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Mathematics

Prove the following identities :

(cosec A - sin A)(sec A - cos A)(tan A + cot A) = 1

Trigonometric Identities

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Answer

Solving L.H.S. of the equation :

(cosec A - sin A)(sec A - cos A)(tan A + cot A)(1sin Asin A)×(1cos Acos A)×(sin Acos A+cos Asin A)(1sin2Asin A)×(1cos2Acos A)×(sin2A+cos2Acos A. sin A)\Rightarrow \text{(cosec A - sin A)(sec A - cos A)(tan A + cot A)} \\[1em] \Rightarrow \Big(\dfrac{1}{\text{sin A}} - \text{sin A}\Big) \times \Big(\dfrac{1}{\text{cos A}} - \text{cos A}\Big) \times \Big(\dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}}\Big) \\[1em] \Rightarrow \Big(\dfrac{1 - \text{sin}^2 A}{\text{sin A}}\Big) \times \Big(\dfrac{1 - \text{cos}^2 A}{\text{cos A}}\Big) \times \Big(\dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos A. sin A}}\Big)

By formula,

1 - sin2 A = cos2 A, 1 - cos2 A = sin2 A and sin2 A + cos2 A = 1.

(cos2Asin A×sin2Acos A×1cos A. sin A)cos2A.sin2Acos2A.sin2A1.\Rightarrow \Big(\dfrac{\text{cos}^2 A}{\text{sin A}} \times \dfrac{\text{sin}^2 A}{\text{cos A}} \times \dfrac{1}{\text{cos A. sin A}}\Big) \\[1em] \Rightarrow \dfrac{\text{cos}^2 A. \text{sin}^2 A}{\text{cos}^2 A. \text{sin}^2 A} \\[1em] \Rightarrow 1.

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

Hence, proved that (cosec A - sin A)(sec A - cos A)(tan A + cot A) = 1.

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