Question

find the value of this question maybe you help me with a correct answer

Answer 1

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Answer 2

Hello,

The question is from trigonometric ratios.

The question can be answered by putting the values of the ratios on the place.

The values of some trigonometric ratios are

$\sin(30) = \frac{1}{2} \\ \sin(60) = \frac{ \sqrt{3} }{2} \\ \\ \tan(60) = \sqrt{3} \\ \tan(45) = 1$
So,
Putting down the values...

$\frac{ \sin(60) + \sin(30) }{ \sin(60) - \sin(30) } \\ = \frac{ \frac{ \sqrt{3} }{2} + \frac{1}{2} }{ \frac{ \sqrt{3} }{2} - \frac{1}{2} } \\ = \frac{ \frac{ \sqrt{3} + 1 }{2} }{ \frac{ \sqrt{3} - 1}{2} } = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1}$
Now
For tangent

$\frac{ \tan(60) + \tan(45) }{ \tan(60) - \tan(45) } \\ = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1}$
So,

$lhs = rhs$
So,
Proved.

The value of the trigonometric problem is
$\frac{ \sqrt{3} + 1}{ \sqrt{3} - 1 }$

Hope this will be helping you ✌️