Identify the figure of. How old was the mother in the photograph & what were the names of the cousins? speech used in silence silences
Answers
Question:
1. How old was the mother in the photograph and what were the names of the cousins?
2. Identify the figure of speech used in silence silences.
Answer:
1. The mother was twelve years old in the photograph. She had been on a seaside holiday with her two younger cousins Betty and Dolly. They were dressed for the beach and looked absolutely carefree and happy.
2. The figure os speech used in silence silences is alliteration as s sound is being repeated here.
More information:
The Photograph is a poem written by the english writer, Shirley Toulson . In this poem she compares the mortality of humans with the perennial nature of the sea.
In the photograph the mother is only twelve years old. Now she has been dead for twelve years. The poet remembers her mothe's joy when she used to look at the photographs and reminisce the old memories.
The loss of her mother is extremely distressing to the poet. But over the years she has come to terms with it.
The poet says that sea is permenant and eternal compared to the temporary life of human beings.
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