Question

cosec2A+cosec4+cosec8A=cotA-cot8A

Answer 1

Answer:

math]\csc\,2A + \csc\,4A + \csc\,8A = \cot\,A - \cot\,8A[/math]

[math]\text{Right hand side}[/math]

[math]\cot\,A - \cot\,8A[/math]

[math]= \cot\,A - \dfrac{\cos\,8A}{\sin\,8A}[/math]

[math]= \cot\,A - \dfrac{2\cos^2\,4A - 1}{\sin\,8A}\quad\because\,\,(\cos\,2\theta = 2\cos^2\,\theta - 1)[/math]

[math]= \cot\,A - \dfrac{2\cos^2\,4A}{\sin\,8A} +\dfrac{ 1}{\sin\,8A}[/math]

[math]= \cot\,A - \dfrac{2\cos^2\,4A}{2\sin\,4A\cos\,4A} +\csc\,8A\quad \because \,\,(\sin\,2\theta = 2\sin\,\theta \cos\,\theta)[/math]

[math]= \cot\,A - \dfrac{\cos\,4A}{\sin\,4A} +\csc\,8A[/math]

[math]= \cot\,A - \dfrac{2\cos^2\,2A - 1}{\sin\,4A} +\csc\,8A[/math]

[math]= \cot\,A - \dfrac{2\cos^2\,2A}{\sin\,4A} +\dfrac{ 1}{\sin\,4A} +\csc\,8A[/math]

[math]= \cot\,A - \dfrac{2\cos^2\,2A}{2\sin\,2A\cos\,2A} +\csc\,4A +\csc\,8A[/math]

[math]= \cot\,A - \dfrac{\cos\,2A}{\sin\,2A} +\csc\,4A +\csc\,8A[/math]

[math]= \cot\,A - \dfrac{2\cos^2\,A - 1}{\sin\,2A} +\csc\,4A +\csc\,8A[/math]

[math]= \cot\,A - \dfrac{2\cos^2\,A}{\sin\,2A} +\dfrac{ 1}{\sin\,2A} +\csc\,4A +\csc\,8A[/math]

[math]= \cot\,A - \dfrac{2\cos^2\,A}{2\sin\,A\cos\,A} +\csc\,2A +\csc\,4A +\csc\,8A[/math]

[math]= \cot\,A -\cot\,A +\csc\,2A +\csc\,4A +\csc\,8A[/math]

[math]= \csc\,2A +\csc\,4A +\csc\,8A[/math]

[math]=\text{Left hand side}[/math]