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    Trigonometry For NEET: Concepts And Paper Analysis

    Trigonometry For NEET: Concepts And Paper Analysis

    Ramraj SainiUpdated on 13 Jun 2022, 03:22 PM IST

    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.

    Trigonometry For NEET:  Concepts And Paper Analysis
    Trigonometry For Solving NEET Questions(Image: Shutterstock)

    1654752671099

    Angle And Trigonometric Ratios

    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

    Value Of Trigonometric Ratio At Different Angle

    Degree

    30°

    45°

    60°

    90°

    180°

    270°

    360°

    Radian

    0

    π/6

    π/4

    π/3

    π/2

    π

    3π/2

    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

    Trigonometric Ratio In Different Quadrants

    1654752674231

    Trigonometric Ratios Of Allied Angle

    Angle

    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

    Graph Of Trigonometric Function

    1654752675843

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    Trigonometry Basic Identities And Rules

    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.

    1654752672518

    1654752671782

    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

    sine rule

    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

    1654752673097

    cosine rule

    c2 = a2 + b2 - 2ab cos C

    b2 = a2 + c2 - 2ac cos B

    a2 = b2 + c2 - 2bc cos A

    Angle Sum Property

    1654752676002

    Analysis Of Previous Five Years NEET Paper

    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.

    Number Of Questions In Which Trigonometry Is Used

    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.

    NEET Syllabus: Subjects & Chapters
    Select your preferred subject to view the chapters

    Previous Year's Questions To Understand The 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.

    1654752667356

    Q-1 (NEET - 2021)

    Find the value of the angle of emergency from the prism.

    The Refractive index of the glass is 1654752671934.

    1. 1654752670140

    2. 1654752670554

    3. 1654752673239

    4. 1654752668587

    Solution:

    2nd Refracting Surface

    1654752674978

    \mu _{1}\sin 30^{o}=\mu _{2}\sin \theta

    \sqrt{3}\times \frac{1}{2}=1\sin \theta

    \theta =60^{o}

    \theta =Angle\: of\: ehergence=60^{o}

    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:

    1. \mathrm{\pi \ rad}

    2. \mathrm{\frac{3\pi}{2} \ rad}

    3. \mathrm{\frac{\pi}{2} \ rad}

    4. zero

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    Solution:

    $$\begin{aligned} &\text { SHM is represented by }\ displacement= \mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}\\ &\text { Then, the velocity and acceleration after differentiation will be }\\ &\mathbf{v}=\mathbf{A} \omega \cos \omega \mathbf{t}\\ &\mathrm{a}=-\mathrm{A} \omega^{2} \sin \omega \mathrm{t}=\mathrm{A} \omega^{2} \sin (\omega \mathrm{t}+\pi) \end{aligned}

    So the phase difference between x & a is \mathrm{\pi \ rad}

    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:

    1. 30 west

    2. 0 \degree

    3. 60 \degree west

    4. 45 \degree west

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    Solution:

    1654752797333

    Boat - River Problem

    20 \sin \theta = 10 \Rightarrow \theta = 30 \degree west

    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

    1654752668733

    1. 1655113753313

    2. a = \frac{g}{sin \theta }

    3. a = \frac{g}{cosec \theta }

    4. a = g tan \theta

    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



    1654752676468

    sum vec{F_{net}}= 0 or sum vec{F_{x}}= 0,sum vec F_{y}= 0, sum vec F_{z}= 0

    Observing Free Body Diagram or FBD of the block

    In given conditions, for Block to be a stationary

    net force along the plane = 0 \Rightarrow mg\sin \theta =ma\cos \theta \Rightarrow a=g\tan \theta

    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:

    1. cot2θ = cot2θ1 + cot2θ2

    2. tan2θ = tan2θ1 + tan2θ2

    3. cot2θ= cot2θ1 - cot2θ2

    4. 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

    1654752674608

    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.

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