Question

Find the value of $\dfrac{{{{\sin }^2}A - {{\sin }^2}B}}{{\sin A\cos A - \sin B\cos B}} = $
A. $\tan \left( {A - B} \right)$
B. $\tan \left( {A + B} \right)$
C. $\cot \left( {A - B} \right)$
D. $\cot \left( {A + B} \right)$

Answer 1

Hint: In order to solve this type of question, first we will consider the given equation. Then we will simplify it. Next, we will apply trigonometric identities in the equation formed above and simplify it even further to get the desired correct answer.

Formula used:
$\left[ {\because {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right]$
$\left[ {\because 2\sin \theta \cos \theta = \sin 2\theta } \right]$
$\left[ {\because \sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)} \right]$
$\left[ {\because \dfrac{{\sin A}}{{\cos A}} = \tan A} \right]$

Complete step by step solution:
Consider,
$\dfrac{{{{\sin }^2}A - {{\sin }^2}B}}{{\sin A\cos A - \sin B\cos B}}$
$ = \dfrac{{\sin \left( {A + B} \right)\sin \left( {A - B} \right)}}{{\sin A\cos A - \sin B\cos B}}$ $\left[ {\because {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right]$
Multiply numerator and denominator by $2,$
$ = \dfrac{{2\sin \left( {A + B} \right)\sin \left( {A - B} \right)}}{{2\sin A\cos A - 2\sin B\cos B}}$
$ = \dfrac{{2\sin \left( {A + B} \right)\sin \left( {A - B} \right)}}{{\sin 2A - \sin 2B}}$ $\left[ {\because 2\sin \theta \cos \theta = \sin 2\theta } \right]$
Simplifying it,
$ = \dfrac{{2\sin \left( {A + B} \right)\sin \left( {A - B} \right)}}{{2\cos \left( {A + B} \right)\sin \left( {A - B} \right)}}$ $\left[ {\because \sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)} \right]$
$ = \dfrac{{\sin \left( {A + B} \right)}}{{\cos \left( {A + B} \right)}}$
$ = \tan \left( {A + B} \right)$ $\left[ {\because \dfrac{{\sin A}}{{\cos A}} = \tan A} \right]$
$\therefore $ The correct option is B.

Note: Choose the suitable trigonometric identities and be very sure while simplifying them. This type of question requires the use of correct application of trigonometric rules to get the correct answer. Sometimes students get confused with the formula $\left[ {\because \sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)} \right]$ and $\left[ {\because \sin A - \sin B = 2\cos \left( {\dfrac{{A - B}}{2}} \right)\sin \left( {\dfrac{{A + B}}{2}} \right)} \right]$. But we need to choose the correct formula.