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cot70 0 +4cos70 0 = √3

cot700+4cos700=√3

Grade:

6 Answers

Latika Leekha
askIITians Faculty 165 Points
6 years ago
Hello student,
cot700 + 4cos700
= (cos 700 )/(sin 700) + 4cos700
= cos 700 (1/sin 700) + 4
We cannot go any further without the help of tables. Taking the help of tables for the sine values, the above expression reduces to
0.34 (1/0.94 + 4)
=1.73 (approx)
= √3
Tapan Kumar Mondal
9 Points
3 years ago
cot(70)+4cos(70)=cos(70)sin(70)+4cos(70)=cos(70)+4cos(70)sin(70)sin(70)=cos(70)+2sin(140)sin(70)=(cos(70)+sin(140))+sin(140)sin(70)=(sin(20)+sin(140))+sin(140)sin(70)=2sin(80)cos(60)+sin(140)sin(70)=2sin(110)cos(30)sin(70)=2∗3–√2=3–√cot⁡(70)+4cos⁡(70)=cos⁡(70)sin⁡(70)+4cos⁡(70)=cos⁡(70)+4cos⁡(70)sin⁡(70)sin⁡(70)=cos⁡(70)+2sin⁡(140)sin⁡(70)=(cos⁡(70)+sin⁡(140))+sin⁡(140)sin⁡(70)=(sin⁡(20)+sin⁡(140))+sin⁡(140)sin⁡(70)=2sin⁡(80)cos⁡(60)+sin⁡(140)sin⁡(70)=2sin⁡(110)cos⁡(30)sin⁡(70)=2∗32=3Note sin(110)=sin(70)sin⁡(110)=sin⁡(70) at the end. cos(60)=12cos⁡(60)=12 and cos(30)=3√2cos⁡(30)=32Identities used: sin(90−x)=cos(x)sin⁡(90−x)=cos⁡(x)sin(a+b)=sin(a)cos(b)+cos(a)sin(b)sin⁡(a+b)=sin⁡(a)cos⁡(b)+cos⁡(a)sin⁡(b)sin(2x)=2sin(x)cos(x)sin⁡(2x)=2sin⁡(x)cos⁡(x)
Anshul soni
11 Points
3 years ago
Cot70+4cos70(Cos70/sin70)+ 4cos70(Cos70+2.2cos70sin70)/sin70(Cos70+2sin140)/sin70{Cos70+2sin(180-40)}/sin70(Cos70+2sin40)/sin70{Cos(90-20)+sin40+sin40)}/sin(90-20){(Sin20+sin40)+sin40}/cos20{2sin30.cos10+sin40}/cos20{Cos10+sin(90-50)}/cos20{Cos10+cos50}/cos20{2cos30.cos20}/cos202cos30√3=Answer
yathartha gupta
71 Points
3 years ago
I think the solution of this should becot700 + 4cos700= (cos 700 )/(sin 700) + 4cos700= cos 700 (1/sin 700) + 4We cannot go any further without the help of tables. Taking the help of tables for the sine values, the above expression reduces to0.34 (1/0.94 + 4)=1.73 (approx)= √3
Vivek
70 Points
3 years ago
It`s best way is to eliminate options if there are present or otherwise take approximation of valuesLike cos70 is some value less than cos60 this how we can solve easily.that is by approx.
Aarushi Ahlawat
41 Points
2 years ago
Here is the solution without using table except of common angles 30 and 90 degrees. Follwing text is in Latex equation format.
=cot(70)+4cos(70)
=tan(20)+4cos(70)
=\frac{sin(20)+4cos(70)cos(20)}{cos(20)}
=\frac{sin(20)+2(cos(70+20)+cos(70-20))}{cos(20)}
=\frac{sin(20)+2(cos(90)+cos(50))}{cos(20)}
=\frac{sin(20)+2(0+cos(30+20))}{cos(20)}
=\frac{sin(20)+2(cos(30)cos(20)-sin(30)sin(20))}{cos(20)}
=\frac{sin(20)+2(\frac{\sqrt{3}}{2}cos(20)-\frac{1}{2}sin(20))}{cos(20)}
=\frac{sin(20)+\sqrt{3}cos(20)-sin(20)}{cos(20)}
=\frac{\sqrt{3}cos(20)}{cos(20)}
=\sqrt{3}
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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