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Pre-Calculus help please!?

0 votes

what is the numerical value of 

[-sin^2 (x) - cos^2 (x) +6] / [csc^2 (x) - cot^2 (x) +2]

asked Nov 18, 2014 in PRECALCULUS by anonymous

1 Answer

+1 vote

[- sin²x - cos²x +6] / [csc²x - cot²x +2]

= [6 - ( sin²x + cos²x )] / [csc²x - cot²x +2]

Substitute trigonometric identity : sin²x + cos²x = 1

= [ 6 - (1) ] / [csc²x - cot²x +2]

Substitute trigonometric identity : csc²x - cot²x = 1

= [ 6 - 1 ] / [1 +2]

= 5/3

Solution : The numerical value of [- sin²x - cos²x +6] / [csc²x - cot²x +2]  is 5/3

answered Nov 18, 2014 by Shalom Scholar

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