Math, asked by DevanshGoel4635, 1 year ago

Cot alpha + cot (60 + Alpha) -Cot (60 -alpha)=3cot alpha

Answers

Answered by ShreyaBhowmick
10
OR
cota+cot(60+a)−cot(60−a)=cota+cos(60+a)sin(60+a)−cos(60−a)sin(60−a)=cota+cos(60+a)sin(60−a)−cos(60−a)sin(60+a)sin(60+a)sin(60−a)=cot a+sin{(60−a)−(60+a)sin(60+a)sin(60−a)=cota −sin2asin260−sin2a=cosasina−sin2a34−sin2a=(34−sin2a)cosa−sin2a(sina)sina(34−sin2a)=(3−4sin2a)cosa−2*4sinacosa sinasina(3−4sin2a)=3cosa−4sin2acosa−8sin2acosa3sina−4sin3a=3cosa−12sin2acosa3sina−4sin3a=3cosa−12(1−cos2a)cosa3sina−4sin3a=12cos3a−9cosa3sina−4sin3a=3(4cos3a−3cosa)3sina−4sin3a=3cos3asin3a=3cot3a
Attachments:
Answered by utcrush18
6

L.H.S. =  

sinα

cos

sin(60  

+α)

cos(60  

+α)

​  

−  

sin(60  

−α)

cos(60  

−α)

​  

 

sin(60  

−α)cos(60  

+α)

=  

sinα

cosα

​  

+  

sin  

2

−60  

−sin  

2

α

−cos(60  

−α)sin(60  

+α)

​  

 

=  

sinα(  

4

3

​  

−sin  

2

α)

cosα(  

4

3

​  

−sin  

2

α)+sinαsin(60  

−α−60  

−α)

​  

 

=  

4sinα(  

4

3−4sin  

2

α

​  

)

cosα(3−4sin  

2

α)+4sinαsin(−2α)

​  

 

=  

3sinα−4sin  

3

α

cosalpha(3−4sin  

2

α)−4sinα⋅sinαcosα

​  

 

=  

sin3α

cosα[3−12(1−cos  

2

α)]

​  

 

=3⋅  

sin3alpha

[4cos  

3

α−3cosalpha]

​  

=3⋅  

sin3alpha

cos3alpha

​  

=3cot3α

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