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Prove that : (sin A + sec A)2 + (cos A + cosec A)2 = (1 + sec A. cosec A)2.

Trigonometric Identities

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Answer

To prove:

(sin A + sec A)2 + (cos A + cosec A)2 = (1 + sec A. cosec A)2.

Solving L.H.S. of the equation :

⇒ (sin A + sec A)2 + (cos A + cosec A)2

⇒ sin2 A + sec2 A + 2 sin A sec A + cos2 A + cosec2 A + 2 cos A cosec A

⇒ sin2 A + cos2 A + sec2 A + cosec2 A + 2 sin A sec A + 2 cos A cosec A

⇒ 1 + sec2 A + cosec2 A + 2 sin A sec A + 2 cos A cosec A

⇒1+1cos2A+1sin2A+2 sin A×1cos A+2 cos A×1sin A⇒1+sin2A+cos2Asin2A cos2A+2 sin Acos A+2 cos Asin A⇒1+1sin2A cos2A+2 sin2A+2 cos2Asin A cos A⇒1+sec2A cosec2A+2(sin2A+cos2A)sin A cos A⇒1+sec2A cosec2A+2sin A cos A⇒1+sec2A cosec2A+2 sec A cosec A⇒(1 + sec A cosec A)2.\Rightarrow 1 + \dfrac{1}{\text{cos}^2 A} + \dfrac{1}{\text{sin}^2 A} + \text{2 sin A} \times \dfrac{1}{\text{cos A}} + \text{2 cos A} \times \dfrac{1}{\text{sin A}} \\[1em] \Rightarrow 1 + \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin}^2 A\text{ cos}^2 A} + \dfrac{\text{2 sin A}}{\text{cos A}} + \dfrac{\text{2 cos A}}{\text{sin A}} \\[1em] \Rightarrow 1 + \dfrac{1}{\text{sin}^2 A\text{ cos}^2 A} + \dfrac{\text{2 sin}^2 A + \text{2 cos}^2 A}{\text{sin A cos A}} \\[1em] \Rightarrow 1 + \text{sec}^2 A\text{ cosec}^2 A + \dfrac{2(\text{sin}^2 A + \text{cos}^2 A)}{\text{sin A cos A}} \\[1em] \Rightarrow 1 + \text{sec}^2 A\text{ cosec}^2 A + \dfrac{2}{\text{sin A cos A}} \\[1em] \Rightarrow 1 + \text{sec}^2 A\text{ cosec}^2 A + \text{2 sec A cosec A} \\[1em] \Rightarrow (\text{1 + sec A cosec A})^2.

Hence, proved that (sin A + sec A)2 + (cos A + cosec A)2 = (1 + sec A. cosec A)2.

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