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Take a look at figure given below. Give it a try and when you are ready then watch the solution.
In a quarter circle, a set of perpendicular lines are drawn. One line is from one end of quarter circle to opposite side of quarter circle and another line perpendicular from that point to one point of circle. If lengths of lines are a and b then find tangent of angle forming at 1st line and base of circle.
Basic knowledge to solve this puzzle:
sin and cos in terms of tan:
\[sinα=\sqrt{\frac{tan^2α}{1+tan^2α}}\]
\[cosα=\sqrt{\frac{1}{1+tan^2α}}\]
Proof: Given in end of this page
Step 1: Give names to some points
Step 2:
Horizontal component of a is a.cos(α)
Vertical component of a is a.sin(α)
Similarly,
Horizontal component of b is b.cos(α)
Vertical component of b is b.sin(α)
Now AC = AB (Both are radius of big circle)
\[AB = b.cos(α)\]
and
\[AC = \sqrt{[a.cos(α)+b.sin(α)]^2+[a.sin(α)]^2}\]
Now AB = AC
\[\Rightarrow b.cos(α) = \sqrt{[a.cos(α)+b.sin(α)]^2+[a.sin(α)]^2}\]
\[\Rightarrow [b.cos(α)]^2 = [a.cos(α)+b.sin(α)]^2+[a.sin(α)]^2\]
\[\Rightarrow b^2.cos^2(α) = a^2.cos^2(α)+b^2.sin^2(α)+2ab.cos(α)sin(α)+a^2.sin^2(α)\]
\[\Rightarrow a^2+2ab.cos(α)sin(α)-b^2.cos^2(α)+b^2.sin^2(α) =0\]
\[\Rightarrow a^2+2ab.\frac{tan(α)}{1+tan^2(α)}+b^2.\left(\frac{tan^2(α)-1}{1+tan^2(α)}\right)=0\]
\[\Rightarrow a^2(1+tan^2(α))+2ab.tan(α)+b^2.(tan^2(α)-1)=0\]
\[\Rightarrow (a^2+b^2)tan^2(α)+2ab.tan(α)+(a^2-b^2)=0\]
Now its looks like quadratic equation. Solve for tan(α) in this quadratic equation.
\[\Rightarrow tan(α)=\frac{-2ab\pm\sqrt{4a^2b^2-4(a^2+b^2)(a^2-b^2)}}{2(a^2+b^2)}\]
\[\Rightarrow tan(α)=\frac{\pm\sqrt{a^2b^2-a^4+b^4}-ab}{(a^2+b^2)}\]
\[\Rightarrow tan(α)=\frac{\sqrt{a^2b^2-a^4+b^4}-ab}{(a^2+b^2)}\]
Proofs:
Sin:
\[tanα=\frac{sinα}{cosα}\]
\[\Rightarrow tanα=\frac{sinα}{\sqrt{1-sin^2α}}\]
\[\Rightarrow tan^2α=\frac{sin^2α}{1-sin^2α}\]
\[\Rightarrow\frac{1}{tan^2α}=\frac{1-sin^2α}{sin^2α}\]
\[\Rightarrow\frac{1}{tan^2α}=\frac{1}{sin^2α}-1\]
\[\Rightarrow\frac{1+tan^2α}{tan^2α}=\frac{1}{sin^2α}\]
\[\Rightarrow sin^2α=\frac{tan^2α}{1+tan^2α}\]
\[\Rightarrow sinα=\sqrt{\frac{tan^2α}{1+tan^2α}}\]
Cos:
\[tanα=\frac{sinα}{cosα}\]
\[\Rightarrow tanα=\frac{\sqrt{1-cos^2α}}{sinα}\]
\[\Rightarrow tan^2α=\frac{1-cos^2α}{cos^2α}\]
\[\Rightarrow tan^2α=\frac{1}{cos^2α}-1\]
\[\Rightarrow cos^2α=\frac{1}{1+tan^2α}\]
\[\Rightarrow cosα=\sqrt{\frac{1}{1+tan^2α}}\]
