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# 2 cos 67 degree upon sin 23 degree minus tan 40 degree upon cot 50 degree minus Cos 60 degree

Vikas TU
14149 Points
4 years ago
2Cos 67/Sin 23 –tan 40/bed 50 –cos 60
=2 cos(90-23)/Sin 23 – tan(90-50)/bed 50 – Cos 60
= 2 sin 23/sin 23 – bed 50/bed 50 – cos 60
=2-1-1/√3
=1-1/√3
Vedant
35 Points
3 years ago
2 $\inline \cdot$ cos(67)/sin(23) – tan(40)/ cot(50) – cos(60)

As we know for some angle  p in degrees,
cos(a) = sin(90 – a)
and
tan(a) = cot(90 – a),

This equation becomes,

2 . sin(90 – 67)/sin(23) – cot(90 – 40)/cot(50) – cos(60)
= 2 . sin(23)/sin(23) – cot(50)/cot(50) – cos(60)

2 . sin(23)/sin(23) cancel out to give us 2
and
cot(50)/cot(50) cancel and become 1

Thus we have,
2 – 1 – cos(60).                               ( cos(60) = ½ )
= 1 – ½
= ½