Answers

2015-09-23T23:03:29+05:30

We have-

              lsinA +mcosA+ n=0

and        l'sinA m'cosA+n'=0

Now taking sinA and cosA as variables we can apply the principal of cross multiplication.

       sinA                      cosA                   1

m                     n                          L                   m

                    n'                         L'                    m'

using cross multiplication

sinA/(mn'-m'n) = cosA/(nl'-n'l) = 1/(lm'-l'm)

⇒sinA=mn'-m'n/lm'-l'm

squaring both the sides,we get

sin²A=(mn'-m'n/lm'-l'm)²                                      ..........(1)

and cos²A=(mn'-m'n/lm'-l'm)²                               ........(2)

adding boththe equations, we get

1=(mn'-m'n)²+(nl'-n'l)²/(lm'-l'm)²                    ........ {sin²A+cos²A=1}

⇒(lm'-l'm)²=(mn'-m'n)²+(nl'-n'l)²

Hence proved.


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