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aleksandrvk [35]

2 years ago

11

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

1 answer:

pav-90 [236]2 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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Amanda [17]

Answer:

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Step-by-step explanation:

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L= $\sqrt{(X2-X1)^{2}+(Y2-Y1)^{2}}$

For value of p:

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Since, Focus is left side of the vertex,

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Vertex: (h,k)=(4,-3)

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