If 3 sin (alpha-beta)=5 cos (alpha+beta) and | vedclass

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

Question

(C) Given $3 \sin (\alpha-\beta) = 5 \cos (\alpha+\beta)$.Using the expansion formulas,we have $3(\sin \alpha \cos \beta - \cos \alpha \sin \beta) = 5

Vedclass · Accepted Answer

$-\frac{1}{4}$

Vedclass · Answer

(C) Given $3 \sin (\alpha-\beta) = 5 \cos (\alpha+\beta)$.Using the expansion formulas,we have $3(\sin \alpha \cos \beta - \cos \alpha \sin \beta) = 5(\cos \alpha \cos \beta - \sin \alpha \sin \beta)$.Dividing both sides by $\cos \alpha \cos \beta$,we get $3(	an \alpha - 	an \beta) = 5(1 - 	an \alpha 	an \beta)$.We need to find $X = \frac{	an(\pi/4 - \alpha)}{	an(\pi/4 - \beta)} = \frac{(1 - 	an \alpha)/(1 + 	an \alpha)}{(1 - 	an \beta)/(1 + 	an \beta)} = \frac{(1 - 	an \alpha)(1 + 	an \beta)}{(1 + 	an \alpha)(1 - 	an \beta)}$.From the given equation,$3 	an \alpha - 3 	an \beta = 5 - 5 	an \alpha 	an \beta$,which implies $3 	an \alpha + 5 	an \alpha 	an \beta = 5 + 3 	an \beta$,so $	an \alpha(3 + 5 	an \beta) = 5 + 3 	an \beta$.Thus,$	an \alpha = \frac{5 + 3 	an \beta}{3 + 5 	an \beta}$.Substituting this into the expression for $X$:$X = \frac{(1 - \frac{5 + 3 	an \beta}{3 + 5 	an \beta})(1 + 	an \beta)}{(1 + \frac{5 + 3 	an \beta}{3 + 5 	an \beta})(1 - 	an \beta)} = \frac{(3 + 5 	an \beta - 5 - 3 	an \beta)(1 + 	an \beta)}{(3 + 5 	an \beta + 5 + 3 	an \beta)(1 - 	an \beta)} = \frac{(2 	an \beta - 2)(1 + 	an \beta)}{(8 + 8 	an \beta)(1 - 	an \beta)} = \frac{-2(1 - 	an \beta)(1 + 	an \beta)}{8(1 + 	an \beta)(1 - 	an \beta)} = -\frac{2}{8} = -\frac{1}{4}$.

If $3 \sin (\alpha-\beta)=5 \cos (\alpha+\beta)$ and $\alpha+\beta \neq \frac{\pi}{2}$,then $\frac{\tan \left(\frac{\pi}{4}-\alpha\right)}{\tan \left(\frac{\pi}{4}-\beta\right)}=$

Similar Questions

$\cos 12^{\circ} \cdot \cos 24^{\circ} \cdot \cos 36^{\circ} \cdot \cos 48^{\circ} \cdot \cos 72^{\circ} \cdot \cos 84^{\circ} = $

If $x = \log \left[ \cot \left( \frac{\pi}{4} + \theta \right) \right]$, then the value of $\sinh x$ is

If $f_n(x) = \frac{1}{2n} [\sin^{2n} x + \cos^{2n} x]$,then $f_1(x) + f_2(x) - f_3(x) =$

If $\cosh x = \frac{4}{3}$,then $3 \cosh x + 3^2 \cosh 2x + 3^3 \cosh 3x = $

If $A$ and $B$ are acute angles satisfying $3 \cos ^2 A + 2 \cos ^2 B = 4$ and $\frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A}$,then $A + 2B =$ (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module