Question #73c4a

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Answer 1 · 2017-08-11T17:04:26

# 4x.#

We will frequently use the Identity : #sin2A=2sinAcosA.#

Also, we need, #costheta=sin(90-theta).#

We have,

#sin24/sin8+cos24/cos8=(sin24cos8+cos24sin8)/(sin8cos8),#

#={sin(24+8)}/(sin8cos8),#

#=sin32/(sin8cos8)=(2sin32)/(2sin8cos8),#

#=(2sin32)/sin16,#

#={2(2sin16cos16)}/sin16,#

#=4cos16=4sin(90-16)=4sin74.#

# rArr sin24/sin8+cos24/cos8=4x.#