All categories

4 years ago

Cos A/1-tan A + sin A/1-cot A =sin A + cos A

1 answer:

5 0

$\frac{cosA}{ \frac{cosA}{cosA}- \frac{sinA}{cosA}}+ \frac{sinA}{ \frac{sinA}{sinA}- \frac{cosA}{sinA} }=sinA+cosA\\ \frac{cosA}{ \frac{cosA-sinA}{cosA}+ \frac{sinA}{ \frac{sinA-cosA}{sinA} } }=sinA+cosA\\ \frac{ cos^{2}A }{cosA-sinA} }- \frac{sin^{2}A }{cosA-sinA}=sinA+cosA\\
 \frac{ cos^{2}A-sin^{2}A }{cosA-sinA}=sinA+cosA\\cos(2A)=(sinA+cosA)(cosA-sinA)\\
cos(2A)=(sinA+cosA)(sinA-cosA)\\cos(2A)=sin^{2}A - cos^{2}A\\cos2A=cos2A \\1=1$

You might be interested in

What is the solution to - 4x + 28 > 12

raketka [301]

Answer:

$\large\boxed{x$

Step-by-step explanation:

$-4x+28>12\qquad\text{subtract 28 from both sides}\\\\-4x>-16\qquad\text{change the signs}\\\\4x$

6 0

4 years ago

Find the hypotenuse of a triangle with a base of 11 cm and height of 9 cm.

kondor19780726 [428]

Answer:

$\sqrt{202}$

Step-by-step explanation:

The image shows the steps to get this answer. I also checked with a calculator.

Hope this helps!

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5C0%7D%28x%2F%28tan%E2%81%A1%28x%29%29%5E%28cot%E2%81%A1%28x%29%5E2%20%

Viefleur [7K]

It looks like the limit you want to compute is

$\displaystyle L = \lim_{x\to0}\left(\frac x{\tan(x)}\right)^{\cot^2(x)}$

Rewrite the limand with an exponential and logarithm:

$\left(\dfrac{x}{\tan(x)}\right)^{\cot^2(x)} = \exp\left(\cot^2(x) \ln\left(\dfrac{x}{\tan(x)}\right)\right) = \exp\left(\dfrac{\ln\left(\dfrac{x}{\tan(x)}\right)}{\tan^2(x)}\right)$

Now, since the exponential function is continuous at 0, we can write

$\displaystyle L = \lim_{x\to0} \exp\left(\dfrac{\ln\left(\dfrac{x}{\tan(x)}\right)}{\tan^2(x)}\right) = \exp\left(\lim_{x\to0}\dfrac{\ln\left(\dfrac{x}{\tan(x)}\right)}{\tan^2(x)}\right)$

Let M denote the remaining limit.

We have $\dfrac x{\tan(x)}\to1$ as $x\to0$ , so $\ln\left(\dfrac x{\tan(x)}\right)\to0$ and $\tan^2(x)\to0$ . Apply L'Hopital's rule:

$\displaystyle M = \lim_{x\to0}\dfrac{\ln\left(\dfrac{x}{\tan(x)}\right)}{\tan^2(x)} \\\\ M = \lim_{x\to0}\dfrac{\dfrac{\tan(x)-x\sec^2(x)}{\tan^2(x)}\times\dfrac{\tan(x)}{x}}{2\tan(x)\sec^2(x)}$

Simplify and rewrite this in terms of sin and cos :

$\displaystyle M = \lim_{x\to0} \dfrac{\dfrac{\tan(x)-x\sec^2(x)}{\tan^2(x)}\times\dfrac{\tan(x)}{x}}{2\tan(x)\sec^2(x)} \\\\ M= \lim_{x\to0}\dfrac{\sin(x)\cos^3(x) - x\cos^2(x)}{2x\sin^2(x)}$

As $x\to0$ , we get another 0/0 indeterminate form. Apply L'Hopital's rule again:

$\displaystyle M = \lim_{x\to0} \frac{\sin(x)\cos^3(x) - x\cos^2(x)}{2x\sin^2(x)} \\\\ M = \lim_{x\to0} \frac{\cos^4(x) - 3\sin^2(x)\cos^2(x) - \cos^2(x) + 2x\cos(x)\sin(x)}{2\sin^2(x)+4x\sin(x)\cos(x)}$

Recall the double angle identity for sin:

sin(2x) = 2 sin(x) cos(x)

Also, in the numerator we have

cos⁴(x) - cos²(x) = cos²(x) (cos²(x) - 1) = - cos²(x) sin²(x) = -1/4 sin²(2x)

So we can simplify M as

$\displaystyle M = \lim_{x\to0} \frac{x\sin(2x) - \sin^2(2x)}{2\sin^2(x)+2x\sin(2x)}$

This again yields 0/0. Apply L'Hopital's rule again:

$\displaystyle M = \lim_{x\to0} \frac{\sin(2x)+2x\cos(2x)-4\sin(2x)\cos(2x)}{2\sin(2x)+4x\cos(2x)+4\sin(x)\cos(x)} \\\\ M = \lim_{x\to0} \frac{\sin(2x) + 2x\cos(2x) - 2\sin(4x)}{4\sin(2x)+4x\cos(2x)}$

Once again, this gives 0/0. Apply L'Hopital's rule one last time:

$\displaystyle M = \lim_{x\to0}\frac{2\cos(2x)+2\cos(2x)-4x\sin(2x)-8\cos(4x)}{8\cos(2x)+4\cos(2x)-8x\sin(2x)} \\\\ M = \lim_{x\to0} \frac{4\cos(2x)-4x\sin(2x)-8\cos(4x)}{12\cos(2x)-8x\sin(2x)}$