Prove the following identities :
tan2 A - sin2 A = tan2 A. sin2 A
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Solving L.H.S. of the equation :
⇒tan2A−sin2A⇒sin2Acos2A−sin2A⇒sin2A×(1cos2A−1)⇒sin2A×(1−cos2Acos2A)\Rightarrow \text{tan}^2 A - \text{sin}^2 A \\[1em] \Rightarrow \dfrac{\text{sin}^2 A}{\text{cos}^2 A} - \text{sin}^2 A \\[1em] \Rightarrow \text{sin}^2 A \times \Big(\dfrac{1}{\text{cos}^2 A} - 1\Big) \\[1em] \Rightarrow \text{sin}^2 A \times \Big(\dfrac{1 - \text{cos}^2 A}{\text{cos}^2 A}\Big)⇒tan2A−sin2A⇒cos2Asin2A−sin2A⇒sin2A×(cos2A1−1)⇒sin2A×(cos2A1−cos2A)
By formula,
1 - cos2A = sin2 A.
⇒sin2A×sin2Acos2A⇒sin2A.tan2A\Rightarrow \text{sin}^2 A \times \dfrac{\text{sin}^2 A}{\text{cos}^2 A} \\[1em] \Rightarrow \text{sin}^2 A. \text{tan}^2 A⇒sin2A×cos2Asin2A⇒sin2A.tan2A
Since, L.H.S. = R.H.S.
Hence, proved that tan2 A - sin2 A = tan2 A. sin2 A.
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sec2 A + cosec2 A = sec2 A . cosec2 A
(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
(cosec A + sin A)(cosec A - sin A) = cot2 A + cos2 A
(cos A + sin A)2 + (cos A - sin A)2 = 2
