prove this by the simplest method or that one with more easy to understand
Attachments:
Answers
Answered by
7
(1+cos alpha)(1+cos beta)(1+cos gamma)=(1-cos alpha)(1-cos beta)(1-cos gamma)
(1+cos alpha)(1+cos beta)(1+cos gamma) /(1-cos alpha)(1-cos beta)(1-cos gamma) =1
2cos square alpha ×2cos square beta × 2cos square gamma / 2sin square alpha ×2sin square beta × 2sin square gamma = 1
cos square alpha ×cos square beta × cos square gamma = sin square alpha × sin square beta × sin square gamma
cos alpha×cos beta×cos gamma = sin square alpha×sin square beta×sin square gamma
(1+cos alpha)(1+cos beta)(1+cos gamma) /(1-cos alpha)(1-cos beta)(1-cos gamma) =1
2cos square alpha ×2cos square beta × 2cos square gamma / 2sin square alpha ×2sin square beta × 2sin square gamma = 1
cos square alpha ×cos square beta × cos square gamma = sin square alpha × sin square beta × sin square gamma
cos alpha×cos beta×cos gamma = sin square alpha×sin square beta×sin square gamma
Sanaya625:
Plz mark my queation as brainliat for new rank
Answered by
5
★ TRIGONOMETRIC RESOLUTIONS ★
Hence , one of the value of each member of this equality is equal to (Sinα)(Sinβ)(Sinγ)
★✩★✩★✩★✩★✩★✩★✩★✩★✩★✩★
Hence , one of the value of each member of this equality is equal to (Sinα)(Sinβ)(Sinγ)
★✩★✩★✩★✩★✩★✩★✩★✩★✩★✩★
Attachments:
Similar questions