Tanong # 14f11

Tanong # 14f11
Anonim

Sagot:

Mangyaring tingnan sa ibaba.

Paliwanag:

Gagamitin natin # cos2x = 1-2sin ^ 2x # at # sin2x = 2sinx * cosx #.

# LHS = (1-cos2x-sinx) / (sin2x-cosx) #

# = (1- (1-2sin ^ 2x) -sinx) / (2sinxcosx-cosx) #

# = (2sin ^ 2x-sinx) / (2sinxcosx-cosx) #

# = (sinx * (2sinx-1)) / (cosx (2sinx-1) #

# = tanx = RHS #