Trig Identity?

[magnanimous]

How would I solve this identity?

(tan)(sin) tan-sin
________ = _______
tan+sin (tan)(sin)

 

[Deleted member 4993]

How would I solve this identity?

(tan)(sin) tan-sin
________ = _______
tan+sin (tan)(sin)

Hint: Convert Left-hand-side (LHS) to sin & cos and simplify. Do the same to RHS

What are your thoughts?

Please share your work with us ...even if you know it is wrong

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[Ishuda]

How would I solve this identity?

(tan)(sin) tan-sin
________ = _______
tan+sin (tan)(sin)

Well since you want a (tan-sin) on the right, I would probable start by multiplying numerator and denominator of the left hand side by that and then follow Subhotosh Khan's approach except for that (tan-sin)

 

[Berationalgetreal]

With derivatives

An easy way to demonstrate this identity is the following:

When the first derivative of a function is always equal to 0, the function is constant. In this case, we want to demonstrate that

\(\displaystyle f(x)\, =\, \dfrac{\tan(x)\, \cdot\, \sin(x)}{\tan(x)\, +\, \sin(x)}\, -\, \dfrac{\tan(x)\, -\, \sin(x)}{\tan(x)\, \cdot\, \sin(x)}\)

Is always equal to 0 (i. e. is constant).

Calculatin' its first derivative we obtain that

\(\displaystyle \dfrac{df}{dx}\, =\, 0,\, \forall\, x\, \Rightarrow\, f(x)\, =\, c\)

In order to establish the value of c, we put a value, for instance pi/3:

c = 1/2 - 1/2 = 0

Therefore, the identity is proved.