Simplifying the Expression (1/a)arcsec(u/a) + c using Trigonometric Identities

(1/. a)arcsec(u/. a)+c =

To simplify the expression, we can start by using the trigonometric identity for the arcsecant function:

arcsec(x) = arccos(1/x)

Applying this identity to (1/a)arcsec(u/a), we get:

(1/a)arcsec(u/a) = (1/a)arccos(a/u)

Next, we can simplify further by multiplying the numerator and denominator of the arccos term by a to get:

(1/a)arccos(a/u) = (1/a)arccos(a/u) * a/a = (1/a)arccos(a^2/au)

Now, let’s rewrite the expression (1/a)arccos(a^2/au) + c in a different form

To simplify the expression, we can start by using the trigonometric identity for the arcsecant function:

arcsec(x) = arccos(1/x)

Applying this identity to (1/a)arcsec(u/a), we get:

(1/a)arcsec(u/a) = (1/a)arccos(a/u)

Next, we can simplify further by multiplying the numerator and denominator of the arccos term by a to get:

(1/a)arccos(a/u) = (1/a)arccos(a/u) * a/a = (1/a)arccos(a^2/au)

Now, let’s rewrite the expression (1/a)arccos(a^2/au) + c in a different form. Since we know that arccos(x) + arcsin(x) = π/2, we can rewrite the expression as:

(1/a)arccos(a^2/au) + c = (1/a)(arccos(a^2/au) + arcsin(a^2/au)) + c

Using the sum of arcsin and arccos identity, the above expression becomes:

(1/a)(arccos(a^2/au) + arcsin(a^2/au)) + c = (1/a)arcsin((au – a^2)/au) + c

Thus, the simplified expression is:

(1/a)arcsec(u/a) + c = (1/a)arcsin((au – a^2)/au) + c

More Answers:

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Exploring the Derivative of the Inverse Sine Function and its Application in Calculus
How to Integrate (1/a)arctan(u/a) + c Using Trigonometric Functions and Integration

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