If sinA = $\frac{1}{2}$ , verify that 2 sinA cosA = $\frac{2 t a n A}{(1 + t a n 2 A)}$

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Solution

We know that

SinA = $\frac{B C}{A C}$ = $\frac{1}{2}$

Let BC = k and AC = 2k

∴ AB = $\sqrt{A C^{2} - A B^{2}}$ ......... (Pythagoras theorm )

= $\sqrt{(2 k)^{2} - k^{2}}$ = $\sqrt{4 k^{2} - k^{2}}$ = $\sqrt{3 k^{2}}$ = $\sqrt{3}$ k

Now cos A = $\frac{A B}{A C}$ = $\frac{\sqrt{3} k}{2 k}$ = $\frac{\sqrt{3}}{2}$

and tanA = $\frac{B C}{A B}$ = $\frac{k}{\sqrt{3} k}$ = $\frac{1}{\sqrt{3}}$

Now 2 sinA cosA = 2. $\frac{1}{2}$ . $\frac{\sqrt{3}}{2}$ = $\frac{\sqrt{3}}{2}$ ......(i)

and $\frac{2 t a n A}{1 + t a n^{2} A}$ = $\frac{2 . \frac{1}{\sqrt{3}}}{1 + {(\frac{1}{\sqrt{3}})}^{2}}$ = $\frac{2 \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$ = $\frac{2 \frac{1}{\sqrt{3}}}{\frac{4}{3}}$

= $\frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{\sqrt{3}}{2}$ ........... (ii)

Hence from (ii) and (i)

2 sinA cosA = $\frac{2 t a n A}{1 + t a n^{2} A}$ .

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