prove that (1/(cos0 cos1 ))+(1/(cos1 cos2))+......+(1/(cos88 cos89))=((cos1)/(sin^2 1))


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Question Number 181592 by mathlove last updated on 27/Nov/22

$p r o v e t h a t$ $\frac{1}{c o s 0 c o s 1} + \frac{1}{c o s 1 c o s 2} + . . . . . . + \frac{1}{c o s 88 c o s 89} = \frac{c o s 1}{{s i n}^{2} 1}$
	Answered by som(math1967) last updated on 27/Nov/22

	$\frac{1}{s i n 1} [\frac{s i n (1 - 0)}{c o s 0 c o s 1} + \frac{s i n (2 - 1)}{c o s 1 c o s 2} +$ $. . . + \frac{s i n (89 - 88)}{c o s 88 c o s 89}]$ $= \frac{1}{s i n 1} [\frac{s i n 1 c o s 0}{c o s 1 c o s 0} + \frac{s i n 0 c o s 1}{c o s 0 c o s 1}$ $+ \frac{s i n 2 c o s 1}{c o s 2 c o s 1} + \frac{s i n 1 c o s 2}{c o s 1 c o s 2} + . . . + \frac{s i n 89 s i n 88}{c o s 88 c o s 89}]$ $= \frac{1}{s i n 1} [t a n 1 - t a n 0 + t a n 2 - t a n 1$ $+ . . . + t a n 88 - t a n 87 + t a n 89 - t a n 88]$ $= \frac{1}{s i n 1} \times t a n 89$ $= \frac{1}{s i n 1} \times \frac{s i n 89}{c o s 89}$ $= \frac{1}{s i n 1} \times \frac{c o s (90 - 89)}{s i n (90 - 89)}$ $= \frac{1}{s i n 1} \times \frac{c o s 1}{s i n 1} = \frac{c o s 1}{{s i n}^{2} 1} [p r o v e d]$
	Commented by mathlove last updated on 27/Nov/22

	$t h a n k s$
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