Question

$\left[\frac{{\sin }^{2}A-{\sin }^{2}B}{\sin A\cos A-\sin B\cos B}\right]=$

• A. $\tan (A–B)$
• B. $\tan (A+B)$
• C. $\cot \left(A-B\right)$
• D. $\cot \left(A+B\right)$

Answer 1

The correct option is B

$\tan (A+B)$

Explanation for the correct answer:

Simplifying the given expression:

$\left[\frac{{\sin }^{2}A-{\sin }^{2}B}{\sin A\cos A-\sin B\cos B}\right]$

Multiplying the numerator and denominator by $2$,

$$\begin{gathered} =\left[\frac{2\left({\sin }^{2}A-{\sin }^{2}B\right)}{2\left(\sin A\cos A-\sin B\cos B\right)}\right] \\ =\frac{2\left(\sin \left(A+B\right)\sin \left(A-B\right)\right)}{\sin 2A-\sin 2B}\mathbf{[}\mathbf{\because }{\mathbf{sin}}^{\mathbf{2}}\mathbf{A}\mathbf{}\mathbf{-}\mathbf{}{\mathbf{sin}}^{\mathbf{2}}\mathbf{B}\mathbf{=}\mathbf{sin}\left(A+B\right)\mathbf{sin}\left(A-B\right)\mathbf{}\mathbf{and}\mathbf{}\mathbf{2}\left(\sin A\cos A-\sin B\cos B\right)\mathbf{=}\mathbf{sin}\mathbf{2}\mathbf{A}\mathbf{-}\mathbf{sin}\mathbf{2}\mathbf{B}\mathbf{]} \end{gathered}$$

Using the formula $\sin x–\sin y=2\cos \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)$

$$\begin{gathered} =\frac{2\sin (A+B)\sin (A–B)}{2\cos \left({\displaystyle \frac{2(A+B)}{2}}\right)\sin \left({\displaystyle \frac{2(A–B)}{2}}\right)} \\ =\frac{2\sin (A+B)\sin (A–B)}{2\cos (A+B)\sin \left(A–B\right)} \\ =\frac{\sin \left(A+B\right)}{\cos \left(A+B\right)} \\ =\tan \left(A+B\right) \end{gathered}$$

Therefore, the correct answer is option (B).