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Mathematics

Prove that :

(tan A+1cos A)2+(tan A−1cos A)2=2(1 + sin2A1 - sin2A)\Big(\text{tan A} + \dfrac{1}{\text{cos A}}\Big)^2 + \Big(\text{tan A} - \dfrac{1}{\text{cos A}}\Big)^2 = 2\Big(\dfrac{\text{1 + sin}^2 A}{\text{1 - sin}^2 A}\Big)

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Answer

Solving L.H.S. of the equation :

⇒(tan A+1cos A)2+(tan A−1cos A)2⇒(sin Acos A+1cos A)2+(sin Acos A−1cos A)2⇒(sin A + 1cos A)2+(sin A - 1cos A)2⇒sin2A+1+2 sin Acos2A+sin2A+1−2 sin Acos2A⇒sin2A+1+2 sin A+sin2A+1−2 sin Acos2A⇒2(1 + sin2A)cos2A\Rightarrow \Big(\text{tan A} + \dfrac{1}{\text{cos A}}\Big)^2 + \Big(\text{tan A} - \dfrac{1}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{\text{sin A}}{\text{cos A}} + \dfrac{1}{\text{cos A}}\Big)^2 + \Big(\dfrac{\text{sin A}}{\text{cos A}} - \dfrac{1}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{\text{sin A + 1}}{\text{cos A}}\Big)^2 + \Big(\dfrac{\text{sin A - 1}}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + 1 + \text{2 sin A}}{\text{cos}^2 A} + \dfrac{\text{sin}^2 A + 1 - \text{2 sin A}}{\text{cos}^2 A} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + 1 + \text{2 sin A} + \text{sin}^2 A + 1 - \text{2 sin A}}{\text{cos}^2 A} \\[1em] \Rightarrow \dfrac{\text{2(1 + sin}^2 A)}{\text{cos}^2 A}

By formula,

cos2 A = 1 - sin2 A

⇒2(1+sin2A1−sin2A)\Rightarrow 2\Big(\dfrac{1 + \text{sin}^2 A}{1 - \text{sin}^2 A}\Big)

Since, L.H.S. = R.H.S.

Hence, proved that

(tan A+1cos A)2+(tan A−1cos A)2=2(1 + sin2A1 - sin2A)\Big(\text{tan A} + \dfrac{1}{\text{cos A}}\Big)^2 + \Big(\text{tan A} - \dfrac{1}{\text{cos A}}\Big)^2 = 2\Big(\dfrac{\text{1 + sin}^2 A}{\text{1 - sin}^2 A}\Big).

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