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Trigonometry is the branch of Mathematics that deals with the study of the relationship between angles and sides and lengths of triangles. Trigonometry is not a part of the NEET syllabus, but the basic idea of Trigonometry is used to solve NEET Physics as well as some Chemistry questions. Therefore, it holds significant importance for NEET aspirants. This article includes trigonometric ratios and their value at different angles, basic identities, an analysis of five previous years questions and an explanation of trigonometric concepts using previous five years NEET Physics papers.
NEET 2026 registration is ongoing, and around 22–24 lakh students are expected to apply by the extended deadline of March 11, 2026. Early reports show that over 23 lakh students have already registered, which may surpass the 2024 record of 24 lakh, though some reports suggest it could slightly drop to 21.5 lakh.

Consider a right angled triangle ABC, The angle B is a right angle,
and the longest side AC is called the hypotenuse, and the other two
sides AB and BC respectively are called adjacent and opposite sides
corresponding to angle A .
Trigonometric ratios are defined as given below-
sinA = Opposite Side /Hypotenuse = BC/ AC
cosecA = Hypotenuse/ Opposite side = AC/BC
cosA = Adjacent Side /Hypotenuse = AB/ AC
secA = Hypotenuse/Adjacent side = AC/AB
tanA = Opposite Side/ Adjacent side = BC/ AB
cotA = Adjacent side Opposite side = AB/BC
tanA = sinA /cosA
cotA = cosA /sinA
Degree | 0° | 30° | 45° | 60° | 90° | 180° | 270° | 360° |
Radian | 0 | π/6 | π/4 | π/3 | π/2 | π | 3π/2 | 2π |
sin | 0 | 1/2 | 1/√2 | √3/2 | 1 | 0 | -1 | 0 |
cos | 1 | √3/2 | 1/√2 | 1/2 | 0 | -1 | 0 | 1 |
tan | 0 | 1/√3 | 1 | √3 | ∞ | 0 | ∞ | 0 |
cot | ∞ | √3 | 1 | 1/√3 | 0 | ∞ | 0 | ∞ |
sec | 1 | 2/√3 | √2 | 2 | ∞ | -1 | ∞ | 1 |
cosec | ∞ | 2 | √2 | 2/√3 | 1 | ∞ | -1 | ∞ |
Following visual representation includes the mnemonic phrase “ASTC” to remember trigonometric ratios where they are positive.
A: All positive
S: sinθ and cosecθ are positive
T: tanθ and cotθ are positive
C: cosθ and secθ are positive

Angle | -θ | 90 - θ (Ⅰ Quadrant) | 90 + θ (Ⅱ Quadrant) | 180 - θ (Ⅲ Quadrant) | 180 + θ (Ⅳ Quadrant) |
sin | -sinθ | cosθ | cosθ | sinθ | -sinθ |
cos | cosθ | sinθ | -sinθ | -cosθ | -cosθ |
tan | -tanθ | cotθ | -cotθ | -tanθ | tanθ |

Complex formulas, identities, and rules of trigonometry are not relevant for NEET aspirants as these are neither directly asked nor indirectly used in any questions. Following the given identity and rules are important for NEET aspirants.


Apart from the above basic trigonometric relations, a few more formulas will be used while deriving equations. And understanding derivations is good for concept building and will make students aware of how a formula is obtained. Some of the trigonometric ideas used in Physics derivations are listed below
If A, B, C are the angles of a triangle and a, b, and c are corresponding opposite sides then
a/sinA = b/sinB = c/sinC

c2 = a2 + b2 - 2ab cos C
b2 = a2 + c2 - 2ac cos B
a2 = b2 + c2 - 2bc cos A

One of the important activities that help students to understand exam patterns is the analysis of the previous year's papers. Students realize important concepts and topics with their horizontal and vertical spread. After analyzing the last five years of NEET paper careers360 extracted important facts that are mentioned below. The smart study can be beneficial for NEET aspirants.
Year | Number Of Questions |
2021 | 3 |
2020 | 5 |
2019 | 4 |
2018 | 4 |
2017 | 2 |
There are an average of 2-5 questions being asked in the NEET exam every year which use trigonometry concepts.
Let's take some NEET previous year's questions or NEET PYQs to understand the application of trigonometry concepts. After analysing these questions, students can understand that mainly the value of trigonometry ratios at different angles are used to solve the problems.

Q-1 (NEET - 2021)
Find the value of the angle of emergency from the prism.
The Refractive index of the glass is
.
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Solution:
2nd Refracting Surface

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Concept Used
The trigonometric ratio at different angles
sin30° = 1/2 and sin60° = √3/2.
Q-2 (NEET - 2020)
The phase difference between displacement and acceleration of a particle in a simple harmonic motion is:
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zero
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Solution:

So the phase difference between x & a is ![]()
Concept Used
We know the property sinθ = - sin(π+θ)
Q-3 (NEET - 2019)
The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes w.r.t north is given by:
30 west
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Solution:

Boat - River Problem
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Concept Used
The value of sin30° = 1/2
Q-4 (NEET - 2018)
A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and θ for the block to remain stationary on the wedge is





Solution:
As we have learned, In Equilibrium of Concurrent Forces -
If all the forces working on a body are acting on the same point then they are said to be concurrent. Wherein

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Observing Free Body Diagram or FBD of the block
In given conditions, for Block to be a stationary
net force along the plane = 0 ![]()
Concept Used
Property: tanθ = sinθ/cosθ
Q-5 (NEET - 2017)
If θ1 and θ2 be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip θ is given by:
cot2θ = cot2θ1 + cot2θ2
tan2θ = tan2θ1 + tan2θ2
cot2θ= cot2θ1 - cot2θ2
tan2θ = tan2θ1 - tan2θ2
Solution:
If θ1 and θ2 are Apparent angles of dip Let α be the angle that one of the planes makes with the magnetic meridian.
tanθ = V/H where θ is the true angle of the dip

Concept Used
sin2θ + cos2θ = 1
cotθ = 1/tanθ
After analyzing the previous year's questions, students become aware that without mastering lengthy trigonometry formulas and concepts, having an understanding of basic concepts is sufficient for the NEET exam. You can easily learn it with the right strategy and consistent efforts.
On Question asked by student community
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Dear Student,
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