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Math15- cos2θsinθ+sinθ=cscθ-sinθ
LHS: Left Hand Side
RHS: Right Hand Side
LHS= cos2θsinθ+sinθcos2θ=cos2θ-sin2θ
=cos2θ-sin2θsinθ+sinθLCD: sinθ sinθ
= cos2θ-sin2θ+sin2θsinθ where-sin2θ+sin2θ=0
cos2θ = 1 - sin2θ
=1-sin2θsinθ
=1sinθ- sin2θsinθ
cscθ = 1/ sinθ
= cscθ – sinθ = RHS
Hence proved LHS = RHS
16- 2tanθ-sin2θ2sin2θ=tanθ
LHS=2tanθ-sin2θ2sin2θ
sin2θ=(1-cos2θ)
=2sinθcosθ-2sinθcosθ2(1-cos2θ)
=2(sinθ-sinθcos2θcosθ) 2(1-cos2θ)
=sinθ1-cos2θcosθ(1-cos2θ)
= tanθ = RHS
Hence proved LHS = RHS
17- 2cosθ - cos2θcosθ=secθLHS= 2cosθ - cos2θcosθ
Cos2θ = Cos2θ – sin2θ LCD: cosθcosθ
= 2cos2θ-(cos2θ-sin2θ)cosθ
= 2cos2θ-cos2θ+sin2θcosθ
= cos2θ+sin2θcosθ
Cos2θ + sin2θ = 1
= 1cosθ
secθ=1cosθ
= secθ = RHS
Hence proved LHS = RHS
18- cos2θ+ cosθ+1sin2θ+sinθ=cotθ
LHS=cos2θ+ cosθ+1sin2θ+sinθ
Cos2θ = Cos2θ – sin2θ and sin2θ =2sinθcosθ= cos2θ-sin2θ+ cosθ+12sinθcosθ+sinθ
sin2θ=1-cos2θ
=cos2θ-(1-cos2θ)+ cosθ+12sinθcosθ+sinθ
=cos2θ-1+cos2θ+ cosθ+12sinθcosθ+sinθ
=2cos2θ+ cosθ2sinθ(1+cosθ)
=2cosθ(cos+ 1)2sinθ(cosθ+1)
=cosθsinθ
=cotθ = RHS
Hence proved LHS = RHS
19- sinθ+sin2θsecθ+2=sinθcosθ
LHS=sinθ+sin2θsecθ+2
Sin2θ = 2sinθcosθsecθ=1cosθ
=sinθ+2sinθcosθ1cosθ+2
=sinθ+2sinθcosθ1+2cosθcosθ, LCD=1cosθ/cosθ
=sinθ(1+2cosθ)1+2cosθcosθ
=sinθ1cosθ
=sinθcosθ = RHS
Hence proved LHS=RHS
20- 1+cos2θ1-cos2θ=cot2θ
LHS= 1+cos2θ1-cos2θ
Cos2θ = Cos2θ – sin2θ
= 1+cos2θ –sin2θ1-(cos2θ –sin2θ)
= 1+cos2θ –sin2θ1-cos2θ+sin2θ
cos2θ = 1 - sin2θ
sin2θ = 1 - cos2θ
= 1+cos2θ –(1- cos2θ)1-(1-sin2θ)+sin2θ
= 2cos2θ 2sin2θ
=cot2θ = RHS
Hence proved LHS = RHS
21- sinπ6-θ+sinπ6+θ=cosθ
LHS=sinπ6-θ+sinπ6+θ
sin ( a + b) = sin(a)cos(b) + cos(a)sin(b)
sin ( a - b) = sin(a)cos(b) - cos(a)sin(b)
=sinπ6cosθ-cosπ6sinθ+sinπ6cosθ+cosπ6sinθ
=2sinπ6cosθ
sinπ6= 12
=22cosθ
= cosθ = RHS
Hence proved LHS = RHS
22- cosπ4-θ-cosπ4+θ=√2sinθ
cos(a + b) = cos(a)cos(b) – sin(a)sin(b)
cos(a - b) = cos(a)cos(b) + sin(a)sin(b)
