Quantitative Aptitude

Trigonometry Guide & Practice

Master trigonometric ratios, identities, height and distance problems with solved examples and free SSC CGL/Defence exam mock tests. Explore dynamic solver blueprints, master fundamental equations, examine step-by-step solved examples, and practice with real exam-grade mock test sets.

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Trigonometry - Set 5 Practice Test

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Quantitative Aptitude

Trigonometry - Set 4 Practice Test

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15 Qs
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Quantitative Aptitude

Trigonometry - Set 5 Practice Test

Jun 2026
15 Qs
22 min
Hard
Quantitative Aptitude

Trigonometry - Set 3 Practice Test

Jun 2026
14 Qs
21 min
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Quantitative Aptitude

Trigonometry - Set 2 Practice Test

Jun 2026
15 Qs
22 min
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Quantitative Aptitude

Trigonometry - Set 1 Practice Test

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13 Qs
19 min
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Quantitative Aptitude

Trigonometry - Set 5 Practice Test

Jun 2026
12 Qs
18 min
Easy
Quantitative Aptitude

Trigonometry - Set 4 Practice Test

Jun 2026
13 Qs
19 min
Easy
Quantitative Aptitude

Trigonometry - Set 3 Practice Test

Jun 2026
15 Qs
22 min
Easy
Quantitative Aptitude

Trigonometry - Set 2 Practice Test

Jun 2026
15 Qs
22 min
Easy
Quantitative Aptitude

Trigonometry - Set 1 Practice Test

Jun 2026
15 Qs
22 min
Easy

1. Fundamentals & Definitions

Angles and Measurements

  • Angle: The measure of rotation of a given ray about its initial point.
    • Vertex: The point of rotation.
    • Initial Side: The original position of the ray before rotation.
    • Terminal Side: The final position of the ray after rotation.
    • Positive Angle: Angle generated when the rotation is anticlockwise.
    • Negative Angle: Angle generated when the rotation is clockwise.
  • Systems of Angle Measurement:
    • Sexagesimal System (Degree Measure):
      • If a rotation from the initial side to the terminal side is 1360th\frac{1}{360}^{\text{th}} of a complete revolution, the angle is defined as one degree (1∘1^\circ).
      • A degree is divided into 6060 minutes (1∘=60′1^\circ = 60').
      • A minute is divided into 6060 seconds (1′=60′′1' = 60'').
    • Circular System (Radian Measure):
      • An angle subtended at the center of a unit circle (radius r=1r = 1) by an arc of length 11 unit is defined as 11 radian (1 rad1\text{ rad}).
      • In a circle of radius rr, an arc of length ll subtends an angle θ\theta (in radians) at the center: θ=lr  ⟹  l=rθ\theta = \frac{l}{r} \implies l = r\theta
    • Relation between Degrees and Radians:
      • π radians=180∘\pi \text{ radians} = 180^\circ
      • 1 radian=180∘π≈57∘16′22′′1 \text{ radian} = \frac{180^\circ}{\pi} \approx 57^\circ 16' 22''
      • 1∘=π180 radians≈0.01745 radians1^\circ = \frac{\pi}{180} \text{ radians} \approx 0.01745 \text{ radians}
      • Conversion Formulas: Radian measure=π180×Degree measure\text{Radian measure} = \frac{\pi}{180} \times \text{Degree measure} Degree measure=180π×Radian measure\text{Degree measure} = \frac{180}{\pi} \times \text{Radian measure}

Trigonometric Ratios (Right-Angled Triangle)

For a right-angled triangle ABCABC, right-angled at BB (∠B=90∘\angle B = 90^\circ) with acute angle ∠A=θ\angle A = \theta:

