Sin(n+1)a+2sina+sin(n-1)a/cos(n-1)a-cos(n+1)a=cota/2
Answers
Answer:
Step-by-step explanation:
n(n+1)A+2sinnA+sin(n−1)A=sin(nA+A)+sin(nA−A)+2sinnA
=2sinnAcosA+2sinnA (∵sin(a+b)+sin(a−b)=2sinacosb)
=2sinnA(1+cosA)
=4sinnAcos
2
2
A
(∵1+cos2A=2cos
2
A)
cos(n−1)A−cos(n+1)A=cos(nA−A)−cos(nA+A)
=2sinnAsinA (∵cos(a−b)−cos(a+b)=2sinasinb)
=4sinnAsin
2
A
cos
2
A
(∵sin2A=2sinAcosA)
cos(n−1)A−cos(n+1)A
sin(n+1)A+2sinnA+sin(n−1)A
=
4sinnAsin
2
A
cos
2
A
4sinnAcos
2
2
A
=
sin
2
A
cos
2
A
=cot
2
A