Q18 of 44 Page 79

Prove that-

(sec230° + cosec245°) (2cos 60° + sin 90° + tan 45°) = 10



cosec45° = √2


∴ cosec245° = 2



sin90° = 1


tan45° = 1


∴ LHS = (sec230° + cosec245°) (2cos 60° + sin 90° + tan 45°)





= 10 = RHS


∴ LHS = RHS


∴(sec230° + cosec245°) (2cos 60° + sin 90° + tan 45°) = 10


Hence Proved


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