  • Base (Adjacent Side, bb): The side adjacent to angle θ\theta (ABAB).
  • Perpendicular (Opposite Side, pp): The side opposite to angle θ\theta (BCBC).
  • Hypotenuse (hh): The side opposite to the right angle (ACAC).
  • Definitions:
    • sin⁡θ=PerpendicularHypotenuse=ph\sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{p}{h}
    • cos⁡θ=BaseHypotenuse=bh\cos\theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{b}{h}
    • tan⁡θ=PerpendicularBase=pb=sin⁡θcos⁡θ\tan\theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{p}{b} = \frac{\sin\theta}{\cos\theta}
    • csc⁡θ=HypotenusePerpendicular=hp=1sin⁡θ\csc\theta = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{h}{p} = \frac{1}{\sin\theta}
    • sec⁡θ=HypotenuseBase=hb=1cos⁡θ\sec\theta = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{h}{b} = \frac{1}{\cos\theta}
    • cot⁡θ=BasePerpendicular=bp=1tan⁡θ\cot\theta = \frac{\text{Base}}{\text{Perpendicular}} = \frac{b}{p} = \frac{1}{\tan\theta}

Unit Circle Trigonometry

Let P(x,y)P(x, y) be any point on a unit circle centered at the origin O(0,0)O(0, 0) such that the ray OPOP makes an angle θ\theta with the positive x-axis.

  • cos⁡θ=x\cos\theta = x
  • sin⁡θ=y\sin\theta = y
  • tan⁡θ=yx(x≠0)\tan\theta = \frac{y}{x} \quad (x \neq 0)
  • Quadrantal Angles: Angles that are integral multiples of π2\frac{\pi}{2} (0,π2,π,3π2,2π0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, etc.). For these, the terminal side coincides with a coordinate axis.

Height & Distance Terminology

  • Line of Sight: The straight line path from the observer's eye to the object being viewed.
  • Angle of Elevation: The angle between the line of sight and the horizontal plane when the object is above the horizontal plane.
  • Angle of Depression: The angle between the line of sight and the horizontal plane when the object is below the horizontal plane.

2. Core Concepts & Formulas

Trigonometric Values of Standard Angles

Angle (Degrees)Angle (Radians)sin⁡θ\sin\thetacos⁡θ\cos\thetatan⁡θ\tan\thetacsc⁡θ\csc\thetasec⁡θ\sec\thetacot⁡θ\cot\theta
0∘0^\circ00001100Undefined11Undefined
30∘30^\circπ6\frac{\pi}{6}12\frac{1}{2}32\frac{\sqrt{3}}{2}13\frac{1}{\sqrt{3}}2223\frac{2}{\sqrt{3}}3\sqrt{3}
45∘45^\circπ4\frac{\pi}{4}12\frac{1}{\sqrt{2}}12\frac{1}{\sqrt{2}}112\sqrt{2}2\sqrt{2}11
60∘60^\circπ3\frac{\pi}{3}32\frac{\sqrt{3}}{2}12\frac{1}{2}3\sqrt{3}23\frac{2}{\sqrt{3}}2213\frac{1}{\sqrt{3}}
90∘90^\circπ2\frac{\pi}{2}1100Undefined11Undefined00
180∘180^\circπ\pi00−1-100Undefined−1-1Undefined
270∘270^\circ3π2\frac{3\pi}{2}−1-100Undefined−1-1Undefined00
360∘360^\circ2π2\pi001100Undefined11Undefined

Special Competitive Exam Angles

Angle (Degrees)sin⁡θ\sin\thetacos⁡θ\cos\thetatan⁡θ\tan\thetacot⁡θ\cot\theta
15∘15^\circ (π12\frac{\pi}{12})3−122\frac{\sqrt{3}-1}{2\sqrt{2}}3+122\frac{\sqrt{3}+1}{2\sqrt{2}}2−32-\sqrt{3}2+32+\sqrt{3}
75∘75^\circ (5π12\frac{5\pi}{12})3+122\frac{\sqrt{3}+1}{2\sqrt{2}}3−122\frac{\sqrt{3}-1}{2\sqrt{2}}2+32+\sqrt{3}2−32-\sqrt{3}
18∘18^\circ (π10\frac{\pi}{10})5−14\frac{\sqrt{5}-1}{4}10+254\frac{\sqrt{10+2\sqrt{5}}}{4}1−25\sqrt{1 - \frac{2}{\sqrt{5}}}5+25\sqrt{5+2\sqrt{5}}
36∘36^\circ (π5\frac{\pi}{5})10−254\frac{\sqrt{10-2\sqrt{5}}}{4}5+14\frac{\sqrt{5}+1}{4}5−25\sqrt{5-2\sqrt{5}}1+25\sqrt{1 + \frac{2}{\sqrt{5}}}
22.5∘22.5^\circ (π8\frac{\pi}{8})2−22\frac{\sqrt{2-\sqrt{2}}}{2}2+22\frac{\sqrt{2+\sqrt{2}}}{2}2−1\sqrt{2}-12+1\sqrt{2}+1

