Question

If $\frac{{\tan }^3\mathrm{\theta }-1}{\mathrm{tan\theta }-1}=A{\sec }^2\mathrm{\theta }+B\tan \mathrm{\theta },\mathrm{then}A+B=\mathrm{\_\_\_\_\_\_\_}.$

Answer 1

$$\begin{gathered} \frac{{\tan }^3\theta -1}{\tan \theta -1} \\ =\frac{\left(\tan \theta -1\right)\left({\tan }^2\theta +\tan \theta +1\right)}{\tan \theta -1}\left[a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\right] \\ ={\tan }^2\theta +\tan \theta +1 \\ ={\sec }^2\theta +\tan \theta \left(1+{\tan }^2\theta ={\sec }^2\theta \right) \end{gathered}$$

Comparing with the given expression, we get

A = 1 and B = 1

∴ A + B = 2

If $\frac{{\tan }^3\mathrm{\theta }-1}{\mathrm{tan\theta }-1}=A{\sec }^2\mathrm{\theta }+B\tan \mathrm{\theta },\mathrm{then}A+B=\underline{2}.$