If sin 32=1 find 2 ?
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Answer:LHS = sin 32° . cos 58° + cos 32° . sin 58°
= sin 32° . cos (90-32)° + cos 32° . sin (90-32)°
= sin 32° . sin 32° + cos 32° . cos 32°
= sin²32° + cos²32°
= 1
= RHS
Therefore, LHS = RHS
sin 32° . cos 58° + cos 32° . sin 58° = 1
Hence proved!
Formuale used :-
sin A = cos (90-A)
cos A = sin (90-A)
sin²A + cos²A = 1
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