At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
Answer:
[tex]2\sin(\theta)[/tex]
Step-by-step explanation:
[tex]\csc(\theta)=\dfrac{1}{\sin(\theta)}[/tex]
[tex]\sin^2(\theta)+\cos^2(\theta)=1\implies \sin^2(\theta)=1-\cos^2(\theta)[/tex]
[tex]2\csc(\theta)- 2cos^2(\theta)\times csc(\theta)[/tex]
[tex]=\dfrac{2}{\sin(\theta)}-\dfrac{ 2cos^2(\theta)}{\sin(\theta)}[/tex]
[tex]=\dfrac{ 2[1-cos^2(\theta)]}{\sin(\theta)}[/tex]
[tex]=\dfrac{ 2sin^2(\theta)}{\sin(\theta)}[/tex]
[tex]=2\sin(\theta)[/tex]
- Theta turned to A
[tex]\\ \rm\Rrightarrow 2cscA-2cos^2A(cscA)[/tex]
[tex]\\ \rm\Rrightarrow 2(1/sinA)-2cos^2A(1/sinA)[/tex]
[tex]\\ \rm\Rrightarrow \dfrac{2}{sinA}-\dfrac{2cos^2A}{sinA}[/tex]
[tex]\\ \rm\Rrightarrow \dfrac{2-2cos^2A}{sinA}[/tex]
[tex]\\ \rm\Rrightarrow \dfrac{2(1-cos^2A)}{sinA}[/tex]
[tex]\\ \rm\Rrightarrow \dfrac{2sin^2A}{sinA}[/tex]
[tex]\\ \rm\Rrightarrow sinA[/tex]
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.