Answer

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**Hint:**Here we will use the correlation between the tangent, sine and the cosine trigonometric function. First of all we will take the left hand side of the equation and then solve it using the correlation of the functions to get the resultant required solution which will be equal to the given right hand side of the equation.

**Complete step-by-step answer:**

Take the left hand side of the given equation,

$LHS = \sin x\cos x\tan x$

Place tangent as the ratio of the sine to the cosine, that is $\tan x = \dfrac{{\sin x}}{{\cos x}}$

$LHS = \sin x.\cos x.\dfrac{{\sin x}}{{\cos x}}$

Common factors from the numerator and the denominator cancel each other, therefore remove cosine from the numerator and the denominator from the above equation.

$LHS = \sin x.\sin x$

Simplify the above equation,

$LHS = {\sin ^2}x$ … (A)

Using the identity of the unit circle, ${\sin ^2}x + {\cos ^2}x = 1$

Make the required term the subject and move the other term on the opposite side. Remember when you move any term from one side to another, the sign of the term also changes, positive term changes to negative and vice-versa.

$\therefore {\sin ^2}x = 1 - {\cos ^2}x$

Place the above value in the equation (A)

$LHS = 1 - {\cos ^2}x$

From the given equation, we can observe and can state that –

$LHS = RHS$

Hence, proved the required solution.

**Note:**Every point on the circle is unit circle from the origin. So, the coordinates of any point are within one of zero as well. Directly the Pythagoras identity is followed by sines and cosines which concludes that $si{n^2}\theta + co{s^2}\theta = 1$. Always remember the correlation between the six trigonometric functions for replacing the given function and forming the equivalent equation for the efficient and an accurate solution.

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