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balandron [24]
3 years ago
5

Evaluate -tan²A + 1 + sec²A

Mathematics
2 answers:
DiKsa [7]3 years ago
8 0

Answer:


Hello :

secA= 1/cosA

tanA = sinA / cosA

- tan²A + 1 + sec²A  = - (sinA / cosA)² + 1 + ( 1/cosA)²

                                 = - sin²A / cos²A  + 1 + 1 /cos²A

                                 = ( - sin²A  + cos²A +1)/cos²A

                                = ( 1 - sin²A + cos²A ) / cos²A

but : 1 - sin²A = cos²A   because : sin²A + cos²A = 1

so : - tan²A + 1 + sec²A  = ( 2 cos²A) / (cos²A)

- tan²A + 1 + sec²A  = 2



lubasha [3.4K]3 years ago
7 0

Answer:


Step-by-step explanation: sec^2A-tan^2A+1

=1+1

=2


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Triangle ABC has vertices A(–1, 0), B(4, 0), and C(2, 6). If ΔABC is translated using the function rule (x, y) → (x – 6, y – 5)
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Answer:

b.  A''(7, 5), B''(2, 5), C''(4, –1)

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Given vertices of ΔABC:

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Translation <u>mapping rule</u>:

(x, y) → (x – 6, y – 5)

Therefore:

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Rotation of 180° clockwise rule:

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Therefore:

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Learn more about transformations here:

brainly.com/question/28354239

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