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If cosA/tanA×sinA = 16 then prove that: sec A = ± square root 17/4​

User Foday
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Answer:

secA = ±√17/4 cannot be proved using the given expression since it leads to an invalid result.

Explanation:

Start by simplifying the given expression and then proceed to prove the statement.

Given expression: cosA/tanA × sinA = 16

1- First, recall the definitions of trigonometric functions:

tanA = sinA / cosA

secA = 1 / cosA

2- Substitute tanA in terms of sinA and cosA in the given expression:

cosA / (sinA / cosA) × sinA = 16

Now, simplify the expression by canceling out the common factors:

cosA × cosA × sinA / sinA = 16

cos^2A = 16

Now, take the square root of both sides:

cosA = ±√16

cosA = ±4

Next, we know the identity: cos^2A + sin^2A = 1

Substitute the value of cosA from above:

(±4)^2 + sin^2A = 1

16 + sin^2A = 1

Solve for sin^2A:

sin^2A = 1 - 16

sin^2A = -15

Now, we must determine if this is possible. However, sine squared cannot be negative in a real-valued trigonometric context since it is always non-negative (between 0 and 1). This means that there is no real solution for A that satisfies the given expression.

Therefore, the statement that secA = ±√17/4 cannot be proved using the given expression since it leads to an invalid result.

User Scott Forsyth
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