Question

show that : sec2 0 x cosec2 0 = sec2 0 + cosec2 0

Answer 1

HEY MATE,
please see my answer
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LHS
$= > \sec^{2} (\theta) \times \cosec ^{2} (\theta)$

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

we know that
$({\sin}^{2}(\theta) + {\cos}^{2}(\theta) = 1)$

$= > \frac{ \sin {}^{2} (\theta) + \cos {}^{2} (\theta) }{ \sin {}^{2} (\theta) \cos {}^{2} (\theta) }$

$= > \frac{ \sin {}^{2} (\theta) }{ \sin {}^{2} (\theta) . \cos {}^{2} (\theta) } + \frac{ \cos {}^{2} ( \theta) }{\sin {}^{2} (\theta) . \cos {}^{2} (\theta)}$

$= > \frac{1}{ \cos {}^{2} (\theta) } + \frac{1}{ \sin {}^{2} (\theta) }$

$= > \sec {}^{2} (\theta) + \cosec {}^{2} (\theta)$
=RHS
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Answer 2

Answer:

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