Class 11th Neet&jee notes

NEET 2023 PAPER ANALYSIS Bio - Easy Chem - Moderate Phy - Easy Cut off expected GEN - 620 +-10 EWS - 620 +-8 OBC - 600 +-8 SC - 470 +-10 ST - 440 +-5

JEE Advanced Eligibility Criteria Percentile General : 90.7788642 GEN - EWS : 75.6229025 OBC NCL : 73.61114226 SC : 51.97760027 ST : 37.2347772 Cutoff 📈

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📚Notes on Trigonometric Equations and Identities📚 A function f(x) is said to be periodic if there exists some T > 0 such that f(x+T) = f(x) for all x in the domain of f(x). In case, the T in the definition of period of f(x) is the smallest positive real number then this ‘T’ is called the period of f(x). Periods of various trigonometric functions are listed below: 1) sin x has period 2π 2) cos x has period 2π 3) tan x has period π 4) sin(ax+b), cos (ax+b), sec(ax+b), cosec (ax+b) all are of period 2π/a 5) tan (ax+b) and cot (ax+b) have π/a as their period 6) |sin (ax+b)|, |cos (ax+b)|, |sec(ax+b)|, |cosec (ax+b)| all are of period π/a 7) |tan (ax+b)| and |cot (ax+b)| have π/2a as their period ➖Sum and Difference Formulae of Trigonometric Ratios 1) sin(a + ß) = sin(a)cos(ß) + cos(a)sin(ß) 2) sin(a – ß) = sin(a)cos(ß) – cos(a)sin(ß) 3) cos(a + ß) = cos(a)cos(ß) – sin(a)sin(ß) 4) cos(a – ß) = cos(a)cos(ß) + sin(a)sin(ß) 5) tan(a + ß) = [tan(a) + tan (ß)]/ [1 - tan(a)tan (ß)] 6)tan(a - ß) = [tan(a) - tan (ß)]/ [1 + tan (a) tan (ß)] 7) tan (π/4 + θ) = (1 + tan θ)/(1 - tan θ) 8) tan (π/4 - θ) = (1 - tan θ)/(1 + tan θ) 9) cot (a + ß) = [cot(a) . cot (ß) - 1]/ [cot (a) +cot (ß)] 10) cot (a - ß) = [cot(a) . cot (ß) + 1]/ [cot (ß) - cot (a)] ➖Double or Triple -Angle Identities 1) sin 2x = 2sin x cos x 2) cos2x = cos2x – sin2x = 1 – 2sin2x = 2cos2x – 1 3) tan 2x = 2 tan x / (1-tan 2x) 4) sin 3x = 3 sin x – 4 sin3x 5) cos3x = 4 cos3x – 3 cosx 6) tan 3x = (3 tan x - tan3x) / (1- 3tan 2x) ➖For angles A, B and C, we have 1) sin (A + B +C) = sinAcosBcosC + cosAsinBcosC + cosAcosBsinC - sinAsinBsinC 2) cos (A + B +C) = cosAcosBcosC- cosAsinBsinC - sinAcosBsinC - sinAsinBcosC 3) tan (A + B +C) = [tan A + tan B + tan C –tan A tan B tan C]/ [1- tan Atan B - tan B tan C –tan A tan C 4) cot (A + B +C) = [cot A cot B cot C – cotA - cot B - cot C]/ [cot A cot B + cot Bcot C +  cot A cotC–1] ➖List of some other trigonometric formulas: 1) 2sinAcosB = sin(A + B) + sin (A - B) 2) 2cosAsinB = sin(A + B) - sin (A - B) 3) 2cosAcosB = cos(A + B) + cos(A - B) 4) 2sinAsinB = cos(A - B) - cos (A + B) 5) sin A + sin  B = 2 sin [(A+B)/2] cos [(A-B)/2] 6) sin A - sin  B = 2 sin [(A-B)/2] cos [(A+B)/2] 7) cosA + cos  B = 2 cos [(A+B)/2] cos [(A-B)/2] 8) cosA - cos  B = 2 sin [(A+B)/2] sin [(B-A)/2] 9) tanA ± tanB = sin (A ± B)/ cos A cos B 10)cot A ± cot B = sin (B ± A)/ sin A sin B ➖Method of solving a trigonometric equation: 1) If possible, reduce the equation in terms of any one variable, preferably x. Then solve the equation as you used to in case of a single variable. 2) Try to derive the linear/algebraic simultaneous equations from the given trigonometric equations and solve them as algebraic simultaneous equations. 3) At times, you might be required to make certain substitutions. It would be beneficial when the system has only two trigonometric functions. ➖Some results which are useful for solving trigonometric equations: 1) sin θ = sina and cosθ = cosa ⇒ θ = 2nπ + a 2) sin θ = 0 ⇒ θ = nπ 3) cosθ = 0 ⇒ θ = (2n + 1)π/2 4) tan θ = 0 ⇒ θ = nπ 5) sinθ = sina⇒ θ = nπ + (-1)na where a ∈ [–π/2, π/2] 6) cosθ= cos a ⇒ θ = 2nπ ± a, where a ∈[0,π] 7) tanθ = tana⇒ θ = nπ+ a, where a ∈[–π/2, π/2] 8) sinθ = 1 ⇒ θ= (4n + 1)π/2 9) sin θ = -1 ⇒ θ = (4n - 1) π /2 10) sin θ = -1 ⇒ θ = (2n +1) π /2 11) |sinθ| = 1⇒ θ =2nπ 12) cosθ = 1 ⇒ θ =(2n + 1) 13) |cosθ| = 1⇒ θ =nπ

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