Question

question which of the following is equivalent to $\tan^2\thetacos(2\theta)$ for all values of $\theta$ for which $\tan^2\thetacos(2\theta)$ is defined? select the correct answer below: $2sin\theta$ $2\tan\theta$ $csc^2\theta - 2$ $cos\theta - sin\theta$ $2sin^2\theta - \tan^2\theta$

Explanation:

Step1: Recall trigonometric identities

We know that \(\tan\theta=\frac{\sin\theta}{\cos\theta}\), so \(\tan^{2}\theta = \frac{\sin^{2}\theta}{\cos^{2}\theta}\), and the double - angle formula for cosine: \(\cos(2\theta)=\cos^{2}\theta-\sin^{2}\theta\) or \(\cos(2\theta)=1 - 2\sin^{2}\theta\) or \(\cos(2\theta)=2\cos^{2}\theta - 1\). Also, the Pythagorean identity \(\csc^{2}\theta=\frac{1}{\sin^{2}\theta}\) and \(\sin^{2}\theta+\cos^{2}\theta = 1\), \(\cot^{2}\theta=\csc^{2}\theta - 1\).

First, substitute \(\tan^{2}\theta=\frac{\sin^{2}\theta}{\cos^{2}\theta}\) and \(\cos(2\theta)=\cos^{2}\theta-\sin^{2}\theta\) into the expression \(\tan^{2}\theta\cos(2\theta)\):

\(\tan^{2}\theta\cos(2\theta)=\frac{\sin^{2}\theta}{\cos^{2}\theta}\times(\cos^{2}\theta - \sin^{2}\theta)\)

Step2: Simplify the expression

\[ \begin{align*} \frac{\sin^{2}\theta}{\cos^{2}\theta}\times(\cos^{2}\theta-\sin^{2}\theta)&=\sin^{2}\theta-\frac{\sin^{4}\theta}{\cos^{2}\theta}\\ &=\sin^{2}\theta-\tan^{2}\theta\sin^{2}\theta\\ &=\sin^{2}\theta(1 - \tan^{2}\theta) \end{align*} \]
Another way: Use \(\cos(2\theta)=1 - 2\sin^{2}\theta\) and \(\tan^{2}\theta=\frac{\sin^{2}\theta}{\cos^{2}\theta}=\frac{\sin^{2}\theta}{1 - \sin^{2}\theta}\)

\(\tan^{2}\theta\cos(2\theta)=\frac{\sin^{2}\theta}{1 - \sin^{2}\theta}\times(1 - 2\sin^{2}\theta)=\frac{\sin^{2}\theta-2\sin^{4}\theta}{1 - \sin^{2}\theta}\)

Or use the identity \(\cos(2\theta)=\frac{1 - \tan^{2}\theta}{1+\tan^{2}\theta}\) (from the formula \(\cos(2\theta)=\frac{\cos^{2}\theta-\sin^{2}\theta}{\cos^{2}\theta+\sin^{2}\theta}=\frac{1 - \tan^{2}\theta}{1+\tan^{2}\theta}\) by dividing numerator and denominator by \(\cos^{2}\theta\))

Then \(\tan^{2}\theta\cos(2\theta)=\tan^{2}\theta\times\frac{1 - \tan^{2}\theta}{1+\tan^{2}\theta}\)

But let's try to simplify the option \(2\sin^{2}\theta-\tan^{2}\theta\):

We know that \(\tan^{2}\theta=\frac{\sin^{2}\theta}{\cos^{2}\theta}\), so \(2\sin^{2}\theta-\tan^{2}\theta=2\sin^{2}\theta-\frac{\sin^{2}\theta}{\cos^{2}\theta}=\sin^{2}\theta(2-\frac{1}{\cos^{2}\theta})=\sin^{2}\theta(\frac{2\cos^{2}\theta - 1}{\cos^{2}\theta})\)

Wait, let's go back to the original expression:

\(\tan^{2}\theta\cos(2\theta)=\frac{\sin^{2}\theta}{\cos^{2}\theta}\times(\cos^{2}\theta-\sin^{2}\theta)=\sin^{2}\theta-\frac{\sin^{4}\theta}{\cos^{2}\theta}\)

Now, let's simplify the option \(2\sin^{2}\theta-\tan^{2}\theta\):

