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S=1/(sin A*sin 2A)+1/(sin 2A*sin3A)+1/(sin 3A*sin 4A)+1/(sin 4A*sin 5A).... or S=cosec A*cosec 2A+cosec 2A*cosec3A+cosec 3A*cosec 4A+ cosec 4A*cosec 5A........ find S

S=1/(sin A*sin 2A)+1/(sin 2A*sin3A)+1/(sin 3A*sin 4A)+1/(sin 4A*sin 5A)....
or 
S=cosec A*cosec 2A+cosec 2A*cosec3A+cosec 3A*cosec 4A+ cosec 4A*cosec 5A........
find S

Grade:12

1 Answers

Rituraj Tiwari
askIITians Faculty 1792 Points
2 years ago
sinA+sin2A+sin3A=1
⇒sinA(1+sin2A)=1−sin2A
⇒sinA(2−cos2A)=cos2A
⇒sin2A(4−4cos2A+cos4A)=cos4A
⇒(1−cos2A)(4−4cos2A+cos4A)=cos4A
⇒4−8cos2A+5cos4A−cos6A=cos4A
∴cos6A−4cos4A+8cos2A=4

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