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

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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.

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