Behavior & Signs of Trigonometric Functions

  • ASTC Quadrant Rule:
    • Quadrant I (0<θ<90∘0 < \theta < 90^\circ): All functions are positive.
    • Quadrant II (90∘<θ<180∘90^\circ < \theta < 180^\circ): Only Sine (and Cosecant) are positive.
    • Quadrant III (180∘<θ<270∘180^\circ < \theta < 270^\circ): Only Tangent (and Cotangent) are positive.
    • Quadrant IV (270∘<θ<360∘270^\circ < \theta < 360^\circ): Only Cosine (and Secant) are positive.
Quadrantsin⁡x\sin x / csc⁡x\csc xcos⁡x\cos x / sec⁡x\sec xtan⁡x\tan x / cot⁡x\cot x
I++++++
II++−-−-
III−-−-++
IV−-++−-
  • Ranges of Trigonometric Functions:
    • −1≤sin⁡θ≤1-1 \le \sin\theta \le 1
    • −1≤cos⁡θ≤1-1 \le \cos\theta \le 1
    • sec⁡θ≥1\sec\theta \ge 1 or sec⁡θ≤−1\sec\theta \le -1
    • csc⁡θ≥1\csc\theta \ge 1 or csc⁡θ≤−1\csc\theta \le -1
    • −∞<tan⁡θ<∞-\infty < \tan\theta < \infty
    • −∞<cot⁡θ<∞-\infty < \cot\theta < \infty

Fundamental Trigonometric Identities

  • Pythagorean Identities:
    1. sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1
    2. 1+tan⁡2θ=sec⁡2θ  ⟹  sec⁡2θ−tan⁡2θ=1  ⟹  (sec⁡θ−tan⁡θ)=1sec⁡θ+tan⁡θ1 + \tan^2\theta = \sec^2\theta \implies \sec^2\theta - \tan^2\theta = 1 \implies (\sec\theta - \tan\theta) = \frac{1}{\sec\theta + \tan\theta}
    3. 1+cot⁡2θ=csc⁡2θ  ⟹  csc⁡2θ−cot⁡2θ=1  ⟹  (csc⁡θ−cot⁡θ)=1csc⁡θ+cot⁡θ1 + \cot^2\theta = \csc^2\theta \implies \csc^2\theta - \cot^2\theta = 1 \implies (\csc\theta - \cot\theta) = \frac{1}{\csc\theta + \cot\theta}

Even-Odd (Negative Angle) Identities

  • sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta
  • cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta
  • tan⁡(−θ)=−tan⁡θ\tan(-\theta) = -\tan\theta
  • csc⁡(−θ)=−csc⁡θ\csc(-\theta) = -\csc\theta
  • sec⁡(−θ)=sec⁡θ\sec(-\theta) = \sec\theta
  • cot⁡(−θ)=−cot⁡θ\cot(-\theta) = -\cot\theta

