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Question

# Determine the smallest positive value of x (in degrees) for which tan (x+100∘)=tan(x+50∘)tan x tan (x−50∘). Prove that cos 7x−cos 8x1+2 cos 5x=cos 2x−cos 3x

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Solution

## We have, tan (x+100∘)=tan(x+50∘)tan x tan (x−50∘) ⇒ tan (x+100∘)tan (x−50∘)=tan(x+50∘)tan x ⇒ sin (x+100∘)cos(x−50∘)cos(x+100∘)sin(x−50∘)=sin(x+50∘)sin xcos(x+50∘)cos x Applying componendo and dividendo, we get sin(x+100∘)cos(x−50∘)+cos(x+100∘)sin(x−50∘)sin(x+100∘)cos(x−50∘)−cos(x+100∘)sin(x−50∘) =sin(x+50∘)sin x+cos(x+50∘)cos xsin(x+50∘)sin x−cos(x+50∘)cos x ⇒ sin(x+100∘+x−50∘)sin(x+100∘−x+50∘)=cos(x+50∘−x)−cos(x+50∘+x) ∵ sin(A±B)=sin A cos B±cos A sin B and cos(A±B)=cos A cos B∓sin A sin B ⇒ ssin (2x+50∘)sin 150∘=cos 50∘−cos (2x+50∘) ⇒ sin(2x+50∘) cos (2x+50∘)=−sin 150∘ cos 50∘ ⇒ 2 sin (2x+50∘)cos (2x+50∘)=−2×12 cos 50∘ [∵ sin 150∘=12] ⇒ sin (4x+100∘)=−cos 50∘ [∵ 2 sin A cos A=sin 2A] ⇒ sin(4x+100∘)=sin(270∘−50∘) ⇒ 4x+100∘=270∘−50∘ ⇒ 4x=270∘−50∘−100∘, 4x=270∘−150∘ ⇒ 4x=120∘ ⇒ x=30∘II part- We have, LHS=cos 7x−cos 8x1+2 cos 5x Multiply numerator and denominator by 2 sin5x2, we get LHS=2 sin5x2 cos 7x−2 sin 5x2 cos 8x2 sin 5x2+4 sin 5x2 cos 5x =(sin19x2−sin9x2)−(sin 21x2−sin 11x2)2 sin 5x2+2 sin15x2−2 sin 5x2 [∵ 2 sin A cos B = sin (A + B) + sin (A - B)] =(sin19x2+sin11x2)−(sin9x2+sin21x2)2 ssin 15x2 =2 sin15x2 cos 2x−2 sin 15x2 cos 3x2 sin 15x2 [∵ sin C+sin D=2 sinC+D2 cos C−D2] =cos2x−cos3x=RHS Hence proved.

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