Question

$(\sec A+\tan A-1)(\sec A-\tan A+1)=$

• A. $2\sin A$
• B. $2\cos A$
• C. $2\sec A$
• D. $2\tan A$

Answer 1

The correct option is D $2\tan A$

$\begin{array}{c}\left(\sec A+\tan A-1\right)\left(\sec A-\tan A+1\right)\\ =\left[\sec A+\left(\tan A-1\right)\right]\left[\sec A-\left(\tan A-1\right)\right]\\ ={\sec }^{2}A-{\left(\tan A-1\right)}^2\because \left(\left(a-b\right)\left(a+b\right)=a^2-b^2\right)\\ ={\sec }^{2}A-\left({\tan }^{2}A-2\tan A+1\right)\\ ={\sec }^{2}A-{\tan }^{2}A+2\tan A-1\\ =1+2\tan A-1\because \left({\sec }^2\theta -{\tan }^2\theta =1\right)\\ =2\tan A\end{array}$

Hence,

Option $D$ is correct .