Prove that sin 48° sec 48° + cos 48° cosec 42° = 2

Asked by Sakshi | 9 months ago |  134

1 Answer

Solution :-

Taking L.H.S = sin 48° sec 48° + cos 48° cosec 42°

= sin 48° sec (90° − 48°) + cos 48° cosec (90° − 48°)

[∵sec (90 – θ) = cosec θ and cosec (90 – θ) = sec θ]

= sin 48°cosec 48° + cos 48°sec 48°  [∵ cosec θ x sin θ = 1 and cos θ x sec θ = 1]

= 1 + 1 = 2

Hence proved

Answered by Sakshi | 9 months ago

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