Question

${\left(\frac{cosA+cosB}{sinA-sinB}\right)}^{2014}+{\left(\frac{sinA+sinB}{cosA-cosB}\right)}^{2014}=...........$

• A. ${2cot}^{2014}\left(\frac{A+B}{2}\right)$
• B. ${2cot}^{2014}\left(\frac{A-B}{2}\right)$
• C. ${2tan}^{2014}\left(\frac{A+B}{2}\right)$
• D. ${2tan}^{2014}\left(\frac{A-B}{2}\right)$

Answer 1

The correct option is B ${2cot}^{2014}\left(\frac{A-B}{2}\right)$

Given

${\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)}^{2014}+{\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)}^{2014}$

$\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$

$\cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$

$\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$

$\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$

$⟹\frac{\cos A+\cos B}{\sin A-\sin B}=\frac{2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}{2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)}$

$⟹\frac{\cos A+\cos B}{\sin A-\sin B}=\cot \left(\frac{A-B}{2}\right)$

$⟹{\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)}^{2014}={\cot }^{2014}\left(\frac{A-B}{2}\right)\cdots \left(1\right)$

$⟹\frac{\sin A+\sin B}{\cos A-\cos B}=\frac{2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}{-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)}$

$⟹\frac{\sin A+\sin B}{\cos A-\cos B}=-\cot \left(\frac{A-B}{2}\right)$

$⟹{\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)}^{2014}={\cot }^{2014}\left(\frac{A-B}{2}\right)\cdots \left(2\right)$

From $\left(1\right),\left(2\right)$

$⟹{\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)}^{2014}+{\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)}^{2014}={\cot }^{2014}\left(\frac{A-B}{2}\right)+{\cot }^{2014}\left(\frac{A-B}{2}\right)$

$⟹{\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)}^{2014}+{\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)}^{2014}=2{\cot }^{2014}\left(\frac{A-B}{2}\right)$