LHS= cosπ4-θ-cosπ4+θ
=cosπ4cosθ+sinπ4sinθ -(cosπ4cosθ-sinπ4sinθ)
=cosπ4cosθ+sinπ4sinθ -cosπ4cosθ+sinπ4sinθ=2sinπ4sinθ
sinπ4=√22=2√22sinθ = √2sinθ = RHS
Hence proved LHS = RHS
23- sinπ2+θ-sinπ2-θ=0
LHS= sinπ2+θ-sinπ2-θ
sin ( a + b) = sin(a)cos(b) + cos(a)sin(b)
sin ( a - b) = sin(a)cos(b) - cos(a)sin(b)
=sinπ2cosθ+cosπ2sinθ-(sinπ2cosθ-cosπ2sinθ)
=sinπ2cosθ+cosπ2sinθ-sinπ2cosθ+cosπ2sinθ
=cosπ2sinθ
cosπ2=0
= 0 = RHS
Hence proved LHS = RHS
24- sinπ4+θ+cosπ4+θ=√2cosθ
LHS= sinπ4+θ+cosπ4+θ
Sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
cos(a + b) = cos(a)cos(b) – sin(a)sin(b)
=sinπ4cosθ+cosπ4sinθ+ cosπ4cosθ-sinπ4sinθ=√22cosθ+ √22sinθ+√22cos- 22sinθ
=2√22cosθ
=2cosθ = RHS
Hence proved LHS = RHS
25- tanπ4+θtan3π4-θ=-1
LHS= tanπ4+θtan3π4-θ
tana+b=tana+tan(b)1- tanatan(b)
tana-b=tana-tan(b)1+ tanatan(b)
=tanπ4 +tanθ1-tanπ4tanθtan3π4-tanθ1+tan3π4tanθ
tanπ4=1 & tan3π4= -1
=1 +tanθ1-tanθ-1(1+tanθ)1-tanθ
= -1 = RHS
Hence proved LHS = RHS
26- sinπ4+θcosπ4-θ=1
Sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
cos(a - b) = cos(a)cos(b) + sin(a)sin(b)
LHS= sinπ4+θcosπ4-θ
=sinπ4cosθ+cosπ4sinθcosπ4cosθ+sinπ4sinθ
=√22cosθ+√22sinθ√22cosθ+√22sinθ
= 1 = RHS
Hence proved LHS = RHS
27-a) sin2A1+cos2A=tanA
LHS=sin2A1+cos2A
Sin2A= 2sinAcosA
Cos2A= cos2A – sin2A
=2sinAcosA1+cos2A-sin2A
Sin2A = 1 – cos2A
=2sinAcosA1+cos2A-(1-cos2A)
=2sinAcosA2cos2A
=2sinA2cosA
= tanA = RHS
Hence proved LHS = RHS
b) sin212θ1+cos212θ=tan12θ
LHS=sin212θ1+cos212θ
Sin212θ= 2sin12θcos12θ
Cos212θ= cos212θ – sin212θ
=2sin12θcos12θ1+cos212θ-sin212θ
Sin212θ = 1 – cos212θ
=2sin12θcos12θ1+cos212θ-(1-cos212θ)
=2sin12θcos12θ2cos212θ
=2sin12θ2cos12θ
= tan12θ = RHS
Hence proved LHS = RHS
c) ±1-cos2θ1+cos2θ=sinθ1+cos2θ
LHS=±1+cos2θ1+cos2θ
LCD: 1+cos2θ1+cos2θ
=±1-cos2θ1+cos2θ ×1+cos2θ1+cos2θ
=± 12-cos2θ21+cos2θ2
= sinθ21+cos2θ
=sinθ1+cos2θ=RHS
Hence proved LHS = RHS
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