Question

If m =(cos∅-sin∅) and n =(cos∅+sin∅), show that √m/n+√n/m =2/√1-tan²∅.​

Answer 1

Answer:

√m/n+√n/m =2/√1-tan²∅

Step-by-step explanation:

m =(cos∅-sin∅)

n =(cos∅+sin∅)

L.H.S = √m/n+√n/m

         = √(cos∅-sin∅)/(cos∅+sin∅)  +  √(cos∅+sin∅)/(cos∅-sin∅)

         = [ (cos∅-sin∅) + (cos∅+sin∅) ] / √(cos∅-sin∅)*(cos∅+sin∅)

         = 2*cos∅ / √(cos²∅ - sin²∅)                     ∵(a + b)*(a - b) = (a² - b²)

         =  2 / [ √(cos²∅ - sin²∅) / cos∅ ]

         =  2 / [ √1-tan²∅ ]   = R.H.S                       ∵ sin∅ / cos∅ = tan∅

HENCE PROVED

Answer 2

$\tt m = (cos\phi - sin\phi) \\ \tt n = (cos\phi + sin\phi) \\ \\ \tt To\:prove \:\: : \sqrt{\frac {m}{n}} + \sqrt{\frac{n}{m}} = \frac{2}{\sqrt{1-{tan}^{2}\phi}} \\ \\ \tt L.H.S \\ \\ \implies \sqrt{\frac {m}{n}} + \sqrt{\frac{n}{m}} \\ \\ \tt Cross\: multiplying \\ \\ \tt \implies \frac{m + n}{\sqrt{mn}} \\ \\ \tt Putting\:values\:of\:m\:and\:n \\ \\ \tt \implies \frac{cos\phi-sin\phi)+ (cos\phi +sin\phi)}{\sqrt{(cos\phi - sin\phi)(cos\phi +sin\phi)}} \\ \\ \tt \implies \frac{2cos\phi + \cancel{sin\phi} - \cancel{sin\phi}}{\sqrt{{cos}^{2}\phi - {sin}^{2}\phi}} \\ \\ \tt \implies \frac{2cos\phi}{\sqrt{{cos}^{2}\phi - {sin}^{2}\phi}} \\ \\ \tt Dividing\: numerator\:and\: denominator\:by \:cos\phi \\ \\ \tt \implies \frac{2}{\sqrt{\frac{{cos}^{2}\phi-{sin}^{2}\phi}{{cos}^{2}\phi}}} \\ \\ \tt \implies \frac{2}{\sqrt{\frac{{cos}^{2}\phi}{{sin}^{2}\phi} - \frac{{sin}^{2}\phi}{{cos}^{2}\phi}}} \\ \\ \tt \implies \frac{2}{\sqrt{1-{tan}^{2}\phi}} \\ \\ \tt \implies R.H.S \\ \\ \underline{\red{\tt Hence\:proved}}$