Prove the following identities :
sec2 A + cosec2 A = sec2 A . cosec2 A
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Solving L.H.S. of the equation :
⇒sec2A+cosec2A⇒1cos2A+1sin2A⇒sin2A+cos2Acos2A.sin2A\Rightarrow \text{sec}^2 A + \text{cosec}^2 A \\[1em] \Rightarrow \dfrac{1}{\text{cos}^2 A} + \dfrac{1}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos}^2 A. \text{sin}^2 A}⇒sec2A+cosec2A⇒cos2A1+sin2A1⇒cos2A.sin2Asin2A+cos2A
As, sin2 A + cos2 A = 1
⇒1cos2A.sin2A⇒sec2A.cosec2A\Rightarrow \dfrac{1}{\text{cos}^2 A. \text{sin}^2 A} \\[1em] \Rightarrow \text{sec}^2 A. \text{cosec}^2 A⇒cos2A.sin2A1⇒sec2A.cosec2A
Since, L.H.S. = R.H.S.
Hence, proved that sec2 A + cosec2 A = sec2 A . cosec2 A.
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cosec4 A - cosec2 A = cot4 A + cot2 A
sec A(1 - sin A)(sec A + tan A) = 1
(1 + tan2A)cot Acosec2A=tan A\dfrac{\text{(1 + tan}^2 A)\text{cot A}}{\text{cosec}^2 A} = \text{tan A}cosec2A(1 + tan2A)cot A=tan A
tan2 A - sin2 A = tan2 A. sin2 A
