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3 years ago

.prove that cos (90-A).sinA + sin(90 - A).cosA = 1

Mathematics

1 answer:

pav-90 [236]3 years ago

4 0

Answer:

See Explanation

Step-by-step explanation:

$cos (90-A).sinA + sin(90 - A).cosA = 1 \\ \\ LHS = cos (90-A).sinA + sin(90 - A).cosA \\ \\ = sinA.sinA + cosA.cosA\\ \\ = sin ^{2} A + cos^{2} A \\ \\ = 1 \\ \\=RHS\\\\ reason \\ cos (90- \theta) = sin \theta \\ sin (90- \theta) = cos \theta \\ sin ^{2} \theta + cos^{2} \theta = 1 \\ \\ thus \: proved \\$

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24 - 4 x 2 + 4b = 20 - 16

AnnZ [28]

Answer:

b = -3

Step-by-step explanation:

Solve for b:

4 b - 4 2 + 24 = 20 - 16

-4×2 = -8:

4 b + -8 + 24 = 20 - 16

Grouping like terms, 4 b - 8 + 24 = 4 b + (24 - 8):

4 b + (24 - 8) = 20 - 16

24 - 8 = 16:

4 b + 16 = 20 - 16

20 - 16 = 4:

4 b + 16 = 4

Subtract 16 from both sides:

4 b + (16 - 16) = 4 - 16

16 - 16 = 0:

4 b = 4 - 16

4 - 16 = -12:

4 b = -12

Divide both sides of 4 b = -12 by 4:

(4 b)/4 = (-12)/4

4/4 = 1:

b = (-12)/4

The gcd of -12 and 4 is 4, so (-12)/4 = (4 (-3))/(4×1) = 4/4×-3 = -3:

Answer: b = -3

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Step-by-step explanation:

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4 years ago

.prove that cos (90-A).sinA + sin(90 - A).cosA = 1 ​

.prove that cos (90-A).sinA + sin(90 - A).cosA = 1