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प्रश्न
Show that : sin 42° sec 48° + cos 42° cosec 48° = 2
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उत्तर
sin 42° sec 48° + cos 42° cosec 48° = 2
Consider sin 42° sec 48° + cos 42° cosec 48°
`=>` sin 42° sec (90° – 42°) + cos 42° cosec (90° – 42°)
`=>` sin 42° cosec 42° + cos 42° sec 42°
`=> sin 42^circ xx 1/(sin42^circ) + cos42^circ xx 1/(cos42^circ)`
`=>` 1 + 1 = 2
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