INTRODUCTION TO TRIGONOMETRY

Trigonometry, Greek words 'tri' (three), 'gon' (sides), 'metron' (measure) se bana hai. Basically, right-angled triangle ke sides aur angles ke beech ka relationship padhte hain isme.

Right-Angled Triangle mein T-Ratios

Ek right-angled triangle \(\triangle ABC\) mein, jahan \(\angle B = 90^\circ\):

• Hypotenuse (कर्ण): \(90^\circ\) ke opposite side (AC).
• Opposite Side (लंब): Given acute angle (e.g., \(\angle A\)) ke opposite side (BC).
• Adjacent Side (आधार): Given acute angle (e.g., \(\angle A\)) se laga hua side (AB), jo hypotenuse nahi hai.

T-Ratios for \(\angle A\):

• Sine of \(\angle A\) (sin A): \( \frac{\text{Opposite Side}}{\text{Hypotenuse}} = \frac{BC}{AC} \)
• Cosine of \(\angle A\) (cos A): \( \frac{\text{Adjacent Side}}{\text{Hypotenuse}} = \frac{AB}{AC} \)
• Tangent of \(\angle A\) (tan A): \( \frac{\text{Opposite Side}}{\text{Adjacent Side}} = \frac{BC}{AB} \)

Reciprocal T-Ratios:

• Cosecant of \(\angle A\) (cosec A): \( \frac{1}{\sin A} = \frac{\text{Hypotenuse}}{\text{Opposite Side}} = \frac{AC}{BC} \)
• Secant of \(\angle A\) (sec A): \( \frac{1}{\cos A} = \frac{\text{Hypotenuse}}{\text{Adjacent Side}} = \frac{AC}{AB} \)
• Cotangent of \(\angle A\) (cot A): \( \frac{1}{\tan A} = \frac{\text{Adjacent Side}}{\text{Opposite Side}} = \frac{AB}{BC} \)

Important Relations:

• \( \tan A = \frac{\sin A}{\cos A} \)
• \( \cot A = \frac{\cos A}{\sin A} \)

Mnemonic for T-Ratios:

Pandit Badri Prasad Har Har Bole, Sona Chandi Tole

• Sin = Pandit / Har (Perpendicular / Hypotenuse)
• Cos = Badri / Har (Base / Hypotenuse)
• Tan = Prasad / Bole (Perpendicular / Base)

Yaad rakho: T-Ratios sirf acute angles (\(0^\circ < \theta < 90^\circ\)) ke liye define hote hain. \(90^\circ\) ke liye kuch ratios undefined hote hain.

T-Ratios ki Value

• T-Ratios ki value triangle ke sides ke length par depend nahi karti, sirf angle ki value par depend karti hai.
• Agar angle same hai, toh chahe triangle chhota ho ya bada, T-Ratios ki value same rahegi (due to similarity of triangles).
• \( \sin A \) aur \( \cos A \) ki value hamesha \(0\) se \(1\) ke beech hoti hai (inclusive). \( \sin A \le 1 \) aur \( \cos A \le 1 \).
• \( \sec A \) aur \( \text{cosec } A \) ki value hamesha \(1\) se badi ya barabar hoti hai. \( \sec A \ge 1 \) aur \( \text{cosec } A \ge 1 \).
• \( \tan A \) aur \( \cot A \) ki value kuch bhi ho sakti hai (real numbers).

Kuch specific angles \((0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ)\) ke T-Ratios ki values yaad rakhna bahut zaroori hai. Ye values geometry aur Pythagoras Theorem use karke derive ki jaati hain.

\(45^\circ\) ke T-Ratios:

Ek isosceles right-angled triangle lo jisme \(AB = BC = a\). Toh \(\angle A = \angle C = 45^\circ\). Pythagoras se, \(AC = \sqrt{a^2 + a^2} = a\sqrt{2}\).

• \( \sin 45^\circ = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}} \)
• \( \cos 45^\circ = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}} \)
• \( \tan 45^\circ = \frac{a}{a} = 1 \)

\(30^\circ\) aur \(60^\circ\) ke T-Ratios:

Ek equilateral triangle \(ABC\) lo jisme har side \(2a\) ho. \(\angle A = \angle B = \angle C = 60^\circ\). Vertex A se BC par altitude AD draw karo. Ye BC ko bisect karega aur \(\angle A\) ko bhi bisect karega. Toh \(BD = DC = a\) aur \(\angle BAD = 30^\circ\). \(\triangle ADB\) ek right-angled triangle hai jisme \(\angle D = 90^\circ\). Pythagoras se, \(AD = \sqrt{(2a)^2 - a^2} = \sqrt{4a^2 - a^2} = \sqrt{3a^2} = a\sqrt{3}\).

For \(\angle B = 60^\circ\):

• \( \sin 60^\circ = \frac{AD}{AB} = \frac{a\sqrt{3}}{2a} = \frac{\sqrt{3}}{2} \)
• \( \cos 60^\circ = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2} \)
• \( \tan 60^\circ = \frac{AD}{BD} = \frac{a\sqrt{3}}{a} = \sqrt{3} \)

For \(\angle BAD = 30^\circ\):

• \( \sin 30^\circ = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2} \)
• \( \cos 30^\circ = \frac{AD}{AB} = \frac{a\sqrt{3}}{2a} = \frac{\sqrt{3}}{2} \)
• \( \tan 30^\circ = \frac{BD}{AD} = \frac{a}{a\sqrt{3}} = \frac{1}{\sqrt{3}} \)

\(0^\circ\) aur \(90^\circ\) ke T-Ratios:

Ye values limiting cases se derive hoti hain, jab angle \(0^\circ\) ya \(90^\circ\) ke paas jaata hai.

