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!