<u>Answer:</u>
The expression ![\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}](https://tex.z-dn.net/?f=%5Cbold%7B%28%5Csec%20x%2B%5Ctan%20x%29%5E%7B2%7D%20%5Ctext%20%7B%20is%20same%20as%20%7D%20%5Cfrac%7B1%2B%5Csin%20x%7D%7B1-%5Csin%20x%7D%7D)
<u>Solution:</u>
From question, given that ![\bold{(\sec x+\tan x)^{2}}](https://tex.z-dn.net/?f=%5Cbold%7B%28%5Csec%20x%2B%5Ctan%20x%29%5E%7B2%7D%7D)
By using the trigonometric identity
the above equation becomes,
![(\sec x+\tan x)^{2} = \sec ^{2} x+2 \sec x \tan x+\tan ^{2} x](https://tex.z-dn.net/?f=%28%5Csec%20x%2B%5Ctan%20x%29%5E%7B2%7D%20%3D%20%5Csec%20%5E%7B2%7D%20x%2B2%20%5Csec%20x%20%5Ctan%20x%2B%5Ctan%20%5E%7B2%7D%20x)
We know that ![\sec x=\frac{1}{\cos x} ; \tan x=\frac{\sin x}{\cos x}](https://tex.z-dn.net/?f=%5Csec%20x%3D%5Cfrac%7B1%7D%7B%5Ccos%20x%7D%20%3B%20%5Ctan%20x%3D%5Cfrac%7B%5Csin%20x%7D%7B%5Ccos%20x%7D)
![(\sec x+\tan x)^{2}=\frac{1}{\cos ^{2} x}+2 \frac{1}{\cos x} \frac{\sin x}{\cos x}+\frac{\sin ^{2} x}{\cos ^{2} x}](https://tex.z-dn.net/?f=%28%5Csec%20x%2B%5Ctan%20x%29%5E%7B2%7D%3D%5Cfrac%7B1%7D%7B%5Ccos%20%5E%7B2%7D%20x%7D%2B2%20%5Cfrac%7B1%7D%7B%5Ccos%20x%7D%20%5Cfrac%7B%5Csin%20x%7D%7B%5Ccos%20x%7D%2B%5Cfrac%7B%5Csin%20%5E%7B2%7D%20x%7D%7B%5Ccos%20%5E%7B2%7D%20x%7D)
![=\frac{1}{\cos ^{2} x}+\frac{2 \sin x}{\cos ^{2} x}+\frac{\sin ^{2} x}{\cos ^{2} x}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B%5Ccos%20%5E%7B2%7D%20x%7D%2B%5Cfrac%7B2%20%5Csin%20x%7D%7B%5Ccos%20%5E%7B2%7D%20x%7D%2B%5Cfrac%7B%5Csin%20%5E%7B2%7D%20x%7D%7B%5Ccos%20%5E%7B2%7D%20x%7D)
On simplication we get
![=\frac{1+2 \sin x+\sin ^{2} x}{\cos ^{2} x}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%2B2%20%5Csin%20x%2B%5Csin%20%5E%7B2%7D%20x%7D%7B%5Ccos%20%5E%7B2%7D%20x%7D)
By using the trigonometric identity
,the above equation becomes
![=\frac{1+2 \sin x+\sin ^{2} x}{1-\sin ^{2} x}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%2B2%20%5Csin%20x%2B%5Csin%20%5E%7B2%7D%20x%7D%7B1-%5Csin%20%5E%7B2%7D%20x%7D)
By using the trigonometric identity ![(a+b)^{2}=a^{2}+2ab+b^{2}](https://tex.z-dn.net/?f=%28a%2Bb%29%5E%7B2%7D%3Da%5E%7B2%7D%2B2ab%2Bb%5E%7B2%7D)
we get ![1+2 \sin x+\sin ^{2} x=(1+\sin x)^{2}](https://tex.z-dn.net/?f=1%2B2%20%5Csin%20x%2B%5Csin%20%5E%7B2%7D%20x%3D%281%2B%5Csin%20x%29%5E%7B2%7D)
![=\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%281%2B%5Csin%20x%29%5E%7B2%7D%7D%7B1-%5Csin%20%5E%7B2%7D%20x%7D)
![=\frac{(1+\sin x)(1+\sin x)}{1-\sin ^{2} x}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%281%2B%5Csin%20x%29%281%2B%5Csin%20x%29%7D%7B1-%5Csin%20%5E%7B2%7D%20x%7D)
By using the trigonometric identity
we get ![1-\sin ^{2} x=(1+\sin x)(1-\sin x)](https://tex.z-dn.net/?f=1-%5Csin%20%5E%7B2%7D%20x%3D%281%2B%5Csin%20x%29%281-%5Csin%20x%29)
![=\frac{(1+\sin x)(1+\sin x)}{(1+\sin x)(1-\sin x)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%281%2B%5Csin%20x%29%281%2B%5Csin%20x%29%7D%7B%281%2B%5Csin%20x%29%281-%5Csin%20x%29%7D)
![= \frac{1+\sin x}{1-\sin x}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%2B%5Csin%20x%7D%7B1-%5Csin%20x%7D)
Hence the expression ![\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}](https://tex.z-dn.net/?f=%5Cbold%7B%28%5Csec%20x%2B%5Ctan%20x%29%5E%7B2%7D%20%5Ctext%20%7B%20is%20same%20as%20%7D%20%5Cfrac%7B1%2B%5Csin%20x%7D%7B1-%5Csin%20x%7D%7D)
The correct answers in order are subtraction, division, given.
Answer:
Hi please make me brainly
Step-by-step explanation:
please
Answer:
Part a) The drawn in the attached figure
Part b)The slant height of the outside edge is
Step-by-step explanation:
Part a) The drawn in the attached figure
Part b) What is the slant height of the outside edge?
we have that
The diameter of the base of the cone is 12 in
so
----> the radius is half the diameter
Applying the Pythagoras Theorem find the slant height x
![x^{2}=r^{2}+h^{2}](https://tex.z-dn.net/?f=x%5E%7B2%7D%3Dr%5E%7B2%7D%2Bh%5E%7B2%7D)
substitute the values
![x^{2}=6^{2}+8^{2}](https://tex.z-dn.net/?f=x%5E%7B2%7D%3D6%5E%7B2%7D%2B8%5E%7B2%7D)
![x^{2}=100](https://tex.z-dn.net/?f=x%5E%7B2%7D%3D100)
Answer:
BEGIN def function isPositive(int parameter)
//test for positivity
if parameter > 0
RETURN true
else
RETURN false
END def function isPositive
Step-by-step explanation:
A number is a positive integer if it is a whole number and exists between 0 and positive infinity e. g 1, 90000, 42434234 e.t.c. It is a negative integer if it is a whole number and exists between negative infinity and 0(0 may be inclusive). e.g 0, -34342, -6984.
As shown on the first line, to define a function, we begin by writing the name of the function - "isPositive" in this case - followed by a pair of parenthesis which contains the parameter(s) on which the function will act on. In this case, the parameter is an integer - positive or negative.
Line 2 shows a comment describing what happens on the next four lines. i.e checking whether or not the parameter's value is positive
Then, we write an if-else statement to check whether or not the parameter's value is a positive integer. If the parameter's value is greater than 0, the parameter's value is thus a positive integer then the function will return true. This is shown on lines 3 and 4. Otherwise, the parameter's value is a negative integer, thus the function returns a false as shown on lines 5 and 6.
The last line in the expression shows the end of the function's definition.
Hope it helps.