Math, asked by scmchackojohn, 10 months ago

Prove that sinA - 2Sin cube A÷2 Cos cube A -Cos A=tanA

Answers

Answered by shivamchaudhary49

3

Step-by-step explanation:

sinA -2sin^3A ÷ 2cos^3A-cosA

sinA and cosA common

sinA(1-2sin^2A)÷cosA(2cos^2-1)

we put 1 = sin^2A+cos^2A

sinA(sin^2A+cos^2A-2sin^2A)÷cosA(2cos^2A-sin^2-cos^2A)

sinA(cos^2A-sin^2A)÷cosA(cos^2A-sin^2A)

we have sinA/cosA

we know that sinA/ cosA = tanA

Hence proved

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