The value of $\dfrac{1+\cot^2 \theta}{1+\tan^2 \theta}$ will be:
Prove that $\dfrac{1+\sec\theta}{\sec\theta}=\dfrac{\sin^2\theta}{1-\cos\theta}$.
If $3 \cot A = 4$, then the value of $\dfrac{1 - \tan^2 A}{1 + \tan^2 A}$ will be: