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

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Answer 1 · 2018-05-12T08:27:25

Please see below.

As $sinA+sinB=2sin((A+B)/2)cos((A-B)/2)$

and $sinA-sinB=2cos((A+B)/2)sin((A-B)/2)$

$(sin75^@+sin15^@)/(sin75^@-sin15^@)$

= $(2sin((75^@+15^@)/2)cos((75^@-15^@)/2))/(2cos((75^@+15^@)/2)sin((75^@-15^@)/2)$

= $(sin45^@cos30^@)/(cos45^@sin30^@)$

= $tan45^@cot30^@$

= $1xxsqrt3$

= $sqrt3$

Answer 2 · 2018-05-12T09:56:17

Please refer to the 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 $sin75^@/sin15^@$ .

Now, $sin75^@/(sin15^@)=sin75^@/cos(90^@-15^@)=sin75^@/cos75^@$ ,

$=tan75^@=tan(45^@+30^@)$ ,

$=(tan45^@+tan30^@)/(1-tan45^@tan30^@)$ ,

$=(1+1/sqrt3)/(1-1/sqrt3)$ .

$rArr sin75^@/(sin15^@)=(sqrt3+1)/(sqrt3-1)$ .

By componendo-dividendo, then,

$(sin75^@+sin15^@)/(sin75^@-sin15^@)$ ,

$={(sqrt3+1)+(sqrt3-1)}/{(sqrt3+1)-(sqrt3-1)}$ ,

$=sqrt3$ , as Respected Shwetank Mauria Sir has already derived!