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

And \(\tan^{2}\theta\cos(2\theta)=\frac{\sin^{2}\theta}{\cos^{2}\theta}\times(\cos^{2}\theta-\sin^{2}\theta)=\frac{\sin^{2}\theta\cos^{2}\theta-\sin^{4}\theta}{\cos^{2}\theta}=\frac{\sin^{2}\theta(\cos^{2}\theta-\sin^{2}\theta)}{\cos^{2}\theta}\)

Wait, maybe we made a wrong turn. Let's use the identity \(\cos(2\theta)=1 - 2\sin^{2}\theta\) and \(\tan^{2}\theta=\sec^{2}\theta - 1=\frac{1}{\cos^{2}\theta}-1\)

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

\(=\frac{1 - 2\sin^{2}\theta-( \cos^{2}\theta-2\sin^{2}\theta\cos^{2}\theta)}{\cos^{2}\theta}\) (since \(1 - 2\sin^{2}\theta=\cos^{2}\theta-\sin^{2}\theta\), no, better to use \(\sin^{2}\theta=1 - \cos^{2}\theta\))

Wait, let's try the option \(2\sin^{2}\theta-\tan^{2}\theta\):

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

Now, \(\tan^{2}\theta\cos(2\theta)=\frac{\sin^{2}\theta}{\cos^{2}\theta}\times(\cos^{2}\theta-\sin^{2}\theta)=\frac{\sin^{2}\theta\cos^{2}\theta-\sin^{4}\theta}{\cos^{2}\theta}=\frac{\sin^{2}\theta(\cos^{2}\theta-\sin^{2}\theta)}{\cos^{2}\theta}\)

We know that \(\cos(2\theta)=\cos^{2}\theta-\sin^{2}\theta = 2\cos^{2}\theta - 1=1 - 2\sin^{2}\theta\)

Let's substitute \(\cos^{2}\theta=1 - \sin^{2}\theta\) into \(2\sin^{2}\theta-\tan^{2}\theta\):

\(2\sin^{2}\theta-\frac{\sin^{2}\theta}{1 - \sin^{2}\theta}=\frac{2\sin^{2}\theta(1 - \sin^{2}\theta)-\sin^{2}\theta}{1 - \sin^{2}\theta}=\frac{2\sin^{2}\theta-2\sin^{4}\theta-\sin^{2}\theta}{1 - \sin^{2}\theta}=\frac{\sin^{2}\theta-2\sin^{4}\theta}{1 - \sin^{2}\theta}\)

Now, \(\tan^{2}\theta\cos(2\theta)=\frac{\sin^{2}\theta}{\cos^{2}\theta}\times(1 - 2\sin^{2}\theta)=\frac{\sin^{2}\theta(1 - 2\sin^{2}\theta)}{1 - \sin^{2}\theta}\) (since \(\cos^{2}\theta = 1-\sin^{2}\theta\))

Which is the same as \(\frac{\sin^{2}\theta-2\sin^{4}\theta}{1 - \sin^{2}\theta}\), so \(2\sin^{2}\theta-\tan^{2}\theta\) is equivalent to \(\tan^{2}\theta\cos(2\theta)\)

Let's check other options:

- Option \(2\sin\theta\): \(\tan^{2}\theta\cos(2\theta)\) is a quadratic in \(\sin\) or \(\cos\) terms, \(2\sin\theta\) is linear, so not equivalent.
- Option \(2\tan\theta\): \(\tan^{2}\theta\cos(2\theta)\) has a \(\tan^{2}\) term and a cosine term, \(2\tan\theta\) is linear in \(\tan\), not equivalent.
- Option \(\csc^{2}\theta - 2\): \(\csc^{2}\theta-2=\frac{1}{\sin^{2}\theta}-2=\frac{1 - 2\sin^{2}\theta}{\sin^{2}\theta}\), while \(\tan^{2}\theta\cos(2\theta)=\frac{\sin^{2}\theta(1 - 2\sin^{2}\theta)}{\cos^{2}\theta}\), not equivalent.
- Option \(\cos\theta-\sin\theta\): It is a linear combination of \(\cos\theta\) and \(\sin\theta\), while \(\tan^{2}\theta\cos(2\theta)\) is a product of a square of tangent and cosine of double angle, not equivalent.

Answer:

\(2\sin^{2}\theta-\tan^{2}\theta\)