Question

If $(\sec A+\tan A-1)(\sec A-\tan A+1)=a\tan A$.Find $a$

Answer 1

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

$\Rightarrow {\sec }^{2}A-{(\tan A-1)}^2\{\because (a-b)(a+b)=a^2-b^2\}$

$\Rightarrow {\sec }^{2}A-({\tan }^{2}A+1-2\tan A)$

$\Rightarrow {\sec }^{2}A-({\sec }^{2}A-2\tan A)\{\because 1+{\tan }^{2}A={\sec }^{2}A\}$

$\Rightarrow {\sec }^{2}A-{\sec }^{2}A+2\tan A$

$\Rightarrow 2\tan A$

But given that-

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

$\therefore a=2$

Hence, the correct answer is 2.