Question

If $\sec A\tan B+\tan A\sec B=91$, then the value of $(\sec A\sec B+\tan A\tan B{)}^2$ is equal to

Answer 1

Given, $\sec A\tan B+\tan A\sec B=91$

Squaring both sides, we get
$\Rightarrow {\sec }^{2}A{\tan }^{2}B+{\tan }^{2}A{\sec }^{2}B+2\tan A\tan B\sec A\sec B=8281$

$\Rightarrow 2\tan A\tan B\sec A\sec B=8281-{\sec }^{2}A{\tan }^{2}B-{\tan }^{2}A{\sec }^{2}B......\left(1\right)$

Consider, $(secAsecB+tanAtanB{)}^2$

$={\sec }^{2}A{\sec }^{2}B+{\tan }^{2}A{\tan }^{2}B+2\tan A\tan B\sec A\sec B$

$={\sec }^{2}A{\sec }^{2}B+{\tan }^{2}A{\tan }^{2}B+8281-{\sec }^{2}A{\tan }^{2}B-{\tan }^{2}A{\sec }^{2}B$ $\left(by\right(1\left)\right)$

$={\sec }^{2}A{\sec }^{2}B-{\sec }^{2}A{\tan }^{2}B-{\tan }^{2}A{\sec }^{2}B+{\tan }^{2}A{\tan }^{2}B+8281$

$={\sec }^{2}A({\sec }^{2}B-{\tan }^{2}B)-{\tan }^{2}A({\sec }^{2}B-{\tan }^{2}B)+8281$

$={\sec }^{2}A-{\tan }^{2}A+8281$

$=1+8281$

$=8282$

Hence, the value is $8282$