Question
If $\sin A=\frac{3}{5}$ then show that $4 \tan A+3 \sin A=6 \cos A$
✓
Answer
$\sin A =\frac{3}{5} \quad \ldots$ (i) [Given]
In $\triangle ABC$
Let $\angle A B C=90^{\circ}$
$\therefore \sin A =\frac{ BC }{ AC }$
(ii) [By definition]
$\therefore \frac{ BC }{ AC }=\frac{3}{5} \quad \ldots . . .[$ From (i) and (ii)]
Let $B C=3 k, A C=5 k$
In $\triangle ABC , \angle B =90^{\circ}$
$\therefore AB ^2+ BC ^2= AC ^2$ [Pythagoras theorem]
$\therefore A B^2+(3 k)^2=(5 k)^2$
$\therefore AB ^2+9 K ^2=25 k ^2$
$\therefore A B^2=25 k^2-9 k^2$
$\therefore A B^2=16 k^2$
$\therefore AB =4 k$ [Taking square root of both sides]
Now, $\tan A =\frac{ BC }{ AB }$ [By definition]
$\therefore \tan A =\frac{3 k }{4 k }=\frac{3}{4}$
$\cos A =\frac{ AB }{ AC }$ [By definition]
$\therefore \cos A =\frac{4 k }{5 k }=\frac{4}{5}$
$\therefore 4 \tan A +3 \sin A =4\left(\frac{3}{4}\right)+3\left(\frac{3}{5}\right)$
$=3+\frac{9}{5}$
$=\frac{15+9}{5}$
$=\frac{24}{5} \ldots \ldots . . \text { (iii) }$
$6 \cos A =6\left(\frac{4}{5}\right)=\frac{24}{5} \ldots . . . \text { (iv) }$
$\therefore 4 \tan A+3 \sin A=6 \cos A \quad \ldots . .[$ From (iii) and (iv)]
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