Question

If cotΘ+cosΘ=x and cotΘ -cosΘ=y then x²-y²=​

Answer 1

Answer:

Please refer the image

Step-by-step explanation:

Answer 2

cotΘ + cosΘ = x    and   cotΘ - cosΘ = y      

cosΘ/sinΘ + cosΘ = x   and   cosΘ/sinΘ - cosΘ = y      

(cosΘ + sinΘ.cosΘ)/sinΘ = x   and   (cosΘ - sinΘ.cosΘ)/sinΘ = y

    Squaring both the side,,

(cosΘ + sinΘ.cosΘ)²/sin²Θ = x²  and   (cosΘ - sinΘ.cosΘ)²/sin²Θ = y²

(cos²Θ + sinΘ.cos²Θ + sin²Θ.cos²Θ)/sin²Θ = x² and (cos²Θ - sinΘ.cos²Θ + sin²Θ.cos²Θ)/sin²Θ = y²

         

now,  x² - y²

= [(cos²Θ + sinΘ.cos²Θ + sin²Θ.cos²Θ)/sin²Θ] -  [(cos²Θ - sinΘ.cos²Θ + sin²Θ.cos²Θ)/sin²Θ]

= (cos²Θ + sinΘ.cos²Θ + sin²Θ.cos²Θ)/sin²Θ - (cos²Θ - sinΘ.cos²Θ + sin²Θ.cos²Θ)/sin²Θ

 = [(cos²Θ + sinΘ.cos²Θ + sin²Θ.cos²Θ) - (cos²Θ - sinΘ.cos²Θ + sin²Θ.cos²Θ)] / sin²Θ

 =  (cos²Θ + sinΘ.cos²Θ + sin²Θ.cos²Θ - cos²Θ + sinΘ.cos²Θ - sin²Θ.cos²Θ) / sin²Θ

=  (sinΘ.cos²Θ + sinΘ.cos²Θ) / sin²Θ

=  (2sinΘ.cos²Θ) / sin²Θ

=   (2cos²Θ) / sinΘ

=  (2cosΘ . cosΘ) / sinΘ

=  2 cotΘ . cosΘ

   

  The value for x² - y² = 2cotΘ.cosΘ .