Question

Plz solve ques. no. 27

Answer 1

★ TRIGONOMETRIC RESOLUTIONS ★

Given system of equations -

( SecA + TanA ) ( SecB + TanB ) ( SecC + TanC ) = ( SecA - TanA ) ( SecB - TanB) ( SecC - TanC ) = K

Aslike both are equivalent , we equated to K

Now ,

(SecA + TanA )( SecB + TanB ) ( SecC + TanC ) = K

And ,

SecA - TanA ( SecB - TanB ) SecC - TanC = K

Multiplying both of them

[ Sec²A - Tan²A ] [ Sec²B - Tan²B ] [ Sec²C - Tan²C ] = K²

According to formula : ( a + b ) ( a - b ) = a² - b²

And , Sec²θ - Tan²θ = 1

we conclude that ,

K² = 1

Hence ,

K = ± 1

Hence , each side length is equal to ± 1

★✩★✩★✩★✩★✩★✩★✩★✩★✩★✩★