Test Index

TG EAMCET 2 May 2025 Shift 2 Paper

© examsnet.com

Question : 24 of 160

Marks: +1, -0

If

3 \sin (\alpha-\beta)=5 \cos (\alpha+\beta)

\alpha+\beta \neq \frac{\pi}{2}

\frac{\tan \left(\frac{\pi}{4}-\alpha\right)}{\tan \left(\frac{\pi}{4}-\beta\right)} =

0
-4
$-\frac{1}{4}$
$\frac{1}{2}$

Solution:

We have,

3 \sin (\alpha-\beta)=5 \cos (\alpha+\beta)

\Rightarrow 3[\sin \alpha \cos \beta - \cos \alpha \sin \beta]

=5[\cos \alpha \cos \beta - \sin \alpha \sin \beta]

On dividing by

\cos \alpha \cos \beta

, we get

3[\tan \alpha - \tan \beta]=5(1-\tan \alpha \tan \beta)

\Rightarrow 3\tan \alpha - 3\tan \beta = 5 - 5\tan \alpha \tan \beta

\Rightarrow 3\tan \alpha + 5\tan \alpha \tan \beta = 5 + 3\tan \beta

\Rightarrow \tan \alpha (3+5\tan \beta) = 5+3\tan \beta

\Rightarrow \tan \alpha = \frac{5+3\tan \beta}{3+5\tan \beta}

Now,

\tan \left(\frac{\pi}{4}-\alpha\right) = \frac{1-\tan \alpha}{1+\tan \alpha}

And

\tan \left(\frac{\pi}{4}-\beta\right) = \frac{1-\tan \beta}{1+\tan \beta}

\therefore \frac{\tan \left(\frac{\pi}{4}-\alpha\right)}{\tan \left(\frac{\pi}{4}-\beta\right)} = \frac{(1-\tan \alpha)(1+\tan \beta)}{(1+\tan \alpha)(1-\tan \beta)}

\therefore 1-\tan \alpha = 1 - \frac{5+3\tan \beta}{3+5\tan \beta} = \frac{2(\tan \beta - 1)}{3+5\tan \beta}

\text{ And } 1+\tan \beta = 1 + \frac{5+3\tan \beta}{3+5\tan \beta} = \frac{8(1+\tan \beta)}{3+5\tan \beta}

\therefore \frac{\tan \left(\frac{\pi}{4}-\alpha\right)}{\tan \left(\frac{\pi}{4}-\beta\right)} = \frac{ \frac{2(\tan \beta-1)}{3+5\tan \beta} \times (1+\tan \beta) }{ 8(1+\tan \beta) } \times (1-\tan \beta)

= \frac{-2}{8} = \frac{-1}{4}

© examsnet.com

Go to Question:

Prev Question

Next Question