:\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}

or

**In terms of sine and cosine: \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}

or

**Reciprocal relation: \tan(\theta) = \frac{1(\theta)}{\cot(\theta)}

or

**Pythagorean identity: 1 + \tan^2 \theta = \sec^2 \theta

where, θ is one of the acute angle.

Other Important Formulas

• tan(θ)=sin(θ)/cos(θ)
• tan(θ)=1/cot(θ)
• tan2(x)=sec2(x)-1
• tan(-x)=-tan(x)
• tan(90o-x)=cot(x)
• tan(x+π)=tan(x)
• tan(π-x)=-tan(x)
• tan(x+y)= \frac{tan(x)+tan(y)}{1-tan(x).tan(y)}
• tan(x-y)= \frac{tan(x)-tan(y)}{1+tan(x).tan(y)}
• tan(2x)=\frac{2tan(x)}{1-tan^2(x)}
• tan(3x)=\frac{3tan(x)-tan^3(x)}{1-3tan^2(x)}
• tan(x/2)=\sqrt{\frac{1-cos(x)}{1+cos(x)}}

Tangent in Various Quadrants:

The tangent function has a period of π, which means its behavior repeats every π units. It is important to know how tangent behaves in each quadrant.

• In **Quadrant I (0° to 90°), both the sine and cosine are positive, so tan⁡(θ)>0.
• In **Quadrant II (90° to 180°), sine is positive but cosine is negative, so tan⁡(θ)<0.
• In **Quadrant III (180° to 270°), both sine and cosine are negative, so tan⁡(θ)>0.
• In **Quadrant IV (270° to 360°), sine is negative but cosine is positive, so tan⁡(θ)<0.

Sample Questions

**Question 1: Find θ for a right-angled triangle if the length of the opposite side and adjacent side w.r.t θ are 3cm and 3√3 cm respectively.

**Solution:

Given

Length of opposite side = 3 cm

Adjacent side length = 3√3 cm

From Tangent rule-

tan(θ) = Opposite side/Adjacent side

= 3/3√3

= 1/√3

tan 30° = 1/√3

tan(θ) = tan 30°

**θ=30°

**Question 2: Find tan θ for the given cot θ = 0.

**Solution:

Given

cot θ = 0

the relation between tanθ and cotθ is inverse i.e.,

tan θ = 1/cot θ

=1/0

**tan θ = ∞

**Question 3: Find tan θ from the given sin θ = 1/2 and cos θ = √3/2.

**Solution:

Given,

sin θ = 1/2

cos θ = √3/2

tan θ = sin θ / cos θ

= (1/2) / (√3/2)

= (1×2) / (2×√3)

**tan θ = 1/√3

**Question 4: Find tan x from the given sec x=2/5.

**Solution:

Given

sec x = 2/5

we know that sec2 x - tan2 x = 1

From that tan2 x = sec2 x - 1

= (2/5)2-1

= (4/25)-1

= (4-25)/25

tan2(x) = -21/25

tan(x) = √(-21/25)

**tan(x) = √(-21)/5

**Question 5: Find the result of tan(60°+45°).

**Solution:

Given

A=60°

B=45°

We know that tan(A+B)= \frac{tan(A)+tan(B)}{1-tan(A).tan(B)}

tan(60°+45°)=\frac{tan(60°)+tan(45°)}{1-tan(60°).tan(45°)}

=(√3+1)/(1-√3×1)

=(√3+1)/(1-√3)

**tan(60°+45°)=-3.732

**Question 6: Calculate tan(x) where x=π-45°

**Solution:

Given

tan(x)=tan(π-45°)

we know that tan(π-θ)=-tan(θ)

tan(π-45°)=-tan(45°)

tan(45°)=1

**tan(π-45°)=-1

Practice Problem Based on Tan Theta Formula

**Question 1. Find tan⁡(15°) using the tangent subtraction formula.

**Question 2. In a right-angled triangle, if the opposite side is 7 cm and the adjacent side is 24 cm, find the angle θ using the tangent function.

**Question 3. If cot⁡(θ)=2, find tan⁡(θ).

**Question 4. Given that sin⁡(θ)=3/5​ and cos⁡(θ)=4/5, find tan⁡(θ).

**Answer:-

1. \tan(15^\circ) \approx 0.2679
2. \tan(\theta) = \frac{7}{24}\approx 16.26^\circ
3. \tan(\theta)=\frac{1}{2}
4. \tan(\theta)=\frac{3}{4}