Question

Given ,4cotA=3 ,the value of sinA+cosA/sinA-cosA

Answer 1

hHi there !

CotA = 3/4 = base/perpendicular

Consider ∆ABC in which B is a right angled triangle and BA = 3 , CB = 4 and by Pythagoras theorem CA = 5

Now, we can find the value of sinA and cosA

SinA = perpendicular / hypotenuse = 4/5

and cosA = base/ Hypotenuse = 3/5

$now \: \: \frac{sina + cosa}{sina - cosa} \\ \\ \frac{ \frac{4}{5} \ + \frac{3}{5} }{ \frac{4}{5} - \frac{3}{5} } = \frac{ \frac{7}{5} }{ \frac{1}{5} }$

7/5× 5/1= 7

Hence value is 7

Thanks ♥️

Answer 2

Answer:

The value of  (sinA+cosA)/(sinA-cosA) = 7

Step-by-step explanation:

Pythagoras theorem:

• Pythagoras has given a formula for a right angled triangle to find the sides of the triangle.

Formula:

          (Hypotenuse)² = (adjacent side)² + (opposite side)²

      Given, 4cotA = 3

                     cotA = 3/4

       By trigonometric ratios, we know

          cotA =  adjacent/opposite

     So, adjacent = 3 and opposite = 4

       In right angled triangle, by Pythagoras theorem

           (Hypotenuse)² = (adjacent side)² + (opposite side)²

       Let hypotenuse = x then

               x² = 3² + 4²

                   = 9+16

                   = 25

                x = √25

                x = 5

           Hypotenuse = 5

   Now we can write the remaining trigonometric ratios

            sinA = opposite/hypotenuse = 4/5

            cosA = adjacent/hypotenuse = 3/5

 Now we find the value of (sinA+cosA)/(sinA-cosA)

           (sinA+cosA)/(sinA-cosA) = [(4/5)+(3/5)]/[(4/5)-(3/5)]

                                                    = [7/5]/[1/5]

                                                    = 7

    Hence, the value of  (sinA+cosA)/(sinA-cosA) = 7

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