$\frac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} = $

For a positive integer $n$,let ${f_n}(\theta ) = \left( {\tan \frac{\theta }{2}} \right)\,(1 + \sec \theta )\,(1 + \sec 2\theta )\,(1 + \sec 4\theta ) \dots (1 + \sec {2^n}\theta ).$ Then

If $\tan \left( \frac{\pi}{4} + \theta \right) + \tan \left( \frac{\pi}{4} - \theta \right) = \lambda \sec 2\theta$,then $\lambda$ =

The value of $\log \tan 1^{\circ} + \log \tan 2^{\circ} + \log \tan 3^{\circ} + \ldots + \log \tan 89^{\circ}$ is equal to

$\tan 7 \frac{1}{2}^{\circ}$ is equal to

If $\tan \alpha$ equals the integral solution of the inequality $4x^2 - 16x + 15 < 0$ and $\cos \beta$ equals the slope of the bisector of the first quadrant,then $\sin(\alpha + \beta)\sin(\alpha - \beta)$ is equal to