Respuesta :

msm555

Answer:

Solution given:

1.

L.h.s.

cscx+cotx

[tex] \frac{1}{cosx}+\frac{cosx}{sinx}[/tex]

[tex]\frac{1+cosx}{sinx}[/tex]

multiplying and dividing by (1-cosx)

[tex]\frac{(1+cosx)(1-cosx)}{sinx(1-cosx)}[/tex]

[tex]\frac{1-cos²x}{sinx(1-cosx)}[/tex]

we have

1-cos²x=sin²x

[tex]\frac{sin²x}{sinx(1-cosx)}[/tex]

[tex]\frac{sinx}{1-cosx}[/tex]

=R.h.s.

proved.

2.

l.h.s

[tex]\frac{tan y}{1-cot y}+\frac{cot y}{1-tany}[/tex]

[tex]\frac{\frac{sin y}{cos y}}{1-\frac{cos x}{siny}}+\frac{\frac{cos y}{sin y}}{1-\frac{siny}{cos y}}[/tex]

[tex]\frac{\frac{sin y}{cos y}}{\frac{siny -cos y}{siny}}+\frac{\frac{cos y}{sin y}}{\frac{cos y-siny}{cos y}}[/tex]

[tex]\frac{sin²y}{cos y(siny-cosy)}+\frac{cos ² y}{sin y(cos y-siny)}[/tex]

[tex]\frac{sin²y}{cos y(siny-cosy)}- \frac{cos ² y}{sin y(-cos y+siny)}[/tex]

[tex] \frac{sin³y -cos³y}{sinycosy(siny -cosy)}[/tex]

[tex] \frac{(sin y- cos y)(sin²y+sinycosy+cos²y)}{sinycosy(siny -cosy)}[/tex]

[tex] \frac{sin²y+cos²y +siny cosy}{sinxcosy}[/tex]

[tex] \frac{1+sinxcosy}{sinycosy}[/tex]

[tex]\frac{1}{sinycosy}+1[/tex]

1+cscysecy

R.h.s.

proved.