Step-by-step explanation:
<h3><u>Given :-</u></h3>
[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]
<h3>
<u>Required To Prove :-</u></h3>
[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)
<h3><u>Proof :-</u></h3>
On taking LHS
[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]
We know that
Tan θ = 1/ Cot θ
and
Cot θ = 1/Tan θ
=> (1+Cot²θ)(1+Tan²θ)
=> (Cosec² θ) (Sec²θ)
Since Cosec²θ - Cot²θ = 1 and
Sec²θ - Tan²θ = 1
=> (1/Sin² θ)(1/Cos² θ)
Since , Cosec θ = 1/Sinθ
and Sec θ = 1/Cosθ
=> 1/(Sin²θ Cos²θ)
We know that Sin²θ+Cos²θ = 1
=> 1/[(Sin²θ)(1-Sin²θ)]
=> 1/(Sin²θ-Sin²θ Sin²θ)
=> 1/(Sin²θ - Sin⁴θ)
=> RHS
=> LHS = RHS
<u>Hence, Proved.</u>
<h3><u>Answer:-</u></h3>
[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)
<h3><u>Used formulae:-</u></h3>
→ Tan θ = 1/ Cot θ
→ Cot θ = 1/Tan θ
→ Cosec θ = 1/Sinθ
→ Sec θ = 1/Cosθ
<h3><u>Used Identities :-</u></h3>
→ Cosec²θ - Cot²θ = 1
→ Sec²θ - Tan²θ = 1
→ Sin²θ+Cos²θ = 1
Hope this helps!!
Answer:
Step-by-step explanation:
x³-6x²+11x-6
put x=1
1³-6×1²+11×1-6=1-6=11-6=0
by synthetic division
1| 1 -6 11 -6
| 1 -5 6
|----------------
| 1 -5 6 |0
x²-5x+6=0
x²-2x-3x+6=0
x(x-2)-3(x-2)=0
(x-2)(x-3)=0
x=1
x-1=0
so x³-6x²+11x-6=(x-1)(x-2)(x-3)
Answer: y-1=6(x+7)
Step-by-step explanation:
The formula for point-slope form is 
. Since we are given the point and slope, we can directly plug them in.
[distribute by -1]
Now, we know that the point-slope form is y-1=6(x+7).
Answer: 2.2 meters
Step-by-step explanation:
The data is:
The initial piece is 6.35 meters long.
We cut 25 pieces of 16.2cm each, so in pieces only we have:
25*16.2cm = 405cm
1m = 100cm
Then: 405cm = (405/100) m = 4.05m
And each time that we cut a piece, we have 4mm of waste.
Then we have 25 times the 4mm of waste, this means that the total waste is:
4mm*25 = 100mm
and 1000mm = 1m
100mm = (100/1000) m = 0.1m
So the total amount that was cut is:
4.05m + 0.1m = 4.15m
The length in the channel iron will be: 6.35m - 4.15m = 2.2 meters
For each <em>x</em> in the interval 0 ≤ <em>x</em> ≤ 5, the shell at that point has
• radius = 5 - <em>x</em>, which is the distance from <em>x</em> to <em>x</em> = 5
• height = <em>x</em> ² + 2
• thickness = d<em>x</em>
and hence contributes a volume of 2<em>π</em> (5 - <em>x</em>) (<em>x</em> ² + 2) d<em>x</em>.
Taking infinitely many of these shells and summing their volumes (i.e. integrating) gives the volume of the region:
