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Maths Notes 11th 12th JEE

Maths Notes 11th 12th JEE

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📈 Análisis del canal de Telegram Maths Notes 11th 12th JEE

El canal Maths Notes 11th 12th JEE (@class_11th_12th_maths_notes_jee) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 21 927 suscriptores, ocupando la posición 9 170 en la categoría Educación y el puesto 18 557 en la región India.

📊 Métricas de audiencia y dinámica

Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 21 927 suscriptores.

Según los últimos datos del 28 octubre, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -276, y en las últimas 24 horas de -4, conservando un alto alcance.

  • Estado de verificación: No verificado
  • Tasa de interacción (ER): El promedio de interacción de la audiencia es 3.74%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 8.47% de reacciones respecto al total de suscriptores.
  • Alcance de las publicaciones: Cada publicación recibe en promedio 821 visualizaciones. En el primer día suele acumular 1 857 visualizaciones.
  • Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 1.
  • Intereses temáticos: El contenido se centra en temas clave como biology, chemistry, neet, jee, guidance.

📝 Descripción y política de contenido

El autor describe el recurso como un espacio para expresar opiniones subjetivas:
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Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 29 octubre, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Educación.

21 927
Suscriptores
-424 horas
-417 días
-27630 días
Archivo de publicaciones
Chapter samajh aa gaya ✅ Notes complete ✅ Lecture bhi dekh liya ✅ Confidence: 100% 😎 Test paper haath mein aate hi: 404 — Answers Not Found 💀 Sach batao, sirf mere saath hota hai ya tum bhi isi phase mein ho?

Basic Maths Short Notes🌸
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Basic Maths Short Notes🌸

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➖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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