Question

Prove that : $\frac{1-{\tan }^2({45}^{\circ }-A)}{1+{\tan }^2({45}^{\circ }-A)}=\sin 2A$

Answer 1

$LHS=\frac{1-{\tan }^2({45}^{\circ }-A)}{1+{\tan }^2({45}^{\circ }-A)}$

$=\frac{1-{\tan }^2({45}^{\circ }-A)}{{\sec }^2({45}^{\circ }-A)}$

$=\frac{1}{{\sec }^2({45}^{\circ }-A)}-\frac{{\tan }^2({45}^{\circ }-A)}{{\sec }^2({45}^{\circ }-A)}$

$={\cos }^2({45}^{\circ }-A)-{\sin }^2({45}^{\circ }-A)$ [$\because \frac{\tan \theta }{\sec \theta }=\frac{\frac{\sin \theta }{\cos \theta }}{\frac{1}{\cos \theta }}=\sin \theta$]

$=\cos \left\{2\right({45}^{\circ }-A\left)\right\}$ [$\because \cos 2A={\cos }^{2}A-{\sin }^{2}A$]

$=\cos ({90}^{\circ }-2A)$

$=\sin 2A=RHS$ [$\because \cos (90-\theta )=\sin \theta$]