PLEASE HELP DUE SOON! Find the area of the shaded region. Write your answer in exact form.

Mathematics

1 answer:

ivann1987 [24]2 years ago

4 0

Can you repost the picture I can’these it

You might be interested in

The sum of 86, 68, and 38 is 192. What else do you know about the sum of 68, 38, and 86?

Sveta_85 [38]

We have 2 given numbers:
=> 86 + 68 + 38 = 192
=> 68 + 38 + 86 = 192 also
This kind of process used commutative property system of equation in where:
a + b = b + a
No matter what sequence you are going to use, the answer would always be the same.
In the given 2 equations, notice that both digits of 2 group of equation contains the same number. The only difference is they were added with different orders. Yet the answers are still the same.

8 0

3 years ago

Verify the identity (tan x + 1)^2 + (tan x-1)^2= 2 sec^2 x

Elina [12.6K]

$(\tan x+1)^2+(\tan x-1)^2=2\sec^2x\\\\\text{use}\ \tan x=\dfrac{\sin x}{\cos x}\\\\L_s=\left(\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\cos x}\right)^2+\left(\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\cos x}\right)^2\\\\=\left(\dfrac{\sin x+\cos x}{\cos x}\right)^2+\left(\dfrac{\sin x-\cos x}{\cos x}\right)^2\\\\=\dfrac{(\sin x+\cos x)^2}{\cos^2x}+\dfrac{(\sin x-\cos x)^2}{\cos^2x}\\\\\text{use}\ (a\pm b)^2=a^2\pm2ab+b^2$

$=\dfrac{\sin^2x+2\sin x\cos x+\cos^2}{\cos^2x}+\dfrac{\sin^2x-2\sin x\cos x+\cos^2}{\cos^2x}\\\\=\dfrac{\sin^2x+2\sin x\cos x+\cos^2+\sin^2x-2\sin x\cos x+\cos^2}{\cos^2x}\\\\=\dfrac{2\sin^2x+2\cos^2x}{\cos^2x}=\dfrac{2(\sin^2x+\cos^2x)}{\cos^2x}\\\\\text{use}\ \sin^2x+\cos^2x=1\\\\=\dfrac{2(1)}{\cos^2x}=2\cdot\dfrac{1}{\cos^2x}=2\left(\dfrac{1}{\cos x}\right)^2\\\\\text{use}\ \sec x=\dfrac{1}{\cos x}\\\\=2(\sec^2x)=2\sec^2x=R_s\\\\L_s=R_s\Rightarrow The\ identity$