Allied and Complementary Angle Formulas

Function−θ-\theta90∘−θ90^\circ - \theta90∘+θ90^\circ + \theta180∘−θ180^\circ - \theta180∘+θ180^\circ + \theta270∘−θ270^\circ - \theta270∘+θ270^\circ + \theta360∘−θ360^\circ - \theta
sin⁡\sin−sin⁡θ-\sin\thetacos⁡θ\cos\thetacos⁡θ\cos\thetasin⁡θ\sin\theta−sin⁡θ-\sin\theta−cos⁡θ-\cos\theta−cos⁡θ-\cos\theta−sin⁡θ-\sin\theta
cos⁡\coscos⁡θ\cos\thetasin⁡θ\sin\theta−sin⁡θ-\sin\theta−cos⁡θ-\cos\theta−cos⁡θ-\cos\theta−sin⁡θ-\sin\thetasin⁡θ\sin\thetacos⁡θ\cos\theta
tan⁡\tan−tan⁡θ-\tan\thetacot⁡θ\cot\theta−cot⁡θ-\cot\theta−tan⁡θ-\tan\thetatan⁡θ\tan\thetacot⁡θ\cot\theta−cot⁡θ-\cot\theta−tan⁡θ-\tan\theta
csc⁡\csc−csc⁡θ-\csc\thetasec⁡θ\sec\thetasec⁡θ\sec\thetacsc⁡θ\csc\theta−csc⁡θ-\csc\theta−sec⁡θ-\sec\theta−sec⁡θ-\sec\theta−csc⁡θ-\csc\theta
sec⁡\secsec⁡θ\sec\thetacsc⁡θ\csc\theta−csc⁡θ-\csc\theta−sec⁡θ-\sec\theta−sec⁡θ-\sec\theta−csc⁡θ-\csc\thetacsc⁡θ\csc\thetasec⁡θ\sec\theta
cot⁡\cot−cot⁡θ-\cot\thetatan⁡θ\tan\theta−tan⁡θ-\tan\theta−cot⁡θ-\cot\thetacot⁡θ\cot\thetatan⁡θ\tan\theta−tan⁡θ-\tan\theta−cot⁡θ-\cot\theta

Compound Angle Formulas

  • sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B
  • cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B
  • tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}
  • cot⁡(A±B)=cot⁡Acot⁡B∓1cot⁡B±cot⁡A\cot(A \pm B) = \frac{\cot A \cot B \mp 1}{\cot B \pm \cot A}

Double Angle Formulas

  • sin⁡2θ=2sin⁡θcos⁡θ=2tan⁡θ1+tan⁡2θ\sin 2\theta = 2\sin\theta\cos\theta = \frac{2\tan\theta}{1+\tan^2\theta}
  • cos⁡2θ=cos⁡2θ−sin⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ=1−tan⁡2θ1+tan⁡2θ\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}
    • 1−cos⁡2θ=2sin⁡2θ  ⟹  sin⁡2θ=1−cos⁡2θ21 - \cos 2\theta = 2\sin^2\theta \implies \sin^2\theta = \frac{1-\cos 2\theta}{2}
    • 1+cos⁡2θ=2cos⁡2θ  ⟹  cos⁡2θ=1+cos⁡2θ21 + \cos 2\theta = 2\cos^2\theta \implies \cos^2\theta = \frac{1+\cos 2\theta}{2}
  • tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta}

Triple Angle Formulas

  • sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin\theta - 4\sin^3\theta
  • cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta
  • tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1-3\tan^2\theta}

Half-Angle Formulas

  • sin⁡θ2=±1−cos⁡θ2\sin\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{2}}
  • cos⁡θ2=±1+cos⁡θ2\cos\frac{\theta}{2} = \pm\sqrt{\frac{1+\cos\theta}{2}}
  • tan⁡θ2=±1−cos⁡θ1+cos⁡θ=1−cos⁡θsin⁡θ=sin⁡θ1+cos⁡θ\tan\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta}

Sum, Difference & Product Conversion Formulas

  • Product to Sum/Difference:
    • 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A \cos B = \sin(A + B) + \sin(A - B)
    • 2cos⁡Asin⁡B=sin⁡(A+B)−sin⁡(A−B)2\cos A \sin B = \sin(A + B) - \sin(A - B)
    • 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A \cos B = \cos(A + B) + \cos(A - B)
    • 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A \sin B = \cos(A - B) - \cos(A + B)
  • Sum/Difference to Product:
    • sin⁡x+sin⁡y=2sin⁡(x+y2)cos⁡(x−y2)\sin x + \sin y = 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)
    • sin⁡x−sin⁡y=2cos⁡(x+y2)sin⁡(x−y2)\sin x - \sin y = 2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)
    • cos⁡x+cos⁡y=2cos⁡(x+y2)cos⁡(x−y2)\cos x + \cos y = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)
    • cos⁡x−cos⁡y=−2sin⁡(x+y2)sin⁡(x−y2)\cos x - \cos y = -2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)

