If tan -1-k-0205 -12- k -2 5 -12- k -1 -2-tan -1 a- then the value of

Tan^ 1[∑_k=0^20(5π12+kπ2)(5π12+(k+1)π2)] =tan^ 1[ nbsp; ∑_k=0^201cos(5π12+kπ2)cos(5π12+(k+1)π2)] =tan^ 1[ nbsp; ∑_k=0^20sin((5π12+(k+1)π2) (5π12+kπ2))co ...

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Byju's Answer

Standard XII

Mathematics

Property 2

If tan -1[∑k=...

Question

If ${tan}^{- 1} [20 \sum k = 0 sec (\frac{5 π}{12} + \frac{k π}{2}) sec (\frac{5 π}{12} + \frac{(k + 1) π}{2})] = - {tan}^{- 1} a$ , then the value of $a$ is

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Solution

${tan}^{- 1} [20 \sum k = 0 sec (\frac{5 π}{12} + \frac{k π}{2}) sec (\frac{5 π}{12} + \frac{(k + 1) π}{2})]$

$= {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 20 \sum k = 0 \frac{1}{cos (\frac{5 π}{12} + \frac{k π}{2}) cos (\frac{5 π}{12} + \frac{(k + 1) π}{2})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$= {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 20 \sum k = 0 \frac{sin ((\frac{5 π}{12} + \frac{(k + 1) π}{2}) - (\frac{5 π}{12} + \frac{k π}{2}))}{cos (\frac{5 π}{12} + \frac{k π}{2}) cos (\frac{5 π}{12} + \frac{(k + 1) π}{2})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$= {tan}^{- 1} [20 \sum k = 0 tan (\frac{5 π}{12} + \frac{(k + 1) π}{2}) - tan (\frac{5 π}{12} + \frac{k π}{2})]$

$= {tan}^{- 1} [(tan (\frac{5 π}{12} + \frac{π}{2}) - tan (\frac{5 π}{12})) + (tan (\frac{5 π}{12} + π) - tan (\frac{5 π}{12} + \frac{π}{2})) + . . . + (tan (\frac{5 π}{12} + \frac{21 π}{2}) - tan (\frac{5 π}{12} + \frac{20 π}{2}))]$

$= {tan}^{- 1} [tan (\frac{5 π}{12} + \frac{21 π}{2}) - tan (\frac{5 π}{12})]$

$= {tan}^{- 1} [- cot (\frac{5 π}{12}) - tan (\frac{5 π}{12})]$

$= {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎣ - ⎛ ⎜ ⎜ ⎜ ⎝ \frac{2}{sin \frac{5 π}{6}} ⎞ ⎟ ⎟ ⎟ ⎠ ⎤ ⎥ ⎥ ⎥ ⎦$

$= {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ - \frac{2}{sin (π - \frac{π}{6})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$= {tan}^{- 1} (- 4)$
$= - {tan}^{- 1} 4$

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If tan−1[20∑k=0sec(5π12+kπ2)sec(5π12+(k+1)π2)]=−tan−1a, then the value of a is

If ${tan}^{- 1} [20 \sum k = 0 sec (\frac{5 π}{12} + \frac{k π}{2}) sec (\frac{5 π}{12} + \frac{(k + 1) π}{2})] = - {tan}^{- 1} a$ , then the value of $a$ is