Answer:
Step-by-step explanation:
1.
cot x sec⁴ x = cot x+2 tan x +tan³x
L.H.S = cot x sec⁴x
=cot x (sec²x)²
=cot x (1+tan²x)² [ ∵ sec²x=1+tan²x]
= cot x(1+ 2 tan²x +tan⁴x)
=cot x+ 2 cot x tan²x+cot x tan⁴x
=cot x +2 tan x + tan³x [ ∵cot x tan x 
=1]
=R.H.S
2.
(sin x)(tan x cos x - cot x cos x)=1-2 cos²x
L.H.S =(sin x)(tan x cos x - cot x cos x)
= sin x tan x cos x - sin x cot x cos x
= sin²x -cos²x
=1-cos²x-cos²x
=1-2 cos²x
=R.H.S
3.
1+ sec²x sin²x =sec²x
L.H.S =1+ sec²x sin²x
=
[
]
=1+tan²x ![[\frac{\textrm{sin x}}{\textrm{cos x}} = \textrm{tan x}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B%5Ctextrm%7Bsin%20x%7D%7D%7B%5Ctextrm%7Bcos%20x%7D%7D%20%3D%20%5Ctextrm%7Btan%20x%7D%5D)
=sec²x
=R.H.S
4.
L.H.S=
= 2 csc x
= R.H.S
5.
-tan²x + sec²x=1
L.H.S=-tan²x + sec²x
= sec²x-tan²x
=
=1
The another way to represent the same line, for solving 
will be: 
<u><em>Explanation</em></u>
Given equation of the line is: 
"Solving " means <u>we need to get</u> <u>alone in left side.</u> So, we will eliminate all other terms except from the left side.
First we will <u>subtract 2x </u>from both sides. So, we will get......
Now, we will <u>divide both sides by 3</u> for getting alone. So.....
Thus, the another way to represent the same line, for solving will be:
Answer:
6, 2, 2/3, 2/9, 2/27, 2/81
Step-by-step explanation:
The nth term of a geometric progression is expressed as;
Tn = ar^n-1
a is the first term
n is the number of terms
r is the common ratio
Given
a = 6
r = 1/3
when n = 1
T1 = 6(1/3)^1-1
T1 = 6(1/3)^0
T1 = 6
when n = 2
T2= 6(1/3)^2-1
T2= 6(1/3)^1
T2 = 2
when n = 3
T3 = 6(1/3)^3-1
T3= 6(1/3)^2
T3= 6 * 1/9
T3 = 2/3
when n = 4
T4 = 6(1/3)^4-1
T4= 6(1/3)^3
T4= 6 * 1/27
T4 = 2/9
when n = 5
T5 = 6(1/3)^5-1
T5= 6(1/3)^4
T5= 6 * 1/81
T5 = 2/27
when n = 6
T6 = 6(1/3)^6-1
T6= 6(1/3)^5
T6= 6 * 1/243
T6 = 2/81
Hence the first six terms are 6, 2, 2/3, 2/9, 2/27, 2/81
