Ummm let’s see.. yah I have no clue sorry buddy your just gonna have to get this one wrong
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!!
(2-7m^6)^2
rearrange terms:
(2-7m^6)^2
(-7m^6 + 2)^2
expand the squares:
(-7m^6 + 2)^2
(-7m^6 + 2)(-7m^6 + 2)
distribute:
(-7m^6 + 2)(-7m^6 + 2)
-7(-7m^6 + 2) • m^6 + 2(-7m^6 + 2)
distribute:
-7(-7m^6 + 2) • m^6 + 2(-7m^6 + 2)
49m^12 - 14m^6 + 2(-7^6 +2)
distribute:
49m^12 - 14m^6 + 2(-7^6 +2)
49m^12 - 14m^6 - 14m^6 + 4
combine like terms:
49m^12 - 14m^6 - 14m^6 + 4
49m^12 - 28m^6 + 4
solution:
49m^12 - 28m^6 + 4
please mark brainliest and hope it helps
67 ÷ 10 x 4 = 26.8 hours for her to eat 67 cookies
Answer: C Fourth Quadrant
Step-by-step explanation:
