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2 years ago

Express secΦ in terms of tanΦ

Mathematics

1 answer:

Sunny_sXe [5.5K]2 years ago

5 0

Answer:

$\sec\theta=\sqrt{1+\tan^2\theta}$

Step-by-step explanation:

The trigonometric identity is :

$\sec^2\theta-\tan^2\theta=1\\\\or\\\\\sec^2\theta=1+\tan^2\theta$

We need to express secθ in terms of tanθ.

The above equation becomes,

$\sec\theta=\sqrt{1+\tan^2\theta}$

Hence, this is the required solution.

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2 years ago

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katrin2010 [14]

Answer: B, 70 inches.

Step-by-step explanation:

Let's start slow and go shape by shape.

Based on the information we have, we know that the length for the rectangle is 8 inches by 7 inches. 8*7 = 56, so we have one of the answers.

Next, we do the triangle. Based on the information we have, we know that the triangle is 4 inches by 7 inches divided by 2. 4*7 = 28 $\frac {28}{2} = 14$

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3 years ago

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

andrey2020 [161]

Answer:

The maximum value is 1/27 and the minimum value is 0.

Step-by-step explanation:

Note that the given function is equal to $(xyz)^2$ then it means that it is positive i.e $f(x,y,z)\geq 0$ .

Consider the function $F(x,y,z,\lambda)=x^2y^2z^2-\lambda (x^2+y^2+z^2-1)$

We want that the gradient of this function is equal to zero. That is (the calculations in between are omitted)

$\frac{\partial F}{\partial x} = 2x(y^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial y} = 2y(x^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial z} = 2z(x^2y^2 - \lambda)=0$

$\frac{\partial F}{\partial \lambda} = (x^2+y^2+z^2-1)=0$

Note that the last equation is our restriction. The restriction guarantees us that at least one of the variables is non-zero. We've got 3 options, either 1, 2 or none of them are zero.

If any of them is zero, we have that the value of the original function is 0. We just need to check that there exists a value for lambda.

Suppose that x is zero. Then, from the second and third equation we have that

$-2y\lambda = -2z\lambda$ . If lambda is not zero, then y =z. But, since $-2y\lambda=0$ and lambda is not zero, this implies that x=y=z=0 which is not possible. This proofs that if one of the variables is 0, then lambda is zero. So, having one or two variables equal to zero are feasible solutions for the problem.

Suppose that only x is zero, then we have the solution set $y^2+z^2=1$ .

If both x,y are zero, then we have the solution set $z^2=1$ . We can find the different solution sets by choosing the variables that are set to zero.

NOw, suppose that none of the variables are zero.

From the first and second equation we have that

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So, in this case, replacing this in the restriction we have $3z^2=1$ , which gives as another solution set. On this set, we have $x^2=y^2=z^2=\frac{1}{3}$ . Over this solution set, we have that the value of our function is $\frac{1}{3^3}= \frac{1}{27}$

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2 years ago

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Natali [406]

Answer: In the equations that you have given, we have a dependent system.

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2 1 This is the coefficient matrix.
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