Question

(sin 2A+sin3A)/(cos2A−cos3A) =cot A/2.​
prove

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\bf \: Prove \: that : \sf \: \dfrac{sin2A + sin3A}{cos2A - cos3A} = cot\dfrac{A}{2}$

$\large\underline{\bold{Answer - }}$

$\begin{gathered}{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$(1). \: \: \boxed{ \bf{ \: sinx + siny = 2sin\bigg(\dfrac{x + y}{2} \bigg)cos \bigg( \dfrac{x - y}{2} \bigg) }}$

$(2). \: \: \boxed{ \bf{ \: cosx - cosy = 2sin\bigg( \dfrac{x + y}{2} \bigg)sin \bigg( \dfrac{y - x}{2} \bigg) }}$

$(3). \: \: \boxed{ \bf{\dfrac{sinx}{cosx} = tanx}}$

$\large\underline{\bold{Solution-}}$

Consider LHS,

$\sf \: \dfrac{sin2A + sin3A}{cos2A - cos3A}$

$= \sf \: \dfrac{\cancel2 \: sin\bigg( \dfrac{3A + 2A}{2} \bigg) \: cos\bigg( \dfrac{3A - 2A}{2} \bigg) }{\cancel2 \: sin\bigg( \dfrac{2A + 3A}{2} \bigg) \: sin\bigg( \dfrac{3A - 2A}{2} \bigg) }$

$= \sf \: \dfrac{\cancel{sin\bigg( \dfrac{5A}{2} \bigg)} \: cos\bigg( \dfrac{A}{2} \bigg) }{\cancel{sin\bigg(\dfrac{5A}{2} \bigg)} \: sin\bigg( \dfrac{A}{2} \bigg) }$

$= \sf \: cot \: \dfrac{A}{2}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

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Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

Product to Sum Formulas :-

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas :-

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles :-

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

sin (A+B) = sin A cos B + cos A sin B

sin (A -B) = sin A cos B – cos A sin B

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles :-

sin2A = 2sinA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A – sin²A = 1 – 2sin²A = 2cos²A – 1 = [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)