Question

Prove that : Sec²∅ - Cot² (90° - ∅) = Cos²(90° - ∅) + Cos²∅.
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Answer 1

Step-by-step explanation:

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Answer 2

$\huge\underline\mathrm{SOLUTION:-}$

$\mathsf {L.H.S = \sec {}^{2} (\theta) - \cot {}^{2} (\theta) (90 - \theta)}$

$\mathsf {= \sec {}^{2} (\theta) - \tan {}^{2} (\theta) }$

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

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

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

$\mathsf {R.H.S\: = \cos {}^{2} (90 - \theta) + \cos {}^{2} \theta)}$

$\mathsf { \sin {}^{2} (\theta) + \cos {}^{2} (\theta) }$

$\mathsf {= 1}$

$\mathsf \red{L.H.S = R.H.S}$

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