cot70⁰+4cos70⁰=√3

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

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)

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

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

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.