Cosx/1-tanx +sinx/1-cotx=sinx+cosx
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Answered by
4
Answer:
LHS=cosx1−tanx+sinx1−cotx =cos2xcosx(1−tanx)+sin2xsinx(1−cotx) =cos2xcosx−sinx+sin2xsinx−cosx =cos2xcosx−sinx−sin2xcosx−sinx =cos2x−sin2
Answered by
3
Answer:
LHS= [cos x / (1-tan x)] + [ sin x / (1-cot x)]
= [cos x / (1-tan x)]+ [sin x / (1-(1/tan x))]
= [cos x / (1-tan x)]+ [sin x / (tan x-1)/tan x]
= [cos x / (1-tan x)]+ [(sin x tan x) / (tan x-1)]
= [cos x / (1-tan x)]- [(sin x tan x) / (1-tan x)]
= (cos x-sin x tan x)/ (1-tan x)
= [cos x-sin x (sin x/cos x)]/ [1-(sin x/cos x)]
= [cos x-(sin²x/cos x)]/ [1-(sin x/cos x)]
= [(cos ²x-sin ²x )/cos x]/ [(cos x-sin x)/cos x]
= (cos ²x-sin ²x )/(cos x-sin x)
= [(cos x+sin x ) (cos x-sin x )]/(cos x-sin x)
= cos x+sin x
= RHS
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Step-by-step explanation:
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