how to solve the question in the photo
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tan = per/base
cot=base/perpendicular
base=1 perpendicular=√3
hypotenuse is 1+3=4
cos =b/h=1/4
sin=p/h=√3/4
(1-1/16)/(2-3/16)
(15/16)/(32-3/16)=15/29
cot=base/perpendicular
base=1 perpendicular=√3
hypotenuse is 1+3=4
cos =b/h=1/4
sin=p/h=√3/4
(1-1/16)/(2-3/16)
(15/16)/(32-3/16)=15/29
Answered by
1
It is given that cot theta is 1/√3
We know that cot 60° = 1/√3
So, theta =60°
To prove - (1-cos²theta)/(2-sin²theta) =3/5
So, putting the value of theta as 60° in LHS, we get
(1-cos²60°)/(2-sin²60°)
=[1-(1/2)²] / [2-(√3/2)²]
=[1-(1/4)] / [2-(3/4)]
=[(4-1)/4] / [(8-3)/4]
=(3/4) / (5/4)
=3/5
=LHS
Hence, proved
We know that cot 60° = 1/√3
So, theta =60°
To prove - (1-cos²theta)/(2-sin²theta) =3/5
So, putting the value of theta as 60° in LHS, we get
(1-cos²60°)/(2-sin²60°)
=[1-(1/2)²] / [2-(√3/2)²]
=[1-(1/4)] / [2-(3/4)]
=[(4-1)/4] / [(8-3)/4]
=(3/4) / (5/4)
=3/5
=LHS
Hence, proved
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