Question

find the value of k
$1 \div 1 +\sin \theta \: + \: 1 \div 1 - \sin \theta \: = k \sec^{2} \theta$
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Answer 1

Answer:

k = 2

Step-by-step explanation:

$\frac{1}{1 + \sin( \theta) } + \frac{1}{1 - \sin( \theta) } = \frac{1 - \sin(\theta) + 1 + \sin(\theta) }{(1 + \sin( \theta))(1 - \sin(\theta)) }$

$= \frac{2}{1 - { \sin( \theta) }^{2} }$

because (a+b)(a-b) = a^2 - b^2

now,

$1 - { \sin( \theta) }^{2} = { \cos( \theta) }^{2}$

therefore expression equals to 2/cos(theta)^2

and

$\frac{1}{ { \cos( \theta) }^{2} } = { \sec( \theta) }^{2}$

hence expression equals

$2 { \sec( \theta) }^{2}$

therefore, k = 2