Answer:
At x= 4 the given expression is undefined.
Step-by-step explanation:
In a given function , if the denominator becomes zero then the function is undefined.
Here, denominator is x-4 so at x = 4 it will become zero. So we can say that at x=4 , the function is undefined
![\bf [cot(\theta )+csc(\theta )]^2=\cfrac{1+cos(\theta )}{1-cos(\theta )} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{doing the left-hand side}}{[cot(\theta )+csc(\theta )]^2}\implies cot^2(\theta )+2cot(\theta )csc(\theta )+csc^2(\theta ) \\\\\\ \cfrac{cos^2(\theta )}{sin^2(\theta )}+2\cdot \cfrac{cos(\theta )}{sin(\theta )}\cdot \cfrac{1}{sin(\theta )}+\cfrac{1}{sin^2(\theta )}\implies \cfrac{cos^2(\theta )}{sin^2(\theta )}+\cfrac{2cos(\theta )}{sin^2(\theta )}+\cfrac{1}{sin^2(\theta )}](https://tex.z-dn.net/?f=%5Cbf%20%5Bcot%28%5Ctheta%20%29%2Bcsc%28%5Ctheta%20%29%5D%5E2%3D%5Ccfrac%7B1%2Bcos%28%5Ctheta%20%29%7D%7B1-cos%28%5Ctheta%20%29%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bdoing%20the%20left-hand%20side%7D%7D%7B%5Bcot%28%5Ctheta%20%29%2Bcsc%28%5Ctheta%20%29%5D%5E2%7D%5Cimplies%20cot%5E2%28%5Ctheta%20%29%2B2cot%28%5Ctheta%20%29csc%28%5Ctheta%20%29%2Bcsc%5E2%28%5Ctheta%20%29%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7Bcos%5E2%28%5Ctheta%20%29%7D%7Bsin%5E2%28%5Ctheta%20%29%7D%2B2%5Ccdot%20%5Ccfrac%7Bcos%28%5Ctheta%20%29%7D%7Bsin%28%5Ctheta%20%29%7D%5Ccdot%20%5Ccfrac%7B1%7D%7Bsin%28%5Ctheta%20%29%7D%2B%5Ccfrac%7B1%7D%7Bsin%5E2%28%5Ctheta%20%29%7D%5Cimplies%20%5Ccfrac%7Bcos%5E2%28%5Ctheta%20%29%7D%7Bsin%5E2%28%5Ctheta%20%29%7D%2B%5Ccfrac%7B2cos%28%5Ctheta%20%29%7D%7Bsin%5E2%28%5Ctheta%20%29%7D%2B%5Ccfrac%7B1%7D%7Bsin%5E2%28%5Ctheta%20%29%7D)
![\bf \cfrac{\stackrel{\textit{perfect square trinomial}}{cos^2(\theta )+2cos(\theta )+1}}{sin^2(\theta )}\implies \boxed{\cfrac{[cos(\theta )+1]^2}{sin^2(\theta )}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{doing the right-hand-side}}{\cfrac{1+cos(\theta )}{1-cos(\theta )}}\implies \stackrel{\textit{multiplying by the denominator's conjugate}}{\cfrac{1+cos(\theta )}{1-cos(\theta )}\cdot \cfrac{1+cos(\theta )}{1+cos(\theta )}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Cstackrel%7B%5Ctextit%7Bperfect%20square%20trinomial%7D%7D%7Bcos%5E2%28%5Ctheta%20%29%2B2cos%28%5Ctheta%20%29%2B1%7D%7D%7Bsin%5E2%28%5Ctheta%20%29%7D%5Cimplies%20%5Cboxed%7B%5Ccfrac%7B%5Bcos%28%5Ctheta%20%29%2B1%5D%5E2%7D%7Bsin%5E2%28%5Ctheta%20%29%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bdoing%20the%20right-hand-side%7D%7D%7B%5Ccfrac%7B1%2Bcos%28%5Ctheta%20%29%7D%7B1-cos%28%5Ctheta%20%29%7D%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20by%20the%20denominator%27s%20conjugate%7D%7D%7B%5Ccfrac%7B1%2Bcos%28%5Ctheta%20%29%7D%7B1-cos%28%5Ctheta%20%29%7D%5Ccdot%20%5Ccfrac%7B1%2Bcos%28%5Ctheta%20%29%7D%7B1%2Bcos%28%5Ctheta%20%29%7D%7D)
![\bf \cfrac{[1+cos(\theta )]^2}{\underset{\textit{difference of squares}}{[1-cos(\theta )][1+cos(\theta )]}}\implies \cfrac{[cos(\theta )+1]^2}{1^2-cos^2(\theta )} \\\\\\ \cfrac{[cos(\theta )+1]^2}{1-cos^2(\theta )}\implies \boxed{\cfrac{[cos(\theta )+1]^2}{sin^2(\theta )}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5B1%2Bcos%28%5Ctheta%20%29%5D%5E2%7D%7B%5Cunderset%7B%5Ctextit%7Bdifference%20of%20squares%7D%7D%7B%5B1-cos%28%5Ctheta%20%29%5D%5B1%2Bcos%28%5Ctheta%20%29%5D%7D%7D%5Cimplies%20%5Ccfrac%7B%5Bcos%28%5Ctheta%20%29%2B1%5D%5E2%7D%7B1%5E2-cos%5E2%28%5Ctheta%20%29%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B%5Bcos%28%5Ctheta%20%29%2B1%5D%5E2%7D%7B1-cos%5E2%28%5Ctheta%20%29%7D%5Cimplies%20%5Cboxed%7B%5Ccfrac%7B%5Bcos%28%5Ctheta%20%29%2B1%5D%5E2%7D%7Bsin%5E2%28%5Ctheta%20%29%7D%7D)
recall that sin²(θ) + cos²(θ) = 1, thus sin²(θ) = 1 - cos²(θ).
Answer:
72x
Step-by-step explanation:
Answer:
A
Step-by-step explanation:
true
Answer:
<h3>-16</h3>
Step-by-step explanation:
PEMDAS- (Parenthesis, Exponents, Multiply, Divide, Add, and Subtract) from left to right.
BOMDAS- (Brackets, Of, Multiply, Divide, Add, and Subtract) from left to right.
(-8)2
-8*2 (First, remove parenthesis.)
-8*2=-16
-16
Therefore, the final answer is -16.
