Question

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

Answer 1

Please see below.

Explanation:

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

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 `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!