Equation or Identity (1)

Here are Julian from British School Manila's solutions to parts 1,2,3 of the question

1. By the addition formulae we have that $\sin(\pi/2-A)=\sin(\pi/2)\cos(A)-\sin(A)\cos(\pi/2)=\cos(A)$ for all angles A, and hence this is an identity.

2. By pythagoras theorem we have that $o^2+a^2=h^2$, where o, a, and h are the opposite, adjacent and hypotenuse sides respectively of a right angled triangle with angle A. Dividing through by $a^2$ gives $\tan^2(A)+1=\sec^2(A)$. Rearranging gives the statement and hence this is also an identity.

3. Since A,B,and C are the angles of a triangle we know they are related by $A+B+C=\pi$, so $\tan(A+B)=\tan(\pi-C)=\tan(-C)=-\tan(C)$ for all possible values of A,B and C, and hence this is also an identity.

Here are Kyle from Kimberley STEM College's solutions to parts 4,5,6 of the question

4. Here  $\cos^2(B)$ can not be negative and $\sin^2(A)$ cannot be greater than 1, so it can be concluded that  $\cos^2(B)=0$ and $\sin^2(A)=1$. This does not hold for all angles A and B so we can already conclude at this point that the statement is an equation. In fact this equation can not be satisfied by any triangle because the angles of a triangle must satisfy $0 < A, B < \pi$ and so this would require $A=\pi/2$ and $B=\pi/2$ which is impossible in a triangle as it would mean $C=0$.

5. By the addition formulae have that $\sin(\pi-A)=\sin(\pi)\cos(A)-\sin(A)\cos(\pi)=\sin(A)$, and so the statement reduces to $\sin(A)=\sin(B)$. We can see at this point that this does not hold for all angles A and B, and so this is an equation. In the region $0 < A, B < \pi $ this is only satisfied when either $A=B$ or $B=\pi-A$. We can reject the latter solution because this cannot happen if A and B are angles of a triangle, so the solution is $A=B$
(So in summary this is only satisfied when A and B are the equal angles in an isosceles triangle).

6. We can simplify the right hand side as follows:
$RHS=\operatorname{cosec}^2(C)-\cot^2(C)=1/\sin^2(C)-1/\tan^2(C)=(1-\cos^2(C))/\sin^2(C)$
$=\sin^2(C)/\sin^2(C)=1$. Also we can write the left hand side entirely in terms of $\cos(A)$:
$LHS=\sin^2(A)-3\cos(A)=1-\cos^2(A)-3\cos(A)$. Hence the equation reduces to $\cos^2(A)+3\cos(A)=0$ which factorises to $\cos(A)(\cos(A)+3)=0$ and so we see that the only solutions are when $\cos(A)=0$ which, since A is an angle of a triangle, means that $A=\pi/2$. Since this doesn't hold for all A, it is an equation.