Question

If $\csc A + \sec A = \csc B + \sec B$ then $\tan A\tan B$ is equal to
a. $\tan \left( {\dfrac{{A + B}}{2}} \right)$
b. $\cot \left( {\dfrac{{A + B}}{2}} \right)$
c. $\sin \left( {\dfrac{{A + B}}{2}} \right)$
d. $\cos \left( {\dfrac{{A + B}}{2}} \right)$

Answer 1

Here we are asked to find the value of the expression $\tan A\tan B$ by using the given expression. The given expression can be expanded using the standard trigonometric identities. Then it can be simplified to find the required value. After finding the value we have to choose the correct option from the given options.
Formula: Trigonometric identities that we will be using in this problem:
$\csc \theta = \dfrac{1}{{\sin \theta }}$
$\sec \theta = \dfrac{1}{{\cos \theta }}$
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
$\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
$\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
$\cos A - \cos B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$

Complete step by step solution:
We aim to find the value of $\tan A\tan B$ using the given expression $\csc A + \sec A = \csc B + \sec B$ .
First, let us consider the given expression $\csc A + \sec A = \csc B + \sec B$ .
Let us re-arrange the given expression for our convenience.
$\sec A - \sec B = \csc B - \csc A$
Let us expand the above expression using the formula $\csc \theta = \dfrac{1}{{\sin \theta }}$and$\sec \theta = \dfrac{1}{{\cos \theta }}$ . Here the angles are $A$ and $B$ . Thus, we get
\sec A - \sec B = \csc B - \csc A$$$$ \Rightarrow \dfrac{1}{{\cos A}} + \dfrac{1}{{\cos B}} = \dfrac{1}{{\sin A}} + \dfrac{1}{{\sin B}}
Now let us simplify the above expression.
$\Rightarrow \dfrac{{\cos B - \cos A}}{{\cos A\cos B}} = \dfrac{{\sin A - \sin B}}{{\sin A\sin B}}$
Now let us take the numerator of the left-hand side to the denominator of the right-hand side and the denominator of the right-hand side to the numerator of the left-hand side.
$\Rightarrow \dfrac{{\sin A\sin B}}{{\cos A\cos B}} = \dfrac{{\sin A - \sin B}}{{\cos B - \cos A}}$
Now let us rearrange the left-hand side of the above equation for our convenience.
$\Rightarrow \dfrac{{\sin A}}{{\cos A}} \times \dfrac{{\sin B}}{{\cos B}} = \dfrac{{\sin A - \sin B}}{{\cos B - \cos A}}$
As we know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ we can write the above equation as
$\Rightarrow \tan A\tan B = \dfrac{{\sin A - \sin B}}{{\cos B - \cos A}}$
Since here the angles are $A$ and $B$ .
Now let us simplify the right-hand side of the above equation.
We already know that from the trigonometric equation, $\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$ and $\cos A - \cos B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$ . Using these identities in the above equation we get
$\Rightarrow \tan A\tan B = \dfrac{{2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)}}{{2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)}}$
On simplifying the above equation, we get
$\Rightarrow \tan A\tan B = \dfrac{{\cos \left( {\dfrac{{A + B}}{2}} \right)}}{{\sin \left( {\dfrac{{A + B}}{2}} \right)}}$
Now using the formula $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ on the right-hand side of the above equation we get
$\Rightarrow \tan A\tan B = \cot \left( {\dfrac{{A + B}}{2}} \right)$
Since here the angle is $\left( {\dfrac{{A + B}}{2}} \right)$
Thus, we got the value of the expression $\tan A\tan B$ as $\cot \left( {\dfrac{{A + B}}{2}} \right)$ . Now let us see the options for a correct answer.
Option (a) $\tan \left( {\dfrac{{A + B}}{2}} \right)$ is an incorrect answer as we got the value $\cot \left( {\dfrac{{A + B}}{2}} \right)$ from our calculation.
Option (b) $\cot \left( {\dfrac{{A + B}}{2}} \right)$ is the correct answer as we got the same value in our calculation above.
Option (c) $\sin \left( {\dfrac{{A + B}}{2}} \right)$ is an incorrect answer as we got the value $\cot \left( {\dfrac{{A + B}}{2}} \right)$ from our calculation.
Option (a) $\cos \left( {\dfrac{{A + B}}{2}} \right)$ is an incorrect answer as we got the value $\cot \left( {\dfrac{{A + B}}{2}} \right)$ from our calculation.
Hence, option (b) $\cot \left( {\dfrac{{A + B}}{2}} \right)$ is the correct answer.

So, the correct answer is “Option b”.

Note: In this problem, we have used the formulas $\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$ and$\cos A - \cos B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$ . These are derived as a result by using the standard formulas,
1. $\sin (A + B) = \sin A\cos B + \cos A\sin B$
2. $\sin (A - B) = \sin A\cos B - \cos A\sin B$
3. $\cos (A + B) = \cos A\cos B - \sin A\sin B$
4. $\cos (A - B) = \cos A\cos B + \sin A\sin B$