Answer: sin²∅
Explanation: If you divide each term by sin squared theta, sin²∅/sin²∅ = 1, cos²∅/sin²∅ = cot²∅, and 1/sin²∅ = csc²∅, which is your result.
Answer:
sin²θ
Step-by-step explanation:
To determine what we need to divide [tex]\sin^2 \theta+\cos^2 \theta=1[/tex] by to yield [tex]1+\cot^2 \theta=\csc^2\theta[/tex], compare equations:
[tex]\textsf{Equation 1}: \quad \sin^2 \theta+\cos^2 \theta=1[/tex]
[tex]\textsf{Equation 2}: \quad 1+\cot^2 \theta=\csc^2\theta[/tex]
If we divide a term by itself, it will always yield 1.
Therefore, divide each term in the first equation by sin²θ:
[tex]\implies \dfrac{\sin^2 \theta}{\sin^2 \theta}+\dfrac{\cos^2 \theta}{\sin^2 \theta}=\dfrac{1}{\sin^2 \theta}[/tex]
[tex]\implies 1+\dfrac{\cos^2 \theta}{\sin^2 \theta}=\dfrac{1}{\sin^2 \theta}[/tex]
[tex]\boxed{\begin{minipage}{4 cm}\underline{Trigonometric Identities}\\\\$\cot \theta=\dfrac{\cos \theta}{\sin \theta}\\\\\csc \theta=\dfrac{1}{\sin \theta}$\\\end{minipage}}[/tex]
Use the trigonometric identities for cot and cosec:
[tex]\implies 1+\dfrac{\cos^2 \theta}{\sin^2 \theta}=\dfrac{1}{\sin^2 \theta}[/tex]
[tex]\implies 1+\left(\dfrac{\cos \theta}{\sin \theta}\right)^2=\left(\dfrac{1}{\sin \theta}\right)^2[/tex]
[tex]\implies 1+(\cot \theta)^2=(\csc \theta)^2[/tex]
[tex]\implies 1+\cot^2 \theta=\csc^2 \theta[/tex]
Thus proving that sin²θ + cos²θ = 1 should be divided by sin²θ to yield
1 + cot²θ = csc²θ.
