Question

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

Answer 1

Hint-In this question, we have to use the trigonometric identities. We can see in question all terms in sec and tan so we can use the most common identity ${\sec ^2}\theta = 1 + {\tan ^2}\theta$ but sometimes we can use as ${\sec ^2}\theta - {\tan ^2}\theta = 1$.

Complete step-by-step solution -
Given, $\sec A\tan B + \tan A\sec B = 91$
Now, squaring both sides
$\Rightarrow {\left( {\sec A\tan B + \tan A\sec B} \right)^2} = {\left( {91} \right)^2}$
Use algebraic identity, ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$

$$\Rightarrow {\left( {\sec A\tan B} \right)^2} + {\left( {\tan A\sec B} \right)^2} + 2 \times \sec A\tan B \times \tan A\sec B = {\left( {91} \right)^2} \\\ \Rightarrow 2 \times \sec A\tan B \times \tan A\sec B = 8281 - {\sec ^2}A{\tan ^2}B - {\tan ^2}A{\sec ^2}B...............\left( 1 \right) \\\$$

Now, we have to find the value of ${\left( {\sec A\sec B + \tan A\tan B} \right)^2}$ so we open the square by using algebraic identity, ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ .
$\Rightarrow {\left( {\sec A\sec B + \tan A\tan B} \right)^2} = {\left( {\sec A\sec B} \right)^2} + {\left( {\tan A\tan B} \right)^2} + 2 \times \sec A\sec B \times \tan A\tan B \\\ \Rightarrow {\left( {\sec A\sec B + \tan A\tan B} \right)^2} = {\sec ^2}A{\sec ^2}B + {\tan ^2}A{\tan ^2}B + 2 \times \sec A\sec B \times \tan A\tan B \\\$
Now, put the value of $2 \times \sec A\sec B \times \tan A\tan B$ From (1) equation.
$\Rightarrow {\left( {\sec A\sec B + \tan A\tan B} \right)^2} = {\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 \\\ \Rightarrow {\left( {\sec A\sec B + \tan A\tan B} \right)^2} = 8281 + {\sec ^2}A\left( {{{\sec }^2}B - {{\tan }^2}B} \right) - {\tan ^2}A\left( {{{\sec }^2}B - {{\tan }^2}B} \right) \\\$
We know, ${\sec ^2}\theta - {\tan ^2}\theta = 1$
$\Rightarrow {\left( {\sec A\sec B + \tan A\tan B} \right)^2} = 8281 + \left( {{{\sec }^2}A - {{\tan }^2}A} \right)$
Again use, ${\sec ^2}\theta - {\tan ^2}\theta = 1$
$\Rightarrow {\left( {\sec A\sec B + \tan A\tan B} \right)^2} = 8281 + 1 \\\ \Rightarrow {\left( {\sec A\sec B + \tan A\tan B} \right)^2} = 8282 \\\$
So, the value of ${\left( {\sec A\sec B + \tan A\tan B} \right)^2}$ is 8282.

Note-In such types of problems we have to express (evaluate) the required term and also the given mention in the question by using the trigonometric identities and then solve both the equations. So after calculation we will get the required answer.