Question

SinA + cosA = √2, prove that tanA + cotA = 2

Answer 1

`sin(A)+cos(A)=√(2)\hspace{10em}tan(A)+cot(A)=2 \\\\[-0.35em] ~\dotfill\\\\ tan(A)+cot(A)=2\implies \cfrac{sin(A)}{cos(A)}+\cfrac{cos(A)}{sin(A)}=2 \\\\\\ \cfrac{sin^2(A)+cos^2(A)}{cos(A)sin(A)}=2\implies \boxed{\cfrac{1}{cos(A)sin(A)}=2} \\\\[-0.35em] ~\dotfill`

`sin(A)+cos(A)=√(2)\implies (~~sin(A)+cos(A)~~)^2=(√(2))^2 \\\\\\ sin^2(A)+2sin(A)cos(A)+cos^2(A)=2 \\\\\\ 2sin(A)cos(A)+sin^2(A)+cos^2(A)=2\implies 2sin(A)+1=2 \\\\\\ 2sin(A)cos(A)=1\implies \boxed{2=\cfrac{1}{cos(A)sin(A)}}`

now, another way to look at this identity will be as a unified system of equations

`\begin{cases} (~~sin(A)+cos(A)~~)^2=2\\\\ ~~ ~tan(A)+cot(A)=2 \end{cases}\implies (~~sin(A)+cos(A)~~)^2=tan(A)+cot(A)`

and we'd end up with the same rigamarole.