what does 'The Third Level' signify
Answers
Answer:
The Third level is a fictitious story written by the author Jack Finned that weaves a psychological journey of the author into the past.
The narrator Charlie is convinced that there are three levels in the Grand central Station not working. His friends are convinced that this is just a figment of his imagination. His friends assures him that this is his mind taking temporary refuge from a world full of insecurity, fear war and worry.
Charlie one day while hurrying home decided to take the subway to reach home faster. But he gets lost through the labyrinths of the station. He finds himself in 1894 where the world was much quieter and the pace of life was much slower.
The next day Charles converts all his money into old currencies and tries to go back, but he couldn't find the third level. He narrates the experience to his psychiatrist who assures him that it is just him mind taking refuge.
After a few week, Charlie receives a letter from his psychiatrist that he had found the third level and was living there happily with his family. He urges the narrator and his wife to keep on searching until he finally finds the third level.
The third level is the world which is somewhere between somewhere between fantasy and reality. It is the world we escape to when we want to escape from the sorrows and tragedies of this world.
Answer:
Answer:
Explanation:
\Large{\underline{\underline{\it{Given:}}}}
Given:
\sf{\dfrac{tan\:A}{sec\:A-1} -\dfrac{sin\:A}{1+cos\:A} =2\:cot\:A}
secA−1
tanA
−
1+cosA
sinA
=2cotA
\Large{\underline{\underline{\it{To\:Prove:}}}}
ToProve:
LHS = RHS
\Large{\underline{\underline{\it{Solution:}}}}
Solution:
→ Taking the LHS of the equation,
\sf{LHS=\dfrac{tan\:A}{sec\:A-1} -\dfrac{sin\:A}{1+cos\:A} }LHS=
secA−1
tanA
−
1+cosA
sinA
→ Applying identities we get
=\sf{\dfrac{\dfrac{sin\:A}{cos\:A} }{\dfrac{1}{cos\:A}-1 } -\dfrac{sin\:A}{1+cos\:A} }=
cosA
1
−1
cosA
sinA
−
1+cosA
sinA
→ Cross multiplying,
=\sf{\dfrac{\dfrac{sin\:A}{cos\:A} }{\dfrac{1-cos\:A}{cos\:A} } -\dfrac{sin\:A}{1+cos\:A} }=
cosA
1−cosA
cosA
sinA
−
1+cosA
sinA
→ Cancelling cos A on both numerator and denominator
=\sf{\dfrac{sin\:A}{1-cos\:A} -\dfrac{sin\:A}{1+cos\:A}}=
1−cosA
sinA
−
1+cosA
sinA
→ Again cross multiplying we get,
=\sf{\dfrac{sin\:A(1+cos\:A)-sin\:A(1-cos\:A)}{(1+cos\:A)(1-cos\:A)}}=
(1+cosA)(1−cosA)
sinA(1+cosA)−sinA(1−cosA)
→ Taking sin A as common,
\sf{=\dfrac{sin\:A[1+cos\:A-(1-cos\:A)]}{(1^{2}-cos^{2}\:A ) }}=
(1
2
−cos
2
A)
sinA[1+cosA−(1−cosA)]
\sf{=\dfrac{sin\:A[1+cos\:A-1+cos\:A]}{sin^{2}\:A } }=
sin
2
A
sinA[1+cosA−1+cosA]
→ Cancelling sin A on both numerator and denominator
\sf{=\dfrac{2\:cos\:A}{sin\:A} }=
sinA
2cosA
\sf=2\times \dfrac{cos\:A}{sin\:A} }
\sf{=2\:cot\:A}=2cotA
=\sf{RHS}=RHS
→ Hence proved.
\Large{\underline{\underline{\it{Identitites\:used:}}}}
Identititesused:
\sf{tan\:A=\dfrac{sin\:A}{cos\:A} }tanA=
cosA
sinA
\sf{sec\:A=\dfrac{1}{cos\:A} }secA=
cosA
1
\sf{(a+b)\times(a-b)=a^{2}-b^{2} }(a+b)×(a−b)=a
2
−b
2
\sf{(1-cos^{2}\:A)=sin^{2} \:A}(1−cos
2
A)=sin
2
A
\sf{\dfrac{cos\:A}{sin\:A}=cot\:A}
sinA
cosA
=cotA