frac{{sin (B + A) + cos (B - A)}}{{sin (B - A | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given expression: $\frac{{\sin (B + A) + \cos (B - A)}}{{\sin (B - A) + \cos (B + A)}}$Using the identity $\cos 	heta = \sin(90^\circ - 	heta)$:

Vedclass · Accepted Answer

$\frac{{\cos A + \sin A}}{{\cos A - \sin A}}$

Vedclass · Answer

(B) Given expression: $\frac{{\sin (B + A) + \cos (B - A)}}{{\sin (B - A) + \cos (B + A)}}$Using the identity $\cos 	heta = \sin(90^\circ - 	heta)$:$= \frac{{\sin (B + A) + \sin(90^\circ - (B - A))}}{{\sin (B - A) + \sin(90^\circ - (B + A))}}$$= \frac{{\sin (B + A) + \sin(90^\circ - B + A)}}{{\sin (B - A) + \sin(90^\circ - B - A)}}$Using the sum-to-product formula $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:Numerator: $2 \sin(\frac{B+A+90^\circ-B+A}{2}) \cos(\frac{B+A-90^\circ+B-A}{2}) = 2 \sin(45^\circ+A) \cos(B-45^\circ)$Denominator: $2 \sin(\frac{B-A+90^\circ-B-A}{2}) \cos(\frac{B-A-90^\circ+B+A}{2}) = 2 \sin(45^\circ-A) \cos(B-45^\circ)$Since $\cos(B-45^\circ) = \cos(45^\circ-B)$:$= \frac{2 \sin(A+45^\circ) \cos(45^\circ-B)}{2 \sin(45^\circ-A) \cos(45^\circ-B)} = \frac{\sin(A+45^\circ)}{\sin(45^\circ-A)}$Using $\sin(x+y) = \sin x \cos y + \cos x \sin y$ and $\sin(x-y) = \sin x \cos y - \cos x \sin y$:$= \frac{\sin A \cos 45^\circ + \cos A \sin 45^\circ}{\sin 45^\circ \cos A - \cos 45^\circ \sin A}$Dividing numerator and denominator by $\sin 45^\circ$ (or multiplying by $\sqrt{2}$):$= \frac{\sin A + \cos A}{\cos A - \sin A} = \frac{\cos A + \sin A}{\cos A - \sin A}$.

$\frac{{\sin (B + A) + \cos (B - A)}}{{\sin (B - A) + \cos (B + A)}} = $

Similar Questions

$ABC$ is a triangle. If $\sin \left(\frac{A+B}{2}\right) = \frac{\sqrt{3}}{2}$,then the value of $\sin \frac{C}{2}$ is

If $\tan \theta = \frac{1}{\sqrt{11}}$ and $0 < \theta < \frac{\pi}{2},$ then the value of $\frac{\operatorname{cosec}^{2} \theta - \sec ^{2} \theta}{\operatorname{cosec}^{2} \theta + \sec ^{2} \theta}$ is

If $A + B + C = \pi ,$ then $\frac{{\cos A}}{{\sin B\sin C}} + \frac{{\cos B}}{{\sin C\sin A}} + \frac{{\cos C}}{{\sin A\sin B}} = $

If $\sin \alpha$ and $\cos \alpha$ are the roots of the equation $ax^2 + bx + c = 0$ $(c \neq 0)$,then:

If $\sec \theta - \tan \theta = \frac{1}{\sqrt{3}}$,the value of $\sec \theta \cdot \tan \theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module