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Asked by ishansheoran7 | 10 May, 2024, 04:42: PM

Let  ( sinθ + cosecθ ) = α

( sinθ + cosecθ )5 = sin5θ + 5C1 sin4θ cosecθ + 5C2 sin3θ cosec2θ + 5C3 sin2θ cosec3θ + 5C4 sinθ cosec4θ + cosec5θ

( sinθ + cosecθ )5 = sin5θ + cosec5θ + 5 ( sin3θ + cosec3θ ) + 10 ( sinθ + cosecθ )

sin5θ + cosec5θ = ( sinθ + cosecθ )  - 5 ( sin3θ + cosec3θ ) - 10 ( sinθ + cosecθ )

sin5θ + cosec5θ = ( sinθ + cosecθ )  - 5 ( sinθ + cosecθ ) ( sin2θ + 3 (sinθ + cosecθ ) + cosec2θ) - 10 ( sinθ + cosecθ )

sin5θ + cosec5θ = ( sinθ + cosecθ )  - 5 ( sinθ + cosecθ ) [ (sinθ + cosecθ )2 - 2 +3 (sinθ + cosecθ )  ] - 10 ( sinθ + cosecθ )

sin5θ + cosec5θ = α5 - 5 α [ α2 + 3 α -2 ] - 10 α

sin5θ + cosec5θ = α5 - 5 α3 - 15 α2 = α [ α4- 5 α2- 15 α ]

Answered by Thiyagarajan K | 10 May, 2024, 08:08: PM
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