For an acute angle x measured in degrees, \sin(x) - \sin(90^\circ - x) = \frac{1}{5}. What… — SAT

For an acute angle $x$ measured in degrees, $\sin(x) - \sin(90^\circ - x) = \frac{1}{5}$ . What is the value of $12(\tan(x) + \tan(90^\circ - x))$ ?

Answer: 25

The value of the expression is 25.

Applying the co-function identity

\sin(90^\circ - x) = \cos(x)

allows the given equation to be written as

\sin(x) - \cos(x) = \frac{1}{5}

. Squaring both sides of this equation yields

\sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) = \frac{1}{25}

. Applying the Pythagorean identity

\sin^2(x) + \cos^2(x) = 1

simplifies this to

1 - 2\sin(x)\cos(x) = \frac{1}{25}

, which gives

\sin(x)\cos(x) = \frac{12}{25}

. The expression to be evaluated is

12(\tan(x) + \tan(90^\circ - x))

. Using the identity

\tan(90^\circ - x) = \cot(x)

, this expression can be rewritten as

12\left(\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}\right) = 12\left(\frac{\sin^2(x) + \cos^2(x)}{\sin(x)\cos(x)}\right) = 12\left(\frac{1}{\sin(x)\cos(x)}\right)

. Substituting the value of

\sin(x)\cos(x)

gives

12 \times \frac{25}{12} = 25

Apply the co-function identity to rewrite the equation.

\sin(x) - \cos(x) = \frac{1}{5}

Since

\sin(90^\circ - x) = \cos(x)

for any angle

x

, we can substitute

\cos(x)

into the given equation.

Square both sides of the rewritten equation.

\sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) = \frac{1}{25}

Squaring both sides allows us to use the Pythagorean trigonometric identity to find the product of sine and cosine.

Substitute the Pythagorean identity and solve for

\sin(x)\cos(x)

\sin(x)\cos(x) = \frac{12}{25}

Substituting

\sin^2(x) + \cos^2(x) = 1

yields

1 - 2\sin(x)\cos(x) = \frac{1}{25}

, which simplifies to

\sin(x)\cos(x) = \frac{12}{25}

Rewrite the target expression in terms of sine and cosine.

\tan(x) + \tan(90^\circ - x) = \frac{1}{\sin(x)\cos(x)}

Using the co-function identity

\tan(90^\circ - x) = \cot(x)

and expressing the tangent and cotangent functions as ratios of sine and cosine yields

\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} = \frac{\sin^2(x) + \cos^2(x)}{\sin(x)\cos(x)} = \frac{1}{\sin(x)\cos(x)}

Substitute the value of

\sin(x)\cos(x)

and multiply by 12.

Substituting

\sin(x)\cos(x) = \frac{12}{25}

into

12(\tan(x) + \tan(90^\circ - x))

gives

12 \times \frac{25}{12} = 25

Applying co-function identities, the Pythagorean identity, and fundamental trigonometric relations to simplify expressions.

Estimated Time:2m 0s

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