source: ncert
topic: trigonometry
class_level: class_10
chapter: ch8
difficulty: intermediate
Class 10 | Ch8: Introduction to Trigonometry TRIGONOMETRIC RATIOS (right triangle, angle theta): sin(theta)=opposite/hypotenuse=P/H. cos(theta)=adjacent/hypotenuse=B/H. tan(theta)=opposite/adjacent=P/B=sin/cos. cosec(theta)=1/sin=H/P. sec(theta)=1/cos=H/B. cot(theta)=1/tan=B/P. STANDARD VALUES TABLE: Angle: 0° 30° 45° 60° 90° sin: 0 1/2 1/sqrt2 sqrt3/2 1 cos: 1 sqrt3/2 1/sqrt2 1/2 0 tan: 0 1/sqrt3 1 sqrt3 undefined cosec: undef 2 sqrt2 2/sqrt3 1 sec: 1 2/sqrt3 sqrt2 2 undef cot: undef sqrt3 1 1/sqrt3 0 TRIGONOMETRIC IDENTITIES: sin^2(theta)+cos^2(theta)=1. 1+tan^2(theta)=sec^2(theta). 1+cot^2(theta)=cosec^2(theta). COMPLEMENTARY ANGLES: sin(90-A)=cos(A). cos(90-A)=sin(A). tan(90-A)=cot(A). cosec(90-A)=sec(A). sec(90-A)=cosec(A). cot(90-A)=tan(A). SOLVED EXAMPLES: Example 1: If sin(theta)=3/5, find all trig ratios. P=3, H=5, B=sqrt(25-9)=4. cos=4/5, tan=3/4, cosec=5/3, sec=5/4, cot=4/3. Example 2: Prove (sin^2(theta)/cos(theta)) + cos(theta) = sec(theta). LHS=sin^2/cos+cos=(sin^2+cos^2)/cos=1/cos=sec. Proved. Example 3: Evaluate 2tan^2(45°)+cos^2(30°)-sin^2(60°). =2(1)^2+(sqrt3/2)^2-(sqrt3/2)^2=2+3/4-3/4=2. Example 4: sin(theta)=cos(theta). Find theta. tan(theta)=1. theta=45°. COMMON MISTAKES: sin^2+cos^2=1 (NOT sin+cos=1). Values at 90°: sin=1, cos=0, tan=undefined (NOT 0). Memorise table using: "Some People Have Curly Brown Hair Through Proper Brushing".