Prove that (tan2x)(2tan4x)(4tan8x)=cotx
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sin(8x) = 2sin(4x)cos(4x) = 4sin(2x)cos(2x)cos(4x) = 8sin(x)cos(x)cos(2x)cos(4x)
Taking logs, log{sin(8x)} = log{8sin(x)} + log{cos(x)} + log(cos(2x)} + log{cos(4x)}
Differentiate, 8cot(8x) = cot(x) − tan(x) − 2tan(2x) − 4tan(4x)
tan(x) + 2tan(2x) + 4tan(4x) + 8cot(8x) = cot(x)
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edited : or just trig using
cot(u)−2cot(2u) = { cos²(u)−(cos²(u)−sin²(u)) } / { sin(u)cos(u) } = tan(u)
u=x : cot(x) − 2cot(2x) = tan(x) … (i)
u=2x : cot(2x) − 2cot(4x) = tan(2x) … (ii)
u=4x : cot(4x) − 2cot(8x) = tan(4x) … (iii)
(i)+2(ii)+4(iii) gives cot(x) − 8cot(8x) = tan(x) + 2tan(2x) + 4tan(4x)
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