Answer to Question #173512 in Trigonometry for Jon jay Mendoza

Question

1. Is sec (−𝜽) equal to sec 𝜽 or - sec 𝜽? How do you know? 
2. Verify the identity 1 - sin² x cot² x = sin² x. Is there more than one way to verify the identity? If so, tell which way you think is easier and why. 
3. Describe what is wrong with the simplification shown.

cos x - cos x sin² x = cos x - cos x (1 + cos² x)

          = cos x - cos x - cos³ x

          = -cos³ x

4. John said 𝒔𝒊𝒏 𝒙 + 𝒄𝒐𝒔 𝒙 = 𝟐 has no solution. Do you agree with John? Explain why or why not?

Verify identity:

1. 𝒄𝒐𝒕²𝜽+𝟏 / 𝒄𝒐𝒕²𝜽 ≡ 𝒔𝒆𝒄²𝜽 
2.  (𝒄𝒔𝒄²𝜽 − 𝟏)𝒔𝒊𝒏²𝜽≡ 𝒄𝒐𝒔²𝜽
3. 𝟏 − 𝒔𝒆𝒄 𝜶 𝒄𝒐𝒔 𝜶 ≡ 𝒕𝒂𝒏 𝜶 𝒄𝒐𝒕 𝜶 − 1
4. 𝒕𝒂𝒏 𝑨+𝒄𝒐𝒕 𝑨 / 𝒔𝒆𝒄 𝑨 𝒄𝒔𝒄 𝑨 ≡ 1
5. 𝟏 + 𝟐 𝒕𝒂𝒏²𝜽 ≡ 𝒔𝒆𝒄⁴𝜽 − 𝒕𝒂𝒏⁴𝜽

Answer 1

Solution.

1.

$\sec{(-\theta)}=\sec{\theta},$

because $\sec\theta=\frac{1}{\cos \theta}$ and $\cos(-\theta)=\cos \theta.$

2.

$1-\sin^2x\cot^2x=1-\sin^2x\frac{\cos^2x}{\sin^2x}=\newline =1-\cos^2x=\sin^2x.$

As for me, this way is easier, because it is a way of proving by definition.

3.

   $\cos x-\cos x\sin^2x=\cos x-\cos x(1-\cos^2x)=\newline =\cos x-\cos x+\cos^3x=\cos^3x.$

4.

I agree with John, because this equation has no solution.

$\sin x+\cos x=2,$

$\sqrt{2}\sin(x+\frac{\pi}{4})=2,$

$\sin(x+\frac{\pi}{4})=\frac{2}{\sqrt{2}},$

and $\frac{2}{\sqrt{2}}>1.$

This equation has no real solutions.

Verify identity:

1.

$\cot^2\theta+\frac{1}{\cot^2\theta}=\cot^2\theta+\tan^2\theta,\newline \sec^2\theta=\frac{1}{\cos^2\theta}=1+\tan^2\theta.$

But $1\neq \cot^2\theta.$ This equality is not an identity.

$\cot^2\theta+\frac{1}{\cot^2\theta}\neq\sec^2\theta.$

2.

$(\csc^2\theta-1)\sin^2\theta=(\frac{1}{\sin^2\theta}-1)\sin^2\theta=\newline=\frac{1-\sin^2\theta}{\sin^2\theta}\sin^2\theta=1-\sin^2\theta=\cos^2\theta.$

3.

$1-\sec\alpha\cos\alpha=0,$ and $\tan\alpha\cot\alpha-1=0.$

So, $1-\sec\alpha\cos\alpha=\tan\alpha\cot\alpha-1.$

4.

$\tan A+\frac{\cot A}{\sec A\csc A}=\tan A+\frac{\cos A\cdot \cos A\sin A}{\sin A}= \newline=\tan A+\cos^2A=\frac{\sin A+\cos^3A}{\cos A}\neq 1.$

This equality is not an identity.

5.

$\sec^4\theta-\tan^4\theta=(\sec^2\theta-\tan^2\theta)(\sec^2\theta+\tan^2\theta)=\newline =\frac{1-sin^2\theta}{\cos^2\theta}\cdot\frac{1+sin^2\theta}{\cos^2\theta}= \frac{\cos^2\theta+\sin^2\theta+\sin^2\theta}{\cos^2\theta}=1+2\tan^2\theta.$