Question

The value of $\dfrac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)}$ equals to
A . $\dfrac{\cos A+\sin A}{\cos A-\sin A}$
B . $\dfrac{\cos A-\sin A}{\cos A-\sin A}$
C . $\dfrac{\cos A+\sin A}{\cos A+\sin A}$
D . None of these

Answer 1

Hint: In this question, we have to find the value of $\dfrac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)}$ by simplifying it. First we use the complimentary angles of trigonometric ratios to change the functions. After conversion, cos value changes to sin value and the question will be in sin form. Then we apply the formula of $\sin A+\sin B$for further solving. By using the formula and simplifying the equation, we are able to get the desired value.

Formula Used:
To solve this question, we use the identity which is described below:-
$\sin ({{90}^{\circ }}-A)=\cos A$
$\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$

Complete Step- by- step Solution:
Given that $\dfrac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)}$ …………………………. (1)
We know the complementary angles of trigonometric ratios which is given by
$\sin ({{90}^{\circ }}-A)=\cos A$
Then equation (1) becomes
$\dfrac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)}$= $\dfrac{\sin (B+A)+\sin ({{90}^{\circ }}-(B-A))}{\sin (B-A)+\sin ({{90}^{\circ }}-(B+A))}$
Simplifying it, we get
$\dfrac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)}$ = $\dfrac{\sin (B+A)+\sin ({{90}^{\circ }}-B+A)}{\sin (B-A)+\sin ({{90}^{\circ }}-B-A)}$…………. (2)
We know the trigonometric identities
$\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$
Equating the equation (2) with the above formula, we get
$\dfrac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)}$ = $\dfrac{2\sin \left( {{45}^{\circ }}+A \right)\cos \left( B-{{45}^{\circ }} \right)}{2\sin ({{45}^{\circ }}-A)\cos (B-{{45}^{\circ }})}$
Further solving the above equation, we get
$\dfrac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)}$ = $\dfrac{\sin \left( {{45}^{\circ }}+A \right)}{\sin ({{45}^{\circ }}-A)}$
We know $\sin (a+b)=\sin a\cos b+\cos a\sin b$
And $\sin (a-b)=\sin a\cos b-\cos a\sin b$
Then $\sin ({{45}^{\circ }}+A)=\sin {{45}^{\circ }}\cos A+\cos {{45}^{\circ }}\sin A$
And $\sin ({{45}^{\circ }}-A)=\sin {{45}^{\circ }}\cos A-\cos {{45}^{\circ }}\sin A$
Hence $\dfrac{\sin ({{45}^{\circ }}+A)=\sin {{45}^{\circ }}\cos A+\cos {{45}^{\circ }}\sin A}{\sin ({{45}^{\circ }}-A)=\sin {{45}^{\circ }}\cos A-\cos {{45}^{\circ }}\sin A}$
Thus $\dfrac{\sin ({{45}^{\circ }}+A)}{\sin ({{45}^{\circ }}-A)}=\dfrac{\dfrac{\sqrt{2}}{2}\cos A+\dfrac{\sqrt{2}}{2}\sin A}{\dfrac{\sqrt{2}}{2}\cos A-\dfrac{\sqrt{2}}{2}\sin A}$
$\dfrac{\sin ({{45}^{\circ }}+A)}{\sin ({{45}^{\circ }}-A)}=\dfrac{\dfrac{\sqrt{2}}{2}(\cos A+\sin A)}{\dfrac{\sqrt{2}}{2}(\cos A-\sin A)}$
Then $\dfrac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)}$ = $\dfrac{\cos A+\sin A}{\cos A-\sin A}$

Hence, the option A is correct.

Note: Whenever you get this type of question, important is to know about the complementary angles of trigonometric ratios and how to write the equation in the form of complementary angles. We know any two angle with their sum of ${{90}^{\circ }}$are called complementary. So the complement of any angle is that value which is obtained by subtracting it from ${{90}^{\circ }}$. According to trigonometric complimentary ratio theorem, trigonometric function of complementary angle is defined as another trigonometric function of the original angle. Then
$\sin ({{90}^{\circ }}-A)=\cos A$
$\cos ({{90}^{\circ }}-A)=\sin A$
$tan({{90}^{\circ }}-A)=\cot A$
$\cot ({{90}^{\circ }}-A)=\tan A$
$\sec ({{90}^{\circ }}-A)=\cos ecA$
$\cos ec({{90}^{\circ }}-A)=\sec A$
These are some trigonometric functions which are used to solve the questions based on trigonometry.