Prove that sinA-sin3A+sin5A-sin7A/cosA-cos3A-cos5A+cos7A= cot2A
1 answer:
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You can do this using synthetic division, which is the easiest way. If x - 2 = 0, then x = 2. That 2 will go outside the "box" and the leading coefficients of the terms in the polynomial will go inside the "box". 2 (1 -3 -10 24). Bring down the first number, the 1. Multiply that 1 by the 2 to get 2. Put that 2 up under the -3 and add to get -1. Multiply that -1 by the 2 to get -2. Put that =-2 up under the -10 and add to get -12. Multiply that -12 by the 2 to get -24. Put the -24 up under the 24 and add to get 0. That means that x - 2 is a factor of the polynomial. What's left, the bolded numbers, are the coefficients of a new polynomial that is one degree less than the polynomial you started with. In other words, when we divide your polynomial by x-2, you get 
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Answer:
x=16
Step-by-step explanation:
We are given that
we have to solve this equation for x
in order to do that we need to isolate x first
hence
subtracting 11 from both sides we get
Now we square on both sides we get
hence x =16
Let point A be and point B
Use the point-slope formula and whichever point you want as , in this case I'll use