Let 0

Question

Let $0<\alpha <1,\beta =\frac{1}{3\alpha }$ and ${	an }^{-1}(1-\alpha )+{	an }^{-1}(1-\beta )=\frac{\pi }{4}$. Then $6(\alpha +\beta )$ is equal to:

Accepted Answer

Let

$A={	an }^{-1}(1-\alpha ),\, B={	an }^{-1}(1-\beta )$

and it is given that

$A+B=\frac{\pi }{4}.$

Also,

$0We have to find $6(\alpha +\beta )$.

Using the tangent addition formula,

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

Since

$A+B=\frac{\pi }{4},$

we get

$	an (A+B)=1.$

Now,

$	an A=1-\alpha ,\, 	an B=1-\beta .$

So,

$\frac{(1-\alpha )+(1-\beta )}{1-(1-\alpha )(1-\beta )}=1.$

Simplifying the numerator:

$(1-\alpha )+(1-\beta )=2-\alpha -\beta .$

Now simplify the denominator:

$1-(1-\alpha )(1-\beta )=1-(1-\alpha -\beta +\alpha \beta )=\alpha +\beta -\alpha \beta .$

Hence,

$\frac{2-\alpha -\beta }{\alpha +\beta -\alpha \beta }=1.$

Therefore,

$2-\alpha -\beta =\alpha +\beta -\alpha \beta .$

So,

$2=2\alpha +2\beta -\alpha \beta .$

Now use

$\beta =\frac{1}{3\alpha }.$

Substituting:

$2=2\alpha +\frac{2}{3\alpha }-\alpha \cdot \frac{1}{3\alpha }.$

Since

$\alpha \cdot \frac{1}{3\alpha }=\frac{1}{3},$

we get

$2=2\alpha +\frac{2}{3\alpha }-\frac{1}{3}.$

Add $\frac{1}{3}$ to both sides:

$\frac{7}{3}=2\alpha +\frac{2}{3\alpha }.$

Multiply by $3\alpha $:

$7\alpha =6{\alpha }\^{2}+2.$

So,

$6{\alpha }\^{2}-7\alpha +2=0.$

Factorising:

$(3\alpha -2)(2\alpha -1)=0.$

Thus,

$\alpha =\frac{2}{3}\, \, \alpha =\frac{1}{2}.$

Now find $\beta =\frac{1}{3\alpha }$.

Case 1:
If

$\alpha =\frac{2}{3},$

then

$\beta =\frac{1}{2}.$

So,

$\alpha +\beta =\frac{2}{3}+\frac{1}{2}=\frac{7}{6}.$

Hence,

$6(\alpha +\beta )=7.$

Case 2:
If

$\alpha =\frac{1}{2},$

then

$\beta =\frac{2}{3}.$

Again,

$\alpha +\beta =\frac{1}{2}+\frac{2}{3}=\frac{7}{6}.$

Hence,

$6(\alpha +\beta )=7.$

Therefore, the required value is

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Solutions · 1