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3 years ago

Prove that: sin8A/1-cos8A=cot4A

Mathematics

1 answer:

nikitadnepr [17]3 years ago

8 0

Step-by-step explanation:

Recall the identity

$\tan \frac{A}{2}= \dfrac{1 - \cos A}{\sin A}$

We can see that

$\dfrac{\sin 8A}{1 - \cos 8A} = \dfrac{1}{\tan 4A}= \cot 4A$

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e-lub [12.9K]

Answer: R = {4, 5, 7, 19}

Step-by-step explanation:

Domain represents the x-values.

Range represents the y-values.

Insert the values of the domain into the given equation.

$\begin {array}{c|l|c}\underline{\ x\ }&\underline{\ 2(2^x)+3 =\ }&\underline{\ r(x)\ }\\ -1&2(2^{-1})+3=&4\\0&2(2^0)+3=&5\\1&2(2^1)+3=&7\\3&2(2^3)+3=&19\\\end{array}$

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3 years ago

3) x2+ y2 - x + 3y - 42 = 0 X+y=4

REY [17]

The solution is $(-1,5)$ and $(7,-3)$

Step-by-step explanation:

The expression is $x^{2} +y^{2} -x+3y-42=0$ and $x+y=4$

Using substitution method we can solve the expression.

Let us substitute $x=4-y$ in $x^{2} +y^{2} -x+3y-42=0$

$(4-y)^{2} +y^{2} -(4-y)+3y-42=0$

Expanding and simplifying the expression, we get,

$\begin{array}{r}{16-8 y+y^{2}+y^{2}-4+y+3 y-42=0} \\{2 y^{2}-4 y-30=0}\end{array}$

Let us use the quadratic equation formula to solve this equation,

$\begin{aligned}y &=\frac{4 \pm \sqrt{16-4(2)(-30)}}{2(2)} \\&=\frac{4 \pm \sqrt{16+240}}{4} \\&=\frac{4 \pm 16}{4} \\y &=1 \pm 4\end{aligned}$

Thus, $y=5$ and $y=-3$

Substituting y-values in the equation $x+y=4$ , we get the value of x.

For $y=5$ ⇒ $x=4-5=-1$

For $y=-3$ ⇒ $x=4+3=7$

Thus, the solution set is $(-1,5)$ and $(7,-3)$

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Prove that: sin8A/1-cos8A=cot4A​

Prove that: sin8A/1-cos8A=cot4A