Question

Prove That
cotx cot2x - cot2x cot3x - cot3x cotx = 1 ​

Answer 1

cot x cot 2x - cot 2x cot 3x - cot 3x cot 4x 

= cot x cot 2x - cot 3x (cot 2x+ cot x)

= cot x cot 2x - cot (2x+x)(cot 2x+ cot x)

[ as cot(A+B)=cotAcotBcotAcotB−1]

= cot x cot 2x-(cotx+cot2xcot2xcotx−1)(cot 2x+ cot x)

= cot x cot 2x- (cot 2x cot x-1)

= cot x cot 2x- cot 2x cot x+1

= 1

Hence, proved

Answer 2

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

• Prove That :- cotx cot2x - cot2x cot3x - cot3x cotx = 1

$\large\underline\blue{\bold{Understanding \: the \: Concept :- }}$

Here the concept of Trigonometric Identities is used.

In this question mainly we will use only 1 identitity.

Firstly, if we look carefully, there are 3 angles x, 2x and 3x, such that largest angle 3x is sum of other two angles.

We shall simplify the equation to a easier form by simply breaking the angles and then applying the identity.

Then we will find the the main equation to be solved. We will keep on simplifying and then finally, we will get our answer.

Let's do it !!

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$\large\underline\blue{\bold{Formula \: used:- }}$

$\large \blue{\bf \:  ⟼ cot(x + y) = \dfrac{cotx \: coty - 1}{coty + cotx} }$

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

$\large \blue{\bf \:As \: cot3x = cot(2x + x)}$

$\bf\implies \:cot3x = \dfrac{cotx \: cot2x - 1}{cot2x + cotx}$

$\large \blue{\bf \:on \: cross \: multiplication \: we \: get}$

$\sf \: ⟼cot3xcot2x + cot3xcotx = cot2xcotx - 1$

$\large \blue{\sf \: ⟼cotxcot2x - cot2xcot3x - cot3xcotx = 1}$

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$\large \red{\bf \: ⟼ Explore \: more } ✍$

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)