Maximum and Minimum of acos⁡θ+bsin⁡θa\cos\theta + b\sin\theta

  • For the expression f(θ)=acos⁡θ+bsin⁡θf(\theta) = a\cos\theta + b\sin\theta:
    • Maximum Value: a2+b2\sqrt{a^2 + b^2}
    • Minimum Value: −a2+b2-\sqrt{a^2 + b^2}
    • Range: [−a2+b2,a2+b2][-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]

Trigonometric Equations & General Solutions

For n∈Zn \in \mathbb{Z}:

  • sin⁡θ=0  ⟹  θ=nπ\sin\theta = 0 \implies \theta = n\pi
  • cos⁡θ=0  ⟹  θ=(2n+1)π2\cos\theta = 0 \implies \theta = (2n+1)\frac{\pi}{2}
  • tan⁡θ=0  ⟹  θ=nπ\tan\theta = 0 \implies \theta = n\pi
  • sin⁡θ=sin⁡α  ⟹  θ=nπ+(−1)nα\sin\theta = \sin\alpha \implies \theta = n\pi + (-1)^n\alpha
  • cos⁡θ=cos⁡α  ⟹  θ=2nπ±α\cos\theta = \cos\alpha \implies \theta = 2n\pi \pm \alpha
  • tan⁡θ=tan⁡α  ⟹  θ=nπ+α\tan\theta = \tan\alpha \implies \theta = n\pi + \alpha
  • sin⁡2θ=sin⁡2α  ⟹  θ=nπ±α\sin^2\theta = \sin^2\alpha \implies \theta = n\pi \pm \alpha
  • cos⁡2θ=cos⁡2α  ⟹  θ=nπ±α\cos^2\theta = \cos^2\alpha \implies \theta = n\pi \pm \alpha
  • tan⁡2θ=tan⁡2α  ⟹  θ=nπ±α\tan^2\theta = \tan^2\alpha \implies \theta = n\pi \pm \alpha

Properties of Triangles (For NDA/CDS/JEE)

In any triangle ABCABC with sides a,b,ca, b, c opposite to angles A,B,CA, B, C:

  • Sine Rule: asin⁡A=bsin⁡B=csin⁡C=2R(R=circumradius)\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \quad (R = \text{circumradius})
  • Cosine Rule:
    • cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}
    • cos⁡B=a2+c2−b22ac\cos B = \frac{a^2 + c^2 - b^2}{2ac}
    • cos⁡C=a2+b2−c22ab\cos C = \frac{a^2 + b^2 - c^2}{2ab}
  • Projection Formula:
    • a=bcos⁡C+ccos⁡Ba = b\cos C + c\cos B
    • b=ccos⁡A+acos⁡Cb = c\cos A + a\cos C
    • c=acos⁡B+bcos⁡Ac = a\cos B + b\cos A
  • Napier's Analogy (Law of Tangents):
    • tan⁡(A−B2)=a−ba+bcot⁡(C2)\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b}\cot\left(\frac{C}{2}\right)
    • tan⁡(B−C2)=b−cb+ccot⁡(A2)\tan\left(\frac{B-C}{2}\right) = \frac{b-c}{b+c}\cot\left(\frac{A}{2}\right)
    • tan⁡(C−A2)=c−ac+acot⁡(B2)\tan\left(\frac{C-A}{2}\right) = \frac{c-a}{c+a}\cot\left(\frac{B}{2}\right)
  • Mollweide's Equation: a−bc=sin⁡(A−B2)cos⁡(C2)\frac{a-b}{c} = \frac{\sin\left(\frac{A-B}{2}\right)}{\cos\left(\frac{C}{2}\right)}

Inverse Trigonometric Functions

FunctionDomainPrincipal Value Branch (Range)
sin⁡−1x\sin^{-1}x[−1,1][-1, 1][−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]
cos⁡−1x\cos^{-1}x[−1,1][-1, 1][0,π][0, \pi]
tan⁡−1x\tan^{-1}xR\mathbb{R}(−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})
csc⁡−1x\csc^{-1}xR∖(−1,1)\mathbb{R} \setminus (-1, 1)[−π2,π2]∖{0}[-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}
sec⁡−1x\sec^{-1}xR∖(−1,1)\mathbb{R} \setminus (-1, 1)[0,π]∖{π2}[0, \pi] \setminus \{\frac{\pi}{2}\}
cot⁡−1x\cot^{-1}xR\mathbb{R}(0,π)(0, \pi)

Typical Exam Weightage

ExamTypical Questions
SSC (CGL / CHSL / MTS)3–4 questions
Defense (NDA / CDS)2–3 questions

Heavily formula-dependent — a small set of standard-angle values and identities covers the large majority of questions asked.

