Question

If 3sinθ+5cosθ=5, (0<θ<900), then the value of 5sinθ−3cosθ will be​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: 3sin\theta + 5cos\theta = 5$

On squaring both sides, we get

$\rm \: (3sin\theta + 5cos\theta)^{2} = {5}^{2}$

We know,

$\boxed{ \rm{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: }}$

So, using this identity we get

$\rm \: {9sin}^{2}\theta + 25 {cos}^{2}\theta + 2 \times 3sin\theta \times 5cos\theta = 25$

We know,

$\boxed{ \rm{ \: {sin}^{2}\theta + {cos}^{2}\theta = 1 \: }}$

So, using this identity, the above can be rewritten as

$\rm \: 9(1 - {cos}^{2}\theta)+25(1 - {sin}^{2}\theta)+ 2\times 3sin\theta \times5cos\theta = 25$

$\rm \: 9 - 9{cos}^{2}\theta+25 - 25{sin}^{2}\theta+ 2\times 3sin\theta \times5cos\theta = 25$

can be further rewritten as

$\rm \: 9 - 9{cos}^{2}\theta - 25{sin}^{2}\theta+ 2\times 3sin\theta \times5cos\theta = 0$

can be re-arranged as

$\rm \:9{cos}^{2}\theta + 25{sin}^{2}\theta - 2\times 3sin\theta \times5cos\theta = 9$

$\rm \: {(3cos\theta)}^{2}+ {(5sin\theta)}^{2} - 2\times 5sin\theta \times3cos\theta = 9$

$\rm \:{(5sin\theta)}^{2} + {(3cos\theta)}^{2} - 2\times 5sin\theta \times3cos\theta = 9$

$\rm \:{(5sin\theta - 3cos\theta )}^{2} = 9$

$\rm \:{(5sin\theta - 3cos\theta )}^{2} = {3}^{2}$

$\rm\implies \:5sin\theta - 3cos\theta \: = \: \pm \: 3$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 } \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

We have

$\rm 3 \sin \theta+5 \cos \theta=5$

$\[ \begin{array}{l} \rm\Rightarrow 3 \sin \theta=5(1-\cos \theta)=5 \times 2 \sin ^{2} \theta / 2 \\ \\ \displaystyle \rm\ \Rightarrow \frac{3}{5}=\frac{2 \sin ^{2} \theta / 2}{\sin \theta}=\frac{2 \sin ^{2} \theta / 2}{2 \sin \theta / 2 \cos \theta / 2} \\ \\\displaystyle \rm \Rightarrow \tan \frac{\theta}{2}=\frac{3}{5} \end{array} \]$

Now,

$\\ \[ 5 \sin \theta-3 \cos \theta =5 \times \frac{2 \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}-3 \frac{\left(1-\tan ^{2} \frac{\theta}{2}\right)}{1+\tan ^{2} \frac{\theta}{2}} \]$

$\\ \[ =5 \times \frac{2 \times \frac{3}{5}}{1+\frac{9}{25}}-\frac{3 \times\left(1-\frac{9}{25}\right)}{1+\frac{9}{25}} \]$

$\\ = 5 \times \frac{ \frac{6}{5} }{ \frac{25 + 9}{25} } - \frac{3 \times ( \frac{25 - 9}{25} )}{ \frac{25 + 9}{25} }$

$\\ = 5 \times \frac{ \frac{6}{5} }{ \frac{34}{25} } - \frac{ \frac{48}{25} }{ \frac{34}{25} }$

$\\ = 5 \times \frac{6}{5} \times \frac{25}{34} - \frac{48}{34}$

$\\ = \frac{150}{34} - \frac{48}{34}$

$\\ = \frac{75}{17} - \frac{ 24}{ 17}$

$\\ = \frac{51}{17}$

$\\ \red{ \boxed{ = 3}}$

Therefore, the value of $\rm 5 \sin \theta-3 \cos \theta$ is 3 .