Question

ques...

Answer 1

Given
\[\sin^{-1}\frac{2a}{1+a^2}+\sin^{-1}\frac{2b}{1+b^2}=2\tan^{-1}x\]
Prove that
\[x=\frac{a+b}{1-ab}\]
What I've done:
Let
\[a=\tan \alpha\]\[b=\tan \beta\]
\[\implies \sin^{-1}(\sin(2 \alpha))+\sin^{-1}(\sin(2 \beta))=2 \alpha+2 \beta=2(\alpha+\beta)=2\tan^{-1}x\]\[\alpha+\beta=\tan^{-1}x\]\[x=\tan(\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha .\tan \beta}=\frac{a+b}{1-ab}\]
Is this ok?

Answer 2

looks good

Answer 3

here is my reasoning:
I have applied the sin() operator to both sides of your equation, namely:
\[\Large \begin{gathered}
\sin \left( {{{\sin }^{ - 1}}\frac{{2a}}{{1 + {a^2}}} + {{\sin }^{ - 1}}\frac{{2b}}{{1 + {b^2}}}} \right) = \hfill \\
\hfill \\
= \sin \left( {2{{\tan }^{ - 1}}x} \right) \hfill \\
\end{gathered} \]
so I got:
\[\Large \begin{gathered}
\frac{{2a}}{{1 + {a^2}}}\frac{{1 - {b^2}}}{{1 + {b^2}}} + \frac{{1 - {a^2}}}{{1 + {a^2}}}\frac{{2b}}{{1 + {b^2}}} = \frac{{2x}}{{1 + {x^2}}} \hfill \\
\hfill \\
\frac{{2\left( {a - b} \right)\left( {1 - ab} \right)}}{{\left( {1 + {a^2}} \right)\left( {1 + {b^2}} \right)}} = \frac{{2x}}{{1 + {x^2}}} \hfill \\
\end{gathered} \]
since:
\[\Large \begin{gathered}
\cos \left( {\arcsin \left( {\frac{{2b}}{{1 + {b^2}}}} \right)} \right) = \frac{{1 - {b^2}}}{{1 + {b^2}}} \hfill \\
\hfill \\
\cos \left( {\arcsin \left( {\frac{{2a}}{{1 + {a^2}}}} \right)} \right) = \frac{{1 - {a^2}}}{{1 + {a^2}}} \hfill \\
\hfill \\
\sin \left( {2\arctan x} \right) = \frac{{2x}}{{1 + {x^2}}} \hfill \\
\end{gathered} \]
now,using your value for the quantity x, I can write:
\[\Large \frac{{2x}}{{1 + {x^2}}} = \frac{{2\left( {a - b} \right)\left( {1 - ab} \right)}}{{\left( {1 + {a^2}} \right)\left( {1 + {b^2}} \right)}}\]

Answer 4

But the problem I think in my solution is
LET
\[a=\tan \alpha\]
and
\[b=\tan \beta\]
a and b are most likely constants and not variables, but can I still let them equal to something?

Answer 5

what is the range of values of constants a and b?

Answer 6

Idk, there is nothing given just
If
\[\sin^{-1}\frac{2a}{1+a^2}+\sin^{-1}\frac{2b}{1+b^2}=2\tan^{-1}x\]
prove that
\[x=\frac{a+b}{1-ab}\]
That's all there is given

Answer 7

I think that your formulas, namely
\[\begin{gathered}
a = \tan \alpha \hfill \\
b = \tan \beta \hfill \\
\end{gathered} \]
can be accepted

Answer 8

I guess
since \[y=\tan^{-1}x\]
is defined
\[\forall x \in \mathbb{R}\]
so whatever the constants a and b(it's not given but they must be real)
\[\tan^{-1}a \tan^{-1}b\]
would be defined