Figures are typical ranges based on recent-year patterns, not a guarantee for any specific upcoming paper — always cross-check against the latest official syllabus and previous-year papers for Trigonometry.

Solved Examples

1Example 1: Easy

Question: If cos⁡θ=45\cos\theta = \frac{4}{5} where 0∘<θ<90∘0^\circ < \theta < 90^\circ, find the value of 3sin⁡θ+2tan⁡θ5cot⁡θ−csc⁡θ\frac{3\sin\theta + 2\tan\theta}{5\cot\theta - \csc\theta}.

View Detailed Solution & Explanation
Step-by-Step Explanation
  1. Find the remaining ratios: Since θ\theta is in the first quadrant, all trigonometric ratios are positive. We are given cos⁡θ=BaseHypotenuse=45\cos\theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{4}{5}. Using the Pythagorean theorem: Perpendicular=Hypotenuse2−Base2=(5x)2−(4x)2=25x2−16x2=3x\text{Perpendicular} = \sqrt{\text{Hypotenuse}^2 - \text{Base}^2} = \sqrt{(5x)^2 - (4x)^2} = \sqrt{25x^2 - 16x^2} = 3x Therefore:

    • sin⁡θ=PerpendicularHypotenuse=35\sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{3}{5}
    • tan⁡θ=PerpendicularBase=34\tan\theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{3}{4}
    • cot⁡θ=BasePerpendicular=43\cot\theta = \frac{\text{Base}}{\text{Perpendicular}} = \frac{4}{3}
    • csc⁡θ=HypotenusePerpendicular=53\csc\theta = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{5}{3}
  2. Substitute these values into the expression: Numerator=3sin⁡θ+2tan⁡θ=3(35)+2(34)=95+32=18+1510=3310\text{Numerator} = 3\sin\theta + 2\tan\theta = 3\left(\frac{3}{5}\right) + 2\left(\frac{3}{4}\right) = \frac{9}{5} + \frac{3}{2} = \frac{18 + 15}{10} = \frac{33}{10} Denominator=5cot⁡θ−csc⁡θ=5(43)−53=203−53=153=5\text{Denominator} = 5\cot\theta - \csc\theta = 5\left(\frac{4}{3}\right) - \frac{5}{3} = \frac{20}{3} - \frac{5}{3} = \frac{15}{3} = 5

  3. Divide Numerator by Denominator: Value=33/105=3350=0.66\text{Value} = \frac{33/10}{5} = \frac{33}{50} = 0.66

2Example 2: Moderate

Question: Evaluate the expression: cos⁡(90∘−A)⋅sec⁡(90∘−A)⋅tan⁡Acsc⁡(90∘−A)⋅sin⁡(90∘−A)⋅cot⁡(90∘−A)+tan⁡(90∘−A)cot⁡A\frac{\cos(90^\circ - A)\cdot\sec(90^\circ - A)\cdot\tan A}{\csc(90^\circ - A)\cdot\sin(90^\circ - A)\cdot\cot(90^\circ - A)} + \frac{\tan(90^\circ - A)}{\cot A}

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Step-by-Step Explanation
  1. Apply Allied/Complementary angle formulas to each component:

    • cos⁡(90∘−A)=sin⁡A\cos(90^\circ - A) = \sin A
    • sec⁡(90∘−A)=csc⁡A\sec(90^\circ - A) = \csc A
    • csc⁡(90∘−A)=sec⁡A\csc(90^\circ - A) = \sec A
    • sin⁡(90∘−A)=cos⁡A\sin(90^\circ - A) = \cos A
    • cot⁡(90∘−A)=tan⁡A\cot(90^\circ - A) = \tan A
    • tan⁡(90∘−A)=cot⁡A\tan(90^\circ - A) = \cot A
  2. Substitute these expressions back into the original equation: First Term=sin⁡A⋅csc⁡A⋅tan⁡Asec⁡A⋅cos⁡A⋅tan⁡A\text{First Term} = \frac{\sin A \cdot \csc A \cdot \tan A}{\sec A \cdot \cos A \cdot \tan A} Second Term=cot⁡Acot⁡A\text{Second Term} = \frac{\cot A}{\cot A}

