If - is -01003 -7- k-2 3 -7- k -1 -2- A cosec B -7- then - B - A - where - is the greatest integer function and A - B - N

∑_k=0^100(3π7+kπ2)(3π7+(k+1)π2) =∑_k=0^100sin(3π7+(k+1)π2 3π7 kπ2)cos(3π7+kπ2)cos(3π7+(k+1)π2) = ∑_k=0^100tan(3π7+(k+1)π2) tan(3π7+kπ2) =tan(3π7+101π2) tan(3 ...

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Byju's Answer

Standard XII

Mathematics

Properties of Modulus

If ∑ is =0100...

Question

If $100 \sum k = 0 sec (\frac{3 π}{7} + \frac{k π}{2}) sec (\frac{3 π}{7} + \frac{(k + 1) π}{2}) = - A cosec (\frac{B π}{7})$ , then $[\frac{B}{A}]$ (where $[.]$ is the greatest integer function and $A, B \in N$ ) is

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Solution

$100 \sum k = 0 sec (\frac{3 π}{7} + \frac{k π}{2}) sec (\frac{3 π}{7} + \frac{(k + 1) π}{2}) = 100 \sum k = 0 \frac{sin (\frac{3 π}{7} + \frac{(k + 1) π}{2} - \frac{3 π}{7} - \frac{k π}{2})}{cos (\frac{3 π}{7} + \frac{k π}{2}) cos (\frac{3 π}{7} + \frac{(k + 1) π}{2})} = 100 \sum k = 0 tan (\frac{3 π}{7} + \frac{(k + 1) π}{2}) - tan (\frac{3 π}{7} + \frac{k π}{2})$
$= tan (\frac{3 π}{7} + \frac{101 π}{2}) - tan (\frac{3 π}{7})$
$= tan (\frac{3 π}{7} + \frac{π}{2} + 50 π) - tan (\frac{3 π}{7})$
$= tan (\frac{3 π}{7} + \frac{π}{2}) - tan (\frac{3 π}{7})$
$= - cot (\frac{3 π}{7}) - tan (\frac{3 π}{7})$
$= - ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{cos (\frac{3 π}{7})}{sin (\frac{3 π}{7})} + \frac{sin (\frac{3 π}{7})}{cos (\frac{3 π}{7})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$= - 2 cosec (\frac{6 π}{7}) \Rightarrow [\frac{B}{A}] = 3$

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If 100∑k=0sec(3π7+kπ2)sec(3π7+(k+1)π2)= −A cosec (Bπ7), then [BA] (where [.] is the greatest integer function and A,B∈N) is

If $100 \sum k = 0 sec (\frac{3 π}{7} + \frac{k π}{2}) sec (\frac{3 π}{7} + \frac{(k + 1) π}{2}) = - A cosec (\frac{B π}{7})$ , then $[\frac{B}{A}]$ (where $[.]$ is the greatest integer function and $A, B \in N$ ) is