Question

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Answer 1

Answer:

Step-by-step explanation:

We have,

$\tan(A)=\dfrac{y}{x}$

Now,

$\tt{\dfrac{x\,cos(A)+y\,sin(A)}{x\,cos(A)-y\,sin(A)}}$

taking cos(A) in common,

$\tt{=\dfrac{cos(A)\left\{x+y\dfrac{sin(A)}{cos(A)}\right\}}{cos(A)\left\{x-y\dfrac{sin(A)}{cos(A)}\right\}}}$

cos(A) will be canceled out,

$\tt{=\dfrac{x+y\dfrac{sin(A)}{cos(A)}}{x-y\dfrac{sin(A)}{cos(A)}}}$

$\tt{=\dfrac{x+y\,tan(A)}{x-y\,tan(A)}}$

$\tt{=\dfrac{x+y\cdot\dfrac{y}{x}}{x-y\cdot\dfrac{y}{x}}}$

$\tt{=\dfrac{x+\dfrac{{y}^{2}}{x}}{x-\dfrac{{y}^{2}}{x}}}$

$\tt{=\dfrac{\dfrac{{x}^{2}+{y}^{2}}{x}}{\dfrac{{x}^{2}-{y}^{2}}{x}}}$

$\tt{=\dfrac{{x}^{2}+{y}^{2}}{{x}^{2}-{y}^{2}}}$

Answer 2

Given that,

$\rm \tan A=\dfrac{y}{x}$

$\;$

We have,

$\rm\dfrac{x\cos A+y\sin A}{x\cos A-y\sin A}$

$\;$

$\rm=\dfrac{x\cos A+y\sin A}{x\cos A-y\sin A}\times\dfrac{\dfrac{1}{\cos A}}{\dfrac{1}{\cos A}}$

$\;$

$\rm=\dfrac{\dfrac{x\cos A}{\cos A}+\dfrac{y\sin A}{\cos A}}{\dfrac{x\cos A}{\cos A}-\dfrac{y\sin A}{\cos A}}$

$\;$

$\boxed{\rm\tan\theta=\dfrac{\sin\theta}{\cos\theta}}$

$\rm=\dfrac{x+y\tan A}{x-y\tan A}$

$\;$

$\rm=\dfrac{x+\dfrac{y^{2}}{x}}{x-\dfrac{y^{2}}{x}}$

$\;$

$\rm=\dfrac{x+\dfrac{y^{2}}{x}}{x-\dfrac{y^{2}}{x}}\times\dfrac{x}{x}$

$\;$

$\rm=\dfrac{x(x+\dfrac{y^{2}}{x})}{x(x-\dfrac{y^{2}}{x})}$

$\;$

$\rm=\dfrac{x^{2}+y^{2}}{x^{2}-y^{2}}$

$\;$

$\rm\therefore\dfrac{x\cos A+y\sin A}{x\cos A-y\sin A}=\dfrac{x^{2}+y^{2}}{x^{2}-y^{2}}$

$\;$

$\Large\textrm{Learn More}$

$\textbf{- Trigonometric Functions}$

$\boxed{\rm\sin\theta=\dfrac{y}{r}}$

$\;$

$\boxed{\rm\cos\theta=\dfrac{x}{r}}$

$\;$

$\boxed{\rm\tan\theta=\dfrac{y}{x}}$

$\;$

$\textbf{- Trigonometric Identities}$

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

$\;$

$\boxed{\rm\tan^{2}\theta+1=\sec^{2}\theta}$

$\;$

$\boxed{\rm\tan\theta=\dfrac{\sin\theta}{\cos\theta}}$