If \pi < \theta < \frac{3\pi}{2} and \tan \theta = \frac{3}{4}, what is the value of the e… — ACT

If $\pi < \theta < \frac{3\pi}{2}$ and $\tan \theta = \frac{3}{4}$ , what is the value of the expression $\frac{\sin^2 \theta}{1 + \cos \theta}$ ?

\frac{9}{5}

The expression

\frac{\sin^2 \theta}{1 + \cos \theta}

can be simplified using the Pythagorean identity

\sin^2 \theta = 1 - \cos^2 \theta

. Factoring the numerator gives

(1 - \cos \theta)(1 + \cos \theta)

, which cancels with the denominator to leave

1 - \cos \theta

. Given

\tan \theta = \frac{3}{4}

in Quadrant III, the reference right triangle has sides 3 and 4 with hypotenuse 5. Since cosine is negative in the third quadrant,

\cos \theta = -\frac{4}{5}

. Evaluating

1 - \left(-\frac{4}{5}\right)

yields

\frac{9}{5}

Simplify the algebraic trigonometric expression using standard fundamental identities.

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

Applying the Pythagorean identity

\sin^2 \theta + \cos^2 \theta = 1

allows factoring and canceling terms.

Determine the value and sign of

\cos \theta

based on the given tangent ratio and quadrant constraint.

\cos \theta = -\frac{4}{5}

Since

\tan \theta = \frac{3}{4} = \frac{\text{opposite}}{\text{adjacent}}

, the hypotenuse is

5

. In Quadrant III (

\pi < \theta < \frac{3\pi}{2}

), cosine is negative.

Substitute the value of

\cos \theta

into the simplified expression.

1 - \left(-\frac{4}{5}\right) = 1 + \frac{4}{5} = \frac{9}{5}

Subtracting a negative value results in addition.

Fundamental Pythagorean Identities and Quadrant Signs of Trigonometric Functions

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