Answer:
Step-by-step explanation:
[tex]\sf 1.1)\dfrac{Sin \ \theta-Cos \ \theta}{Sin \ \theta + Cos \ \theta}=\dfrac{(Sin \ \theta-Cos \ \theta)}{((Sin \ \theta + Cos \ \theta)}*\dfrac{(Sin \ \theta-Cos \ \theta)}{(Sin \ \theta-Cos \ \theta)}[/tex]
[tex]\sf = \dfrac{(Sin \ \theta-Cos \ \theta)^2}{Sin^2 \ \theta-Cos^2 \ \theta}\\\\=\dfrac{Sin^2 \ \theta+Cos^2 \ \theta-2Sin \ \theta \ Cos\ \theta}{(Sin \ \theta-Cos \ \theta)}\\\\\\\bf Identity: \ (a +b)^2= a^2 + b^2 - 2ab\\\\\\=\dfrac{1-2Sin \ \theta \ Cos \ \theta}{(Sin \ \theta-Cos \ \theta)} = LHS[/tex]
[tex]\sf 1.2) LHS = tan^2 \ x - Sin^2 \ x = \dfrac{Sin^2 \ x}{Cos^2 \ x}-Sin^2 \ x[/tex]
[tex]\sf =\dfrac{Sin^2 \ x}{Cos^2 \ x}-\dfrac{Sin^2 \ x*Cos^2 \ x}{1*Cos^2 \ x}\\\\\\ = \dfrac{Sin^2 \ x - Sin^2 \ x*Cos^2 \ x}{Cos^2 \ x}\\\\\\= \dfrac{Sin^2 \ x *(1 -Cos^2 \ x)}{Cos^2 \ x}\\\\=\dfrac{Sin^2 \ x*Sin^2 \ x}{Cos^2 \ x} \\\\ \bf 1 - Cos^2 \ x = Sin^2 \ x\\\\= \dfrac{Sin^2 \ x}{Cos^2 \ x}*Sin^2 \ x\\\\=tan^2 \ x * Sin^2 \ x = RHS[/tex]
[tex]\sf 1.3) LHS = \dfrac{1-Cos \ x}{Sin \ x}-\dfrac{Sin \ x}{1+Cos \ x} =\dfrac{(1-Cos \ x)(1+Cos \ x)}{Sin \ x*(1+Cos \ x)}-\dfrac{Sin \ x*Sin \ x}{(1+Cos \ x)*Sin \ x}\\[/tex]
[tex]\sf =\dfrac{1 - Cos^2 \ x}{Sin \ x*(1+Cos \ x)}-\dfrac{Sin^2 \ x}{Sin \ x*(1+Cos \ x)}\\\\=\dfrac{Sin^2 \ x}{Sin \ x*(1+Cos \ x)} - \dfrac{Sin^2 \ x}{Sin \ x*(1+Cos \ x)}\\\\=\dfrac{Sin^2 x - Sin^2 \ x}{Sin \ x*(1+Cos \ x)} \\\\= 0 = RHS[/tex]
[tex]\sf 1.4) LHS = Sin x - \dfrac{1}{Sin \ x + Cos \ x}+Cos \ x \\[/tex]
[tex]\sf = \dfrac{Sin \ x *(Sin \ x + Cos \ x) - 1 + Cos \ x * (Sin \ x + Cos \ x)}{Sin \ x + Cos \ x }\\\\\\= \dfrac{Sin \ x * Sin \ x + Sin \ x*Cos \ x -1 + Cos \ x*Sin \ x + Cos \ x*Cos \ x}{Sin \ x + Cos \ x}\\\\\\=\dfrac{Sin^2 \x + Sin \ x \ Cos \ x - 1 + Cos \ x \ Sin \ x + Cos^2 \x}{Sin \ x + Cos \ x}\\\\=\dfrac{Sin^2 \ x + Cos^2 \ x - 1 + Sin \ xCos \x +Sin \ x Cos \ x}{Sin \ x + Cos \ x}\\\\= \dfrac{1 - 1 +2Sin \ x Cos \ x}{Sin \ x + Cos \ x}\\\\= \dfrac{2Sin \ x Cos \ x}{Sin \ x + Cos \ x}[/tex]
