Question

prove that , sec square 0 + cosec square 0 = sec square 0 × cosec square 0​

Answer 1

Step-by-step explanation:

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Answer 2

FORMULA TO BE IMPLEMENTED

${\sin }^{2} \theta + {\cos }^{2} \theta = 1$

TO PROVE

${\sec }^{2} \theta + {\cosec }^{2} \theta = {\sec }^{2} \theta \times {\cosec }^{2} \theta$

PROOF

We are aware of the Trigonometric identity that

${\sin }^{2} \theta + {\cos }^{2} \theta = 1$

$\sf \:{Dividing \: both \: sides \: by \: \: {\sin }^{2} \theta \times {\cos }^{2} \theta \: we \: get }$

$\displaystyle \: \frac{ {\sin }^{2} \theta }{{\sin }^{2} \theta \times {\cos }^{2} \theta} + \frac{ {\cos }^{2} \theta }{{\sin }^{2} \theta \times {\cos }^{2} \theta} = \frac{1}{ \frac{ {\sin }^{2} \theta }{{\sin }^{2} \theta \times {\cos }^{2} \theta} }$

$\implies \: \displaystyle \: \frac{ 1 }{ {\cos }^{2} \theta} + \frac{1 }{{\sin }^{2} \theta } = \frac{1}{{\sin }^{2} \theta \times {\cos }^{2} \theta}$

$\implies \: {\sec }^{2} \theta + {\cosec }^{2} \theta = {\sec }^{2} \theta \times {\cosec }^{2} \theta$