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Mathematics

Prove that:

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

Trigonometric Identities

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Answer

Solving L.H.S.,

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

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

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