Solve #tan2A=cot(A-18)#
#0

Question

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

Charles Autterson · Accepted Answer

#A = 13.95 + k60^@# First, convert 18 radians into degrees. 18 radians --> #((180^@)(18))/3.14 = 1031^@85# tan 2A = cot (A - 18) (1) Use identity: cot a = tan (90 - a) Call a = (A - 18) cot (A - 18) = cot (A - 1031.85) = cot (A - 311.85) = cot (A - 18) = tan (90 - (A - 311.85)) = tan (401.85 - A) cot (A - 18) = tan (41.85 - A) Equation (1) becomes: #tan 2A = tan (41^@85 - A)# Unit circle and property of tan function give --> #2A = 41^@85 - A + k180^@# #3A = 41^@85 + k180^@#. #A = 13^@95 + k60^@# Check with calculator. k = 1 --> A = 13.95 + 60 = 73.95 --> tan 2A = tan 147.5 = - 0.63 cot (73.95 - 1031.85) = cot (- 237.9 - 720) = - 0.63 . Proved

Emery Adkins · Answer

undefined 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.

Solve #tan2A=cot(A-18)# #0&lt;theta&lt;90# ?

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