Question

prove that 1-tan square A divided by 1+tan square A is equal to 2 cos square A -1 ​

Answer 1

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Answer 2

LHS:

$\displaystyle\frac{1-\tan^2A}{1+\tan^2A} \\ \\ \\ \frac{1-\tan^2A}{\sec^2A}\ \ \ \ \ [\because\ \sec^2\theta-\tan^2\theta=1\ \ \ ; \ \ \ \sec^2\theta=1+\tan^2\theta] \\ \\ \\ \frac{1}{\sec^2A}-\frac{\tan^2\theta}{\sec^2\theta} \\ \\ \\ \cos^2A-\sin^2A\ \ \ \ \ \left[\because\ \frac{\sec\theta}{\csc\theta}=\tan\theta\ \ ; \ \ \frac{\sec\theta}{\tan\theta}=\csc\theta\ \ ; \ \ \frac{\tan\theta}{\sec\theta}=\sin\theta\right] \\ \\ \\ \cos^2A-(1-\cos^2A) \\ \\ \\ \cos^2A-1+\cos^2A \\ \\ \\ 2\cos^2A-1$

:RHS

Hence proved!