Summary Table (Most Important!):

Agar do angles ka sum \(90^\circ\) ho, toh unhe complementary angles kehte hain. Ek right-angled triangle \(ABC\) mein, jahan \(\angle B = 90^\circ\), acute angles \(\angle A\) aur \(\angle C\) complementary hote hain, yaani \(\angle A + \angle C = 90^\circ\). Iska matlab \(\angle C = 90^\circ - \angle A\).

Relations:

• \( \sin (90^\circ - A) = \cos A \)
• \( \cos (90^\circ - A) = \sin A \)
• \( \tan (90^\circ - A) = \cot A \)
• \( \cot (90^\circ - A) = \tan A \)
• \( \sec (90^\circ - A) = \text{cosec } A \)
• \( \text{cosec } (90^\circ - A) = \sec A \)

Proof (for \(\sin (90^\circ - A) = \cos A\)): \(\triangle ABC\) mein, \(\angle B = 90^\circ\). For \(\angle A\):

• \( \sin A = \frac{BC}{AC} \)
• \( \cos A = \frac{AB}{AC} \)

For \(\angle C = (90^\circ - A)\):

• \( \sin C = \frac{AB}{AC} \implies \sin (90^\circ - A) = \frac{AB}{AC} \)
• \( \cos C = \frac{BC}{AC} \implies \cos (90^\circ - A) = \frac{BC}{AC} \)

Compare karne par, \( \sin (90^\circ - A) = \cos A \) aur \( \cos (90^\circ - A) = \sin A \). Similarly, other relations bhi prove kiye ja sakte hain.

Application:

Ye relations expressions ko simplify karne aur unknown angles find karne mein help karte hain.

Ek equation jisme trigonometric ratios involve hote hain aur jo angle ki har value ke liye true ho, use Trigonometric Identity kehte hain. Ye Pythagoras Theorem se derive hoti hain.

Fundamental Trigonometric Identities:

1. \( \sin^2 A + \cos^2 A = 1 \)

• Is identity se do aur forms derive ho sakte hain:
• \( \sin^2 A = 1 - \cos^2 A \)
• \( \cos^2 A = 1 - \sin^2 A \)

1. \( 1 + \tan^2 A = \sec^2 A \) (for \(0^\circ \le A < 90^\circ\))

• Is identity se do aur forms derive ho sakte hain:
• \( \sec^2 A - \tan^2 A = 1 \)
• \( \sec^2 A - 1 = \tan^2 A \)

1. \( 1 + \cot^2 A = \text{cosec}^2 A \) (for \(0^\circ < A \le 90^\circ\))

• Is identity se do aur forms derive ho sakte hain:
• \( \text{cosec}^2 A - \cot^2 A = 1 \)
• \( \text{cosec}^2 A - 1 = \cot^2 A \)

Derivation of Identities (Pythagoras Theorem se):

Ek right-angled triangle \(ABC\) lo, jahan \(\angle B = 90^\circ\). Pythagoras Theorem ke according: \(AB^2 + BC^2 = AC^2\)

Identity 1: \( \sin^2 A + \cos^2 A = 1 \)

• Equation ko \(AC^2\) se divide karo:

\( \frac{AB^2}{AC^2} + \frac{BC^2}{AC^2} = \frac{AC^2}{AC^2} \) \( \left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = 1 \)

• Hum jaante hain \( \cos A = \frac{AB}{AC} \) aur \( \sin A = \frac{BC}{AC} \).
• Toh, \( (\cos A)^2 + (\sin A)^2 = 1 \implies \cos^2 A + \sin^2 A = 1 \).

Identity 2: \( 1 + \tan^2 A = \sec^2 A \)

• Equation \(AB^2 + BC^2 = AC^2\) ko \(AB^2\) se divide karo (assuming \(AB \ne 0\), so \(A \ne 90^\circ\)):

\( \frac{AB^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AC^2}{AB^2} \) \( 1 + \left(\frac{BC}{AB}\right)^2 = \left(\frac{AC}{AB}\right)^2 \)

• Hum jaante hain \( \tan A = \frac{BC}{AB} \) aur \( \sec A = \frac{AC}{AB} \).
• Toh, \( 1 + (\tan A)^2 = (\sec A)^2 \implies 1 + \tan^2 A = \sec^2 A \).

Identity 3: \( 1 + \cot^2 A = \text{cosec}^2 A \)

• Equation \(AB^2 + BC^2 = AC^2\) ko \(BC^2\) se divide karo (assuming \(BC \ne 0\), so \(A \ne 0^\circ\)):

\( \frac{AB^2}{BC^2} + \frac{BC^2}{BC^2} = \frac{AC^2}{BC^2} \) \( \left(\frac{AB}{BC}\right)^2 + 1 = \left(\frac{AC}{BC}\right)^2 \)

• Hum jaante hain \( \cot A = \frac{AB}{BC} \) aur \( \text{cosec } A = \frac{AC}{BC} \).
• Toh, \( (\cot A)^2 + 1 = (\text{cosec } A)^2 \implies 1 + \cot^2 A = \text{cosec}^2 A \).

Solving Problems using Identities:

• Most problems mein, sabhi T-Ratios ko \(\sin\) aur \(\cos\) mein convert karna helpful hota hai.
• Algebraic identities \((a+b)^2, (a-b)^2, a^2-b^2\) ka use bhi frequently hota hai.
• Rationalization (denominator se square root hatana) bhi common technique hai.
• Left Hand Side (LHS) ko Right Hand Side (RHS) ke equal prove karna ya vice-versa.