Question

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Answer 1

Answer:

0

Step-by-step explanation:

Let us simplify the expression in ∑ first.

k = 0

$\sec( \frac{7\pi}{12} + \frac{k\pi}{2} )\sec( \frac{7\pi}{12} + \frac{(k + 1)\pi}{2} )$

$= \sec( \frac{7\pi}{12} ) \sec( \frac{7\pi}{12} + \frac{\pi}{2} )$

$= - \sec( \frac{7\pi}{12} ) \csc( \frac{7\pi}{12} )$

k = 1

$\sec( \frac{7\pi}{12} + \frac{k\pi}{2} )\sec( \frac{7\pi}{12} + \frac{(k + 1)\pi}{2} )$

$= \sec( \frac{7\pi}{12} + \frac{\pi}{2} )\sec( \frac{7\pi}{12} + \pi)$

$= \csc( \frac{7\pi}{12} ) \sec( \frac{7\pi}{12} )$

k = 2

$\sec( \frac{7\pi}{12} + \frac{k\pi}{2} )\sec( \frac{7\pi}{12} + \frac{(k + 1)\pi}{2} )$

$= \sec( \frac{7\pi}{12} + \pi)\sec( \frac{7\pi}{12} + \frac{3\pi}{2} )$

$= - \sec( \frac{7\pi}{12} ) \csc( \frac{7\pi}{12} )$

k = 3

$\sec( \frac{7\pi}{12} + \frac{k\pi}{2} )\sec( \frac{7\pi}{12} + \frac{(k + 1)\pi}{2} )$

$= \sec( \frac{7\pi}{12} + \frac{3\pi}{2} )\sec( \frac{7\pi}{12} + 2\pi )$

$= \csc( \frac{7\pi}{12} )\sec( \frac{7\pi}{12})$

The pattern will repeat like this. Now

$∑[\sec( \frac{7\pi}{12} + \frac{k\pi}{2} )\sec( \frac{7\pi}{12} + \frac{(k + 1)\pi}{2} )]$

$= (-1+1-1+1-1+1-1+1-1+1-1) \sec( \frac{7\pi}{12} ) \csc( \frac{7\pi}{12} )$

$= - \sec( \frac{7\pi}{12} ) \csc( \frac{7\pi}{12} )$

The given expression becomes

$\sec {}^{ - 1} ( \frac{ - \sec( \frac{7\pi}{12} ) \csc( \frac{7\pi}{12} ) }{4 } )$

$= \sec {}^{ - 1} ( \frac{ - 1}{2.2 \sin( \frac{7\pi}{12} ) \cos( \frac{7\pi}{12} ) } )$

$= \sec {}^{ - 1} ( \frac{ - 1}{2 \sin( \frac{7\pi}{6} ) } )$

$= \sec {}^{ - 1} ( - \frac{1}{2} \csc(\pi + \frac{\pi}{6} ) )$

$= \sec {}^{ - 1} (\frac{1}{2} \csc( \frac{\pi}{6} ) )$

$= \sec {}^{ - 1} ( \frac{1}{2} \times 2 )$

$= \sec {}^{ - 1} (1)$

$= 0$

Answer 2

Answer:

Sec-1 [ 1/4 Σ sec ( 7π/12 + kπ/12) sec ( 7π/12 + (k+1)π/2) ]

= Sec-1 [ 1/4 Σ sec ( 7π/12 + kπ/12) cosec ( 7π/12 + kπ/2) ]

= Sec-1 [ -1/4 Σ 2/ sin ( 7π/6 + kπ)

= Sec-1 [ 1/2 Σ 1/(-1) ^(k+1)sin (π/6)

= Sec-1 [ Σ 1/(-1) ^(k+1)]

= Sec-1 (1)

= 0