You are Manas Chatterjee, the music teacher of your school. Your school is going to organize the annual function in November for which you require musical instruments for orchestra for your very big group. Write a letter to the Melody Music Company, 45 Ajmal Khan Road, Karol Bagh, Delhi, placing the order for musical instruments.
Answers
Hi,
Here is your answer
ABC International School
Delhi
29 July 2020
The Manager
Melody Music Company
45 Ajmal Khan Road
Karol Bagh
Delhi
Sub: To place an order for musical instruments
Sir
I, Manas Chatterjee the music teacher of ABC International School would like to place an order for the following musical instruments.
Sl.No Items Quantity
1. Guitar 5
2. Piano 4
3. Guitar 5
Please give the concession that you provide when placing a bulk order. All the items should he of high grade and should be packed properly. The items should reach the school within the next ten working days. Payment will be done on delivery. Any damage during transportation will be your responsibility. No payment will be offered if goods are found to be damaged.
With regards
Sd/-
Manas Chatterjee
Hope this helps you.
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