Solving equations involving inverse trigonometric functions

Question

My math teacher said me if while solving trigonometric equations involving inverse trigonometric functions you are not able to get enough information about the inputs and if you have to solve for specific ranges of input or if nothing is given then directly use the below formula without thinking of anything and just check for extraneous solutions $$\tan^{-1}(x)+\tan^{-1}(y)=\tan^{-1}\left(\frac{x+y}{1-xy}\right)$$ Though the above formula is only valid for $xy$ being less than 1 And guess what! I saw the textbook doing the same in many illustration in the above picture I have shared 2 of them in illustration 7.62 look at the 3rd line and also in illustration 7.63 look at 3rd line In both places they have directly used the above formula without mentioning anything also my teacher didn't go in much details

What I want to ask that is it even correct to do so? or am I missing something? Won't it cause us to miss some solutions? or is it some standard convention or something? Because the book I'm using is quite famous in India and many people use it , it's name is 'Cengage Trigonometry for JEE Advance '

Answer 1

You don't need to use the addition formula for $\tan^{-1}$. Instead, use the addition formula for $\tan$ itself.

For the first problem:

\begin{align*} \sin^{-1}x&=\tan^{-1}2x-\tan^{-1}x\\ \implies \tan(\sin^{-1}x)&=\tan\left(\tan^{-1}2x-\tan^{-1}x\right)\\ \implies \frac{x}{\sqrt{1-x^2}}&=\frac{2x-x}{1-2x\cdot (-x)}\\ &\vdots \end{align*}

Note that we require $|x|\le 1$ for the problem to be meaningful.

For the second problem:

\begin{align*} \tan^{-1}6x&=\tan^{-1}\left(\frac{3x^2+1}{x}\right)-\tan^{-1}\left(\frac{1-3x^2}{x}\right)\\ \implies 6x&=\tan\left(\tan^{-1}\left(\frac{3x^2+1}{x}\right)-\tan^{-1}\left(\frac{1-3x^2}{x}\right)\right)\\ \implies 6x&=\frac{\frac{3x^2+1}{x}-\frac{1-3x^2}{x}}{1-\frac{3x^2+1}{x}\cdot\left(-\frac{1-3x^2}{x}\right)}\\ &\vdots \end{align*}

In these workings, you only need: \begin{align*} \tan(A+B)&\equiv\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}\\ \tan(\tan^{-1}(x))&\equiv x\\ \tan(\sin^{-1}x)&\equiv\frac{x}{\sqrt{1-x^2}} \end{align*}

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This method should give all solutions therefore, but it may also give some non-solutions too as the direction of implication does not hold in reverse as the tangent function is not injective. For this reason, solutions should be checked in the original equation as a last step.

Answer 2

I too thought the same while solving these questions from Cengage.

What I made out is, just try using the other variations of these formulae; at some step you'll see that the RHS becomes out of range of LHS or using it would totally cancel ±π from the whole equation

Example 1: Use that ±π formula in 2nd step of 7.62, RHS becomes out of range of sin inverse. Even in the 2nd step of 7.63, doing the same would result in RHS being out of range of tan inverse in LHS

Example 2: Let the term inside cot inverse in 7.63 be -ve. So use that π + tan inverse in both the cot inverse conversions to tan inverse (because if the term in cot inverse is -ve, its reciprocal will also be -ve; so if you have to use that formula, you'll have use it for both terms of cot inverse or none of them which means you can't add π only one of them). So π cancels out from LHS and RHS

Sometimes, using the other variations might take the value of x out of its possible range as deduced from the original expression given in the Q, basically giving an INVALID SOLUTION

In this way, use similar logics in other Qs as well. But in a good Question of JEE Advanced, multiple variations of these formulae may be applicable. So always check for these logics carefully before proceeding in any step

PS: I don't know how to use LaTeX, so couldn't mention exact formulae for reference. Kindly bear with me. I hope you understood what I'm trying to say.

Best of luck for JEE