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Mathematics

Prove that :

sin⁡A−sin⁡3Acos⁡3A−cos⁡A×(sec⁡A−cosec⁡A)=cosec⁡A(cot⁡A−1)\dfrac{\sin A - \sin^3 A}{\cos^3 A - \cos A} \times (\sec A - \cosec A) = \cosec A(\cot A - 1)

Trigonometric Identities

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Answer

Solving L.H.S,

⇒sin⁡A−sin⁡3Acos⁡3A−cos⁡A×(sec⁡A−cosec⁡A)⇒sin⁡A(1−sin⁡2A)cos⁡A(cos⁡2A−1)×(1cos⁡A−1sin⁡A)⇒sin⁡A(1−sin⁡2A)−cos⁡A(1−cos⁡2A)×(1cos⁡A−1sin⁡A)⇒sin⁡A(cos⁡2A)−cos⁡A(sin⁡2A)×(sin⁡A−cos⁡Acos⁡Asin⁡A)⇒−cos⁡Asin⁡A×(sin⁡A−cos⁡Acos⁡Asin⁡A)⇒−(sin⁡A−cos⁡Asin⁡2A)⇒cos⁡A−sin⁡Asin⁡2A⇒1sin⁡A(cos⁡A−sin⁡Asin⁡A)⇒1sin⁡A(cos⁡Asin⁡A−sin⁡Asin⁡A)⇒cosec⁡A(cot⁡A−1).\Rightarrow \dfrac{\sin A - \sin^3 A}{\cos^3 A - \cos A} \times (\sec A - \cosec A) \\[1em] \Rightarrow \dfrac{\sin A(1 - \sin^2 A)}{\cos A(\cos^2 A - 1)} \times \Big(\dfrac{1}{\cos A} - \dfrac{1}{\sin A}\Big) \\[1em] \Rightarrow \dfrac{\sin A(1 - \sin^2 A)}{-\cos A(1- \cos^2 A)} \times \Big(\dfrac{1}{\cos A} - \dfrac{1}{\sin A}\Big) \\[1em] \Rightarrow \dfrac{\sin A (\cos^2 A)}{-\cos A(\sin^2 A)} \times \Big(\dfrac{\sin A - \cos A}{\cos A \sin A}\Big) \\[1em] \Rightarrow -\dfrac{\cos A}{\sin A} \times \Big(\dfrac{\sin A - \cos A}{\cos A \sin A}\Big) \\[1em] \Rightarrow -\Big(\dfrac{\sin A - \cos A}{\sin^2 A}\Big) \\[1em] \Rightarrow \dfrac{\cos A - \sin A}{\sin^2 A} \\[1em] \Rightarrow \dfrac{1}{\sin A} \Big(\dfrac{\cos A - \sin A}{\sin A}\Big) \\[1em] \Rightarrow \dfrac{1}{\sin A} \Big(\dfrac{\cos A}{\sin A} - \dfrac{\sin A}{\sin A}\Big) \\[1em] \Rightarrow \cosec A (\cot A - 1).

Since,

L.H.S = R.H.S

Hence, proved that

sin⁡A−sin⁡3Acos⁡3A−cos⁡A×(sec⁡A−cosec⁡A)=cosec⁡A(cot⁡A−1)\dfrac{\sin A - \sin^3 A}{\cos^3 A - \cos A} \times (\sec A - \cosec A) = \cosec A(\cot A - 1).

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