Mathematics: If you can. i will give brainiest to best answer. this question is a calculus/trigonometry question. i ask that you look over my work, but if you don t have the time, i ask that you me answer the questions i don t know.

If you can. i will give brainiest to best answer. - id2527678, selemontaddesep6bv1p

Ответ:

shenothomas688

03.01.2022

7 is incorrect. The answer should be -4. Here's how I'd derive it:

$\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}$

With $\pi$ , we should expect $\cos x$ . If $\tan x=\dfrac8{15}$ , then

$\sec x=-\sqrt{1+\tan^2x}=-\dfrac{17}{15}\implies\cos x=-\dfrac{15}{17}$

Also,

$\pi$

so we should expect $\tan\dfrac x2$ and

$\tan\dfrac x2=-\sqrt{\dfrac{1-\cos x}{1+\cos x}}=-4$

You seem to be taking

$\tan\dfrac x2=\dfrac{1+\cos x}{-\sin x}$

but this is not an identity.

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8 is incorrect. Just a silly mistake, you swapped the order of the terms in the numerator. It should be

$\dfrac{\sqrt6-\sqrt2}4$

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9. Use the identity from (7). $\dfrac{7\pi}{12}$ lies in the second quadrant, so

$\tan\dfrac{7\pi}{12}=-\sqrt{\dfrac{1-\cos\frac{7\pi}6}{1+\cos\frac{7\pi}6}}=-\sqrt{7+4\sqrt3}=-2-\sqrt3$

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12.

$\dfrac{\cot x-1}{1-\tan x}=\dfrac{\frac{\cos x}{\sin x}-1}{1-\frac{\sin x}{\cos x}}$

$=\dfrac{\cos x(\cos x-\sin x)}{\sin x(\cos x-\sin x)}$

$=\dfrac{\cos x}{\sin x}$

$=\dfrac{\csc x}{\sec x}$

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13.

$\dfrac{1+\tan x}{\sin x+\cos x}=\dfrac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}$

$=\dfrac{\cos x+\sin x}{\cos x(\sin x+\cos x)}$

$=\dfrac1{\cos x}$

$=\sec x$

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14.

$\sin2x(\cot x+\tan x)=2\sin x\cos x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right)$

$=2\sin x\cos x\dfrac{\cos^2x+\sin^2x}{\sin x\cos x}$

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15.

$\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}$

$=\dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}$

$=\cos2\theta$

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Ответ:

pedroramirezr2

27.12.2022

Mathematics

Let's solve this problem step-by-step.

STEP-BY-STEP EXPLANATION:

We will be using simultaneous equations to solve this problem.

Let's first establish the two equations we will be using.

Let x = smaller number

Let y = larger number

Equation No. 1 -

x + y = 0

Equation No. 2 -

14x + 9y = 1

To begin with, we will make ( x ) the subject in the first equation.

Equation No. 1 -

x + y = 0

x = 0 - y

x = - y

Next we will substitute the value of ( x ) from the first equation into the second equation to solve for ( y ).

Equation No. 2 -

14x + 9y = 1

14 ( - y ) + 9y = 1

- 14y + 9y = 1

- 5y = 1

y = 1 / - 5

y = - 1 / 5

Then we will substitute the value of ( y ) from the second equation into the first equation to solve for ( x ).

Equation No. 1 -

x = - y

x = - ( - 1 / 5 )

x = 1 / 5

FINAL ANSWER:

Therefore, the two numbers 1 / 5 and - 1 / 5.

Hope this helps! :)
Have a lovely day! <3

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If you can. i will give brainiest to best answer. this question is a calculus/trigonometry question. i ask that you look over my work, but if you don't have the time, i ask that you me answer the questions i don't know.

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