Prove That ? (Sin75°+Sin15°)/(Sin75°-Sin15°)=√3

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Prove That ? (Sin75°+Sin15°)/(Sin75°-Sin15°)=√3

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Answer 1 · 2018-05-12T08:27:25

Please see below.

Explanation:

As $sin A + sin B = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2})$ $sinA+sinB=2sin((A+B)/2)cos((A-B)/2)$

and $sin A - sin B = 2 cos (\frac{A + B}{2}) sin (\frac{A - B}{2})$ $sinA-sinB=2cos((A+B)/2)sin((A-B)/2)$

$\frac{{sin 75}^{\circ} + {sin 15}^{\circ}}{{sin 75}^{\circ} - {sin 15}^{\circ}}$ $(sin75^@+sin15^@)/(sin75^@-sin15^@)$

= $\frac{2 sin (\frac{75^{\circ} + 15^{\circ}}{2}) cos (\frac{75^{\circ} - 15^{\circ}}{2})}{2 cos (\frac{75^{\circ} + 15^{\circ}}{2}) sin (\frac{75^{\circ} - 15^{\circ}}{2})}$ $(2sin((75^@+15^@)/2)cos((75^@-15^@)/2))/(2cos((75^@+15^@)/2)sin((75^@-15^@)/2)$

= $\frac{{sin 45}^{\circ} {cos 30}^{\circ}}{{cos 45}^{\circ} {sin 30}^{\circ}}$ $(sin45^@cos30^@)/(cos45^@sin30^@)$

= ${tan 45}^{\circ} {cot 30}^{\circ}$ $tan45^@cot30^@$

= $1 \times \sqrt{3}$ $1xxsqrt3$

= $\sqrt{3}$ $sqrt3$

Answer 2 · 2018-05-12T09:56:17

Please refer to the Explanation.

Explanation:

Here is another Method to prove the Result :

Observe that the Exp. on the L.H.S. can be obtained using

componendo-dividendo on $\frac{{sin 75}^{\circ}}{{sin 15}^{\circ}}$ $sin75^@/sin15^@$ .

Now, $\frac{{sin 75}^{\circ}}{{sin 15}^{\circ}} = \frac{{sin 75}^{\circ}}{cos (90^{\circ} - 15^{\circ})} = \frac{{sin 75}^{\circ}}{{cos 75}^{\circ}}$ $sin75^@/(sin15^@)=sin75^@/cos(90^@-15^@)=sin75^@/cos75^@$ ,

$= {tan 75}^{\circ} = tan (45^{\circ} + 30^{\circ})$ $=tan75^@=tan(45^@+30^@)$ ,

$= \frac{{tan 45}^{\circ} + {tan 30}^{\circ}}{1 - {tan 45}^{\circ} {tan 30}^{\circ}}$ $=(tan45^@+tan30^@)/(1-tan45^@tan30^@)$ ,

$= \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}$ $=(1+1/sqrt3)/(1-1/sqrt3)$ .

$\Rightarrow \frac{{sin 75}^{\circ}}{{sin 15}^{\circ}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$ $rArr sin75^@/(sin15^@)=(sqrt3+1)/(sqrt3-1)$ .

By componendo-dividendo, then,

$\frac{{sin 75}^{\circ} + {sin 15}^{\circ}}{{sin 75}^{\circ} - {sin 15}^{\circ}}$ $(sin75^@+sin15^@)/(sin75^@-sin15^@)$ ,

$= \frac{(\sqrt{3} + 1) + (\sqrt{3} - 1)}{(\sqrt{3} + 1) - (\sqrt{3} - 1)}$ $={(sqrt3+1)+(sqrt3-1)}/{(sqrt3+1)-(sqrt3-1)}$ ,

$= \sqrt{3}$ $=sqrt3$ , as Respected Shwetank Mauria Sir has already derived!

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Prove That ? (Sin75°+Sin15°)/(Sin75°-Sin15°)=√3

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