Why we not check conditions while solving questions?

Question

Note:Down ward problem is just an example to express my question(I know the both solution of problem are insufficient but the first solution is in my 10+2 book and second one is mine which is incomplete. I knew that. I am just illustrating simple way of book and complex proof that we all know.)

Note:I am from India.

"Why 10+2 book gives such less detailed proofs?"

In my book of formula given without condition which have wrote down

$$\tan(A+B)=\frac{(\tan A+\tan B)}{1-\tan A\tan B}$$

OK one will solve questions using this frequently..and answer is correct also....

Now let $A=(2k+1)\pi/2$ ,similarly $B=(2l+1)\pi/2$. $A$ may or may not be equal to $B$.

which implies $\tan(A+B)=0$. However, the formula gives infinity by infinity.

Now while solving questions we just ignore about this fact why?

For example..

Prove $$\tan(45^\circ+A)\tan(135^\circ+A)=-1$$

Solution

$${1+\tan A\over1-\tan A}\cdot{\tan A-1\over\tan A+1}=-1$$

According to solution

$A$ is not equal to an odd multiple of $90^\circ$.

But by putting that in question

Case 1:$A=(4n+1)90^\circ$

$$LHS=(-1)(1)=-1$$

Case 2 : $A=(4n-1)90^\circ$ $$LHS=(-1)(1)=-1$$

So it is necessary to add those cases in the solution but in simple method we can't do that just left it without those cases.Also in book it is not written. Nor do we write those cases in exams.Nor do we think about.

Note: this is question only for example we are generally doing this thing in almost all examples my question refers to the all such solution

I am proving this thing at 10 + 2 level. And in our books such detail proof are not given.

Answer 1

We do not ignore it and the solution you give is flawed because it includes the assumptions that $\tan A$ exists and that $\tan A \ne \pm 1$. Also the statement $\tan (A+45) \tan (A+135)=-1$ is not true for all $A$ because $\tan (A+45)$ and $\tan (A+135)$ exist iff $(A+45)/90$ is not an odd integer.

Answer 2

Let us consider an algebraic formula with the structure LHS=RHS. Its proof necessitates a first step which is to check that the domains of definition (i.e., of validity) of both sides are identical.

If this step is not properly done, you can obtain very erroneous conclusions.

Here, in your example, the basis is that the domain of definition (validity) of $\tan(X)$ is $X \in \mathbb{R} - \{(2k+1)\dfrac{\pi}{2}, k \in \mathbb{Z} \}$.

Now, in your formula :

• on the LHS: domain of validity: all ordered couples $(A,B)$ such that $A+B \neq (2k+1)\dfrac{\pi}{2}$ (condition (*)).

• on the RHS : due to numerator : $A \neq (2k+1)\dfrac{\pi}{2}$ and $B \neq (2k+1)\dfrac{\pi}{2}$ (conditions (**));

due to denominator : $\tan(A)\tan(B) \neq 1$ ; this last condition is equivalent to $\dfrac{\sin(A)}{\cos(A)}\dfrac{\sin(B)}{\cos(B)} \neq 1 \ (***) \ \Longleftrightarrow \ \sin(A)\sin(B) \neq \cos(A)\cos(B) \ \Longleftrightarrow$

$\cos(A)\cos(B)-\sin(A)\sin(B) \neq 0 \ \Longleftrightarrow \cos(A+B) \neq 0 \ \Longleftrightarrow A+B \neq (2k+1)\dfrac{\pi}{2}$ which is the same as condition (*).

Remark: note that in (**) above, we hadn't to assume an extra restriction because $\cos(A) \neq 0$ and $\cos(B) \neq 0$ is a consequence of conditions ().

Conclusion: Strictly speaking, conditions on the RHS are more restrictive than on the LHS, because of ( * ) that do not prevail on the LHS. In such a case, the conditions under which an equality is valid are the union of them, i.e., conditions (*)+(**).