  3. Simplify each term:

    • For the first term, recall that sin⁡A⋅csc⁡A=1\sin A \cdot \csc A = 1 and sec⁡A⋅cos⁡A=1\sec A \cdot \cos A = 1. First Term=1⋅tan⁡A1⋅tan⁡A=1\text{First Term} = \frac{1 \cdot \tan A}{1 \cdot \tan A} = 1
    • For the second term, a quantity divided by itself is 11: Second Term=cot⁡Acot⁡A=1\text{Second Term} = \frac{\cot A}{\cot A} = 1
  4. Add the simplified terms together: Result=1+1=2\text{Result} = 1 + 1 = 2

3Example 3: Hard

Question: The angle of elevation of a cloud from a point hh meters above a lake is α\alpha, and the angle of depression of its reflection in the lake is β\beta. Prove that the height of the cloud above the lake is h(tan⁡β+tan⁡αtan⁡β−tan⁡α)h\left(\frac{\tan\beta + \tan\alpha}{\tan\beta - \tan\alpha}\right).

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Step-by-Step Explanation
  1. Formulate the geometric layout:

    • Let the horizontal surface of the lake be represented by line XYXY.
    • Let the point of observation be AA, which is at height hh above the lake, so AY⊥XYAY \perp XY and AY=hAY = h.
    • Let the cloud be at point CC, and its height above the lake surface be HH, so CX⊥XYCX \perp XY and CX=HCX = H.
    • The reflection of the cloud in the lake will be at point C′C' vertically below the surface of the lake by the same distance HH, so XC′=HXC' = H.
    • Draw a horizontal line ADAD from AA to the line of height of the cloud CXCX, meeting it at DD.
      • Then DX=AY=hDX = AY = h.
      • CD=CX−DX=H−hCD = CX - DX = H - h.
      • The depth of the reflection from the horizontal line ADAD is DC′=DX+XC′=h+HDC' = DX + XC' = h + H.
    • Let the horizontal distance AD=dAD = d.
  2. Use Trigonometric ratios on the two right-angled triangles:

    • In right-angled triangle ADCADC (observing the cloud): tan⁡α=CDAD=H−hd  ⟹  d=H−htan⁡α— (Equation 1)\tan\alpha = \frac{CD}{AD} = \frac{H - h}{d} \implies d = \frac{H - h}{\tan\alpha} \quad \text{--- (Equation 1)}
    • In right-angled triangle ADC′ADC' (observing the reflection): tan⁡β=DC′AD=H+hd  ⟹  d=H+htan⁡β— (Equation 2)\tan\beta = \frac{DC'}{AD} = \frac{H + h}{d} \implies d = \frac{H + h}{\tan\beta} \quad \text{--- (Equation 2)}
  3. Equate the horizontal distance dd from Equation 1 and Equation 2: H−htan⁡α=H+htan⁡β\frac{H - h}{\tan\alpha} = \frac{H + h}{\tan\beta} (H−h)tan⁡β=(H+h)tan⁡α(H - h)\tan\beta = (H + h)\tan\alpha Htan⁡β−htan⁡β=Htan⁡α+htan⁡αH\tan\beta - h\tan\beta = H\tan\alpha + h\tan\alpha

  4. Solve for the cloud height HH: Htan⁡β−Htan⁡α=htan⁡β+htan⁡αH\tan\beta - H\tan\alpha = h\tan\beta + h\tan\alpha H(tan⁡β−tan⁡α)=h(tan⁡β+tan⁡α)H(\tan\beta - \tan\alpha) = h(\tan\beta + \tan\alpha) H=h(tan⁡β+tan⁡αtan⁡β−tan⁡α)H = h\left(\frac{\tan\beta + \tan\alpha}{\tan\beta - \tan\alpha}\right) This matches the target expression.