Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Simplify to a single trig function with no denominator.

csc^(2)theta*cos^(2)theta
Answer:

theta

Simplify to a single trig function with no denominator.\newlinecsc2θcos2θ \csc ^{2} \theta \cdot \cos ^{2} \theta \newlineAnswer:

Full solution

Q. Simplify to a single trig function with no denominator.\newlinecsc2θcos2θ \csc ^{2} \theta \cdot \cos ^{2} \theta \newlineAnswer:
  1. Recall definition of cosecant function: Recall the definition of the cosecant function. The cosecant function is the reciprocal of the sine function, so csc(θ)=1sin(θ)\text{csc}(\theta) = \frac{1}{\sin(\theta)}.
  2. Rewrite csc2θ\csc^2\theta: Rewrite csc2θ\csc^2\theta as (1/sin(θ))2(1/\sin(\theta))^2. Using the definition from Step 11, we can rewrite csc2θ\csc^2\theta as (1/sin(θ))2(1/\sin(\theta))^2.
  3. Apply power to the fraction: Apply the power to the fraction.\newlineWhen we raise a fraction to a power, we raise both the numerator and the denominator to that power. So, (1/sin(θ))2(1/\sin(\theta))^2 becomes 12/sin2θ1^2/\sin^{2}\theta, which simplifies to 1/sin2θ1/\sin^{2}\theta.
  4. Multiply the expressions: Multiply the expressions.\newlineNow we multiply 1sin2θ\frac{1}{\sin^{2}\theta} by cos2θ\cos^{2}\theta. This gives us (1sin2θ)cos2θ\left(\frac{1}{\sin^{2}\theta}\right) * \cos^{2}\theta.
  5. Use Pythagorean identity: Use the Pythagorean identity for sine and cosine.\newlineThe Pythagorean identity states that sin2θ+cos2θ=1\sin^{2}\theta + \cos^{2}\theta = 1. We can rearrange this to express sin2θ\sin^{2}\theta in terms of cos2θ\cos^{2}\theta: sin2θ=1cos2θ\sin^{2}\theta = 1 - \cos^{2}\theta.
  6. Substitute sin2θ\sin^2\theta: Substitute sin2θ\sin^2\theta in the expression.\newlineSubstitute sin2θ\sin^2\theta with 1cos2θ1 - \cos^2\theta in the expression from Step 44. This gives us 11cos2θ×cos2θ\frac{1}{1 - \cos^2\theta} \times \cos^2\theta.
  7. Simplify the expression: Simplify the expression.\newlineSince we have a common factor of cos2θ\cos^{2}\theta in the numerator and the denominator, we can simplify the expression to just cos2θ\cos^{2}\theta. This is because (1/(1cos2θ))cos2θ=cos2θ/(1cos2θ)(1cos2θ)=cos2θ(1/(1 - \cos^{2}\theta)) \cdot \cos^{2}\theta = \cos^{2}\theta / (1 - \cos^{2}\theta) \cdot (1 - \cos^{2}\theta) = \cos^{2}\theta.

More problems from Product property of logarithms