Solve #tan2A=cot(A-18)# #0<theta<90# ?

To solve the equation $\tan(2A) = \cot(A - 18^\circ)$ for $0 < \theta < 90^\circ$, first, we need to express $\cot(A - 18^\circ)$ in terms of $\tan(2A)$. Recall that $\cot(\theta) = \frac{1}{\tan(\theta)}$.

So, $\cot(A - 18^\circ) = \frac{1}{\tan(A - 18^\circ)}$.

Now, using the double angle identity for tangent, $\tan(2A) = \frac{2\tan(A)}{1 - \tan^2(A)}$.

Now, substitute these expressions into the equation $\tan(2A) = \cot(A - 18^\circ)$ to get:

$\frac{2\tan(A)}{1 - \tan^2(A)} = \frac{1}{\tan(A - 18^\circ)}$.

Cross multiply to clear the fractions:

$2\tan(A) \tan(A - 18^\circ) = 1 - \tan^2(A)$.

Expand and rearrange the terms to get a quadratic equation:

$2\tan^2(A) - 2\tan(A)\tan(18^\circ) - 1 + \tan^2(A) = 0$.

This is a quadratic equation in terms of $\tan(A)$. Solve it using standard methods for quadratic equations. After finding the solutions for $\tan(A)$, remember that $0 < \theta < 90^\circ$, so you need to check whether the solutions satisfy this condition.

Once you find the valid solutions for $\tan(A)$, you can use inverse tangent to find the values of $A$ within the given range.