You are atul a resident of A-2, Rohini,New delhi. There is no bus-stop within the radius of 2.5km from the apartments causing a lot of incovenience to the residents. Write a letter to the editor drawing attention of the concerned authorities.You may take help of the following.
Answers
Answer:
Explanation:
Format of letter to editor:
Sender's Address
Date
The Editor
Name of newspaper
Place
Sub:
Sir
Introduction
Body
Conclusion
With regards/Yours truly/Yours faithfully
Sd/-
Name
Required letter:
A-2, Rohini
New Delhi
2 September 2020
The Editor
The Daily times
New Delhi
Sub: Views on lack of bus stops in our locality
Sir
Through the esteemed columns of your national daily, I would like to present my views on the lack of bus stops in our locality.
I am Atul, a resident of A-2, Rohini, New Delhi. I would like to bring the notice of the authorities that there are no bus stops within the radius of 2.5 km from our locality. The citizens have to walk over 3 km to reach the nearby bustop. This is really inconvenient and hectic for senior citizens and small children. It pains me to see the old and sick walking in the scorching heat of sun to catch the bus on time. The other day a child lost his life because too much time was wasted finding a suitable means of transport when the child's life could have been saved. I request the concerned authorities to provide a bus stop for our locality. All the citizes would be highly grateful and it would be well appreciated.
Kindly publish the aforesaid matter in your national daily to bring about a change in our locality as it really matters.
Yours truly
Sd/-
Atul
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