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प्रश्न
Prove that sin 48° sec 42° + cos 48° cosec 42° = 2
उत्तर
We will simplify the left-hand side
`sin 48° sec 42° + cos 48° cosec 42° = sin 48^@. sec(90^@ - 48^@) + cos 48^@ cosec (90^@ - 48^@)`
`= sin 48^@. sec 48^@ + cos 48^@. cosec 42^@ = sin 48^@ . sec(90^@ - 48^@) + cos 48^@.cosec (90^@ - 48^@)`
`= sin 48^@.cos 48^@ + cos 48^@. sin 48^@`
= 1 + 1
= 2
Proved
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