Prove that :cosx/(sinx+cosy)+cosy/(siny-cosx)=cosx/(sinx-cosy)+cosy/(siny+cosx)
Soln,
cosx/(sinx+cosy)+cosy/(siny-cosx)=cosx/(sinx-cosy)+cosy/(siny+cosx)
ARRANGING THE TERMS,
cosx/(sinx+cosy)-cosx/(sinx-cosy)=cosy/(siny+cosx)-cosy/(siny-cosx)
TAKING L.H.S.,
L.H.S=cosx/(sinx+cosy)-cosx/(sinx-cosy)
=cosx(sinx-cosy)-cosx(sinx+cosy)/(sinx+cosy)(sinx-cosy) [TAKING LCM]
=cosx*sinx -cosy*cosx- cosx*sinx -cosy*cosx/(sinx+cosy)(sinx-cosy)
=-cosy*cosx-cosy*cosx/(sinx+cosy)(sinx-cosy)
=-2cosy*cosx/(sinx+cosy)(sinx-cosy)
L.H.S=-2cosy*cosx/sin^2x-cos^2y
TAKING R.H.S.,
R.H.S=cosy/(siny+cosx)-cosy/(siny-cosx)
=cosy(siny-cosx)-cosy(siny+cosx)/(siny+cosx)(siny-cosx) [TAKING LCM]
=cosy*siny -cosx*cosy- cosy*siny -cosx*cosy/(siny+cosx)(siny-cosx)
=-2cosx*cosy/sin^2y-cos^2x
=-2cosx*cosy/1-cos^2y-(1-sin^2x)
=-2cosx*cosy/1 -cos^2y- 1+sin^2x
=-2cosx*cosy/-cos^2y+sin^2x
=-2cosx*cosy/sin^2x-cos^2y
R.H.S=-2cosx*cosy/sin^2x-cos^2y
L.H.S=R.H.S [-2cosx*cosy/sin^2x-cos^2y=-2cosx*cosy/sin^2x-cos^2y]
SOLVED