# If 6sinA + 8cosA = 10 , how to prove that TanA = 3/4 ?

Mar 29, 2018

See the explanation below

#### Explanation:

$6 \sin A + 8 \cos A = 10$

Dividing both sides by $10$

$\frac{3}{5} \sin A + \frac{4}{5} \cos A = 1$

Let $\cos \alpha = \frac{3}{5}$ and $\sin \alpha = \frac{4}{5}$

$\cos \alpha = \cos \frac{\alpha}{\sin} \alpha = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$

Therefore,

$\sin A \cos \alpha + \sin \alpha \cos A = \sin \left(A + \alpha\right) = 1$

So,

$A + \alpha = \frac{\pi}{2}$, $\mod \left[2 \pi\right]$

$A = \frac{\pi}{2} - \alpha$

$\tan A = \tan \left(\frac{\pi}{2} - \alpha\right) = \cot \alpha = \frac{3}{4}$

$\tan A = \frac{3}{4}$

$Q E D$

Mar 29, 2018

see below.

#### Explanation:

$\mathmr{and} , 6 \sin A - 10 = - 8 \cos A$
$\mathmr{and} , {\left(6 \sin A - 10\right)}^{2} = {\left(- 8 \cos A\right)}^{2}$
$\mathmr{and} , 36 {\sin}^{2} A - 2 \cdot 6 \sin A \cdot 10 + 100 = 64 {\cos}^{2} A$
$\mathmr{and} , 36 {\sin}^{2} A - 120 \sin A + 100 = 64 {\cos}^{2} A$
$\mathmr{and} , 36 {\sin}^{2} A - 120 \sin A + 100 = 64 \left(1 - {\sin}^{2} A\right)$
$\mathmr{and} , 36 \sin A - 120 \sin A + 100 = 64 - 64 S {\in}^{2} A$
$\mathmr{and} , 100 {\sin}^{2} A - 120 S \in A + 36 = 0$
$\mathmr{and} , {\left(10 \sin A - 6\right)}^{2} = 0$
$\mathmr{and} , 10 \sin A - 6 = 0$
$\mathmr{and} , S \in A = \frac{6}{10}$
$\mathmr{and} , S \in A = \frac{3}{5} = \frac{p}{h}$
Using Pythagoras theorem, we get
${b}^{2} = {h}^{2} - {p}^{2}$
$\mathmr{and} , {b}^{2} = {5}^{2} - {3}^{2}$
$\mathmr{and} , {b}^{2} = 25 - 9$
$\mathmr{and} , {b}^{2} = 16$
$\mathmr{and} , b = 4$
$s o , T a n A = \frac{p}{b} = \frac{3}{4}$

Mar 29, 2018

see solution

#### Explanation:

$6 \sin A + 8 \cos A = 10$

dividing both sides by $\sqrt{{6}^{2} + {8}^{2}}$=$10$

$\frac{6 \sin A}{10} + 8 \cos \frac{A}{10} = \frac{10}{10} = 1$

$\cos \alpha \sin A + \sin \alpha \cos A$=1

where $\tan \alpha = \frac{4}{3}$ or $\alpha = 53 \mathrm{de} g r e e$

this transforms to

$\sin \left(\alpha + A\right) = \sin 90$

$\alpha + A = 90$

$A = 90 - \alpha$

taking $\tan$both sides

$\tan A = \tan \left(90 - \alpha\right)$

$\tan A = \cot \alpha$
$\tan A = \frac{3}{4}$

Mar 29, 2018

$6 \sin A + 8 \cos A = 10$

$\implies 3 \sin A + 4 \cos A = 5$

$\implies \left(\frac{3}{5}\right) \sin A + \left(\frac{4}{5}\right) \cos A = 1$

$\implies \left(\frac{3}{5}\right) \sin A + \left(\frac{4}{5}\right) \cos A = {\left(\sin A\right)}^{2} + {\left(\cos A\right)}^{2}$
$\left[\textcolor{red}{{\sin}^{2} A + {\cos}^{2} A = 1}\right]$

$\implies \left(\frac{3}{5}\right) \sin A + \left(\frac{4}{5}\right) \cos A = \sin A \cdot \sin A + \cos A \cdot \cos A$

$\implies \sin A = \frac{3}{5} \mathmr{and} \cos A = \frac{4}{5}$

Hence, tanA = sinA/cosA =(3/5) / (4/5)=(3/5) × ( 5/4) = 3/4