Question

The $\left( \sec A+\tan A-1 \right)\left( \sec A-\tan A+1 \right)-2\tan A$ is equal to
A. $\sec A$
B. $2\sec A$
C. 0
D. 2

Answer 1

Hint: We use the identity formula of $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ assuming $a=\sec A;b=\tan A-1$. Then we simplify the whole expression and use the trigonometric formula of ${{\sec }^{2}}A={{\tan }^{2}}A+1$ to find the final solution.

Complete step by step answer:
We first multiply the $\left( \sec A+\tan A-1 \right)\left( \sec A-\tan A+1 \right)$ part by using the identity formula of $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$. In this case we assume $a=\sec A;b=\tan A-1$. Therefore,
$\left( \sec A+\tan A-1 \right)\left( \sec A-\tan A+1 \right) ={{\left( \sec A \right)}^{2}}-{{\left( \tan A-1 \right)}^{2}} $
Now we simplify the whole expression
$\left( \sec A+\tan A-1 \right)\left( \sec A-\tan A+1 \right)-2\tan A ={{\left( \sec A \right)}^{2}}-{{\left( \tan A-1 \right)}^{2}}-2\tan A \\
\Rightarrow \left( \sec A+\tan A-1 \right)\left( \sec A-\tan A+1 \right)-2\tan A ={{\sec }^{2}}A-{{\tan }^{2}}A-1+2\tan A-2\tan A \\
\Rightarrow \left( \sec A+\tan A-1 \right)\left( \sec A-\tan A+1 \right)-2\tan A ={{\sec }^{2}}A-{{\tan }^{2}}A-1 \\ $
Now we use the trigonometric formula of ${{\sec }^{2}}A={{\tan }^{2}}A+1$. We get
${{\sec }^{2}}A-{{\tan }^{2}}A-1 ={{\tan }^{2}}A+1-{{\tan }^{2}}A-1 \\
\therefore {{\sec }^{2}}A-{{\tan }^{2}}A-1 =0 $
Therefore, $\left( \sec A+\tan A-1 \right)\left( \sec A-\tan A+1 \right)-2\tan A=0$.

Hence, the correct option is C.

Note:The identity formula of ${{\sec }^{2}}A={{\tan }^{2}}A+1$ is derived from the relation between ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ and dividing it with ${{\cos }^{2}}A$. The assumption of $a=\sec A;b=\tan A-1$ can also be taken as $a=\sec A;b=1-\tan A$ as the square form omits the negative sign.