Question

show that Tan 20 tan 40 80 is equal to under root 3​

Answer 1

Appropriate Question :-

Show that

$\rm :\longmapsto\: tan 20\degree \: tan 40\degree \: tan 60\degree \: = \sqrt{3}$

Identities Used :-

We know that

$\boxed{\bf\: \:sinx \: sin(60\degree \: - x) \: sin(60\degree \: + x) = \dfrac{1}{4} sin3x}$

$\boxed{\bf\: \:cosx \: cos(60\degree \: - x) \: cos(60\degree \: + x) = \dfrac{1}{4} sin3x}$

Now, Consider,

$\rm :\longmapsto\: tan 20\degree \: tan 40\degree \: tan 60\degree \:$

$\sf \: = \: \dfrac{sin20\degree \:}{cos20\degree \:} \dfrac{sin40\degree \:}{cos40\degree \:} \dfrac{sin80\degree \:}{cos80\degree \:}$

$\sf \: = \: \dfrac{sin20\degree \:sin40\degree \:sin80\degree \: }{cos20\degree \:cos40\degree \:cos80\degree \:}$

$\sf \: = \: \dfrac{sin20\degree \:sin(60\degree - 20\degree) \:sin(60\degree + 20\degree)}{cos20\degree \:cos(60\degree - 20\degree) \:cos(60\degree + 20\degree)}$

$\sf \: = \: \dfrac{\dfrac{1}{4}sin(3 \times 20\degree) }{\dfrac{1}{4}cos(3 \times 20\degree) }$

$\sf \: = \: \dfrac{sin60\degree \:}{cos60\degree \:}$

$\sf \: = \: tan 60\degree \:$

$\sf \: = \: \sqrt{3}$

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

Additional Information :-

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)