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Mathematics

Prove that :

cosec A + 1cosec A - 1\dfrac{\text{cosec A + 1}}{\text{cosec A - 1}} = (sec A + tan A)2

Trigonometric Identities

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Answer

To prove:

cosec A + 1cosec A - 1\dfrac{\text{cosec A + 1}}{\text{cosec A - 1}} = (sec A + tan A)2

Solving L.H.S. of the equation :

⇒1sin A+11sin A−1⇒1 + sin Asin A1 - sin Asin A⇒1 + sin A1 - sin A\Rightarrow \dfrac{\dfrac{1}{\text{sin A}} + 1}{\dfrac{1}{\text{sin A}} - 1} \\[1em] \Rightarrow \dfrac{\dfrac{\text{1 + sin A}}{\text{sin A}}}{\dfrac{\text{1 - sin A}}{\text{sin A}}} \\[1em] \Rightarrow \dfrac{\text{1 + sin A}}{\text{1 - sin A}}

Multiplying numerator and denominator by (1 + sin A), we get :

⇒1 + sin A1 - sin A×1 + sin A1 + sin A⇒(1 + sin A)21 - sin2A\Rightarrow \dfrac{\text{1 + sin A}}{\text{1 - sin A}} \times \dfrac{\text{1 + sin A}}{\text{1 + sin A}} \\[1em] \Rightarrow \dfrac{(\text{1 + sin A})^2}{\text{1 - sin}^2 A}

By formula,

1 - sin2 A = cos2 A

⇒(1 + sin A)2cos2A⇒(1 + sin Acos A)2⇒(1cos A+sin Acos A)2⇒(sec A + tan A)2.\Rightarrow \dfrac{\text{(1 + sin A)}^2}{\text{cos}^2 A} \\[1em] \Rightarrow \Big(\dfrac{\text{1 + sin A}}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow (\text{sec A + tan A})^2.

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

Hence, proved that cosec A + 1cosec A - 1\dfrac{\text{cosec A + 1}}{\text{cosec A - 1}} = (sec A + tan A)2.

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