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Question: Find the magnitude of angle \( A \) , if \( \tan A - 2\cos A\tan A + 2\cos A - 1 = 0 \)...

Find the magnitude of angle AA , if tan⁡A−2cos⁡Atan⁡A+2cos⁡A−1=0\tan A - 2\cos A\tan A + 2\cos A - 1 = 0

Explanation

Solution

Hint : The given equation is a trigonometric equation involving tangent and cosine functions, we will simplify the given equation by factorizing it and solve the equation for the value of AA . After that we will get the value for the solution of the given equation for which it is true.
Magnitude of an angle is the amount by which an angle can be rotated to find the position of that angle. In our case the magnitude of our angle AA is the value of AA for which tan⁡A−2cos⁡Atan⁡A+2cos⁡A−1=0\tan A - 2\cos A\tan A + 2\cos A - 1 = 0 is satisfied. So, we have to find that.

Complete step-by-step answer :
The given trigonometric equation is: tan⁡A−2cos⁡Atan⁡A+2cos⁡A−1=0\tan A - 2\cos A\tan A + 2\cos A - 1 = 0
We will do algebraic manipulations here,
tan⁡A−2cos⁡Atan⁡A+2cos⁡A−1=0\tan A - 2\cos A\tan A + 2\cos A - 1 = 0
Taking tan⁡A\tan A common from the first two terms we get:
tan⁡A(1−2cos⁡A)+2cos⁡A−1=0\tan A(1 - 2\cos A) + 2\cos A - 1 = 0
Now by taking −1- 1 common from the remaining terms we get:
tan⁡A(1−2cos⁡A)+(−1)(−2cos⁡A+1)=0\tan A(1 - 2\cos A) + ( - 1)( - 2\cos A + 1) = 0
Reducing the above equation we get:
tan⁡A(1−2cos⁡A)−1(1−2cos⁡A)=0\tan A(1 - 2\cos A) - 1(1 - 2\cos A) = 0
⇒(tan⁡A−1)(1−2cos⁡A)=0\Rightarrow (\tan A - 1)(1 - 2\cos A) = 0
The above equation holds good if any of the following holds,
tan⁡A−1=0\tan A - 1 = 0 or 1−2cos⁡A=01 - 2\cos A = 0
⇒tan⁡A=1\Rightarrow \tan A = 1 or 1=2cos⁡A1 = 2\cos A
⇒tan⁡A=1\Rightarrow \tan A = 1 or cos⁡A=12\cos A = \dfrac{1}{2}
⇒A=π4\Rightarrow A = \dfrac{\pi }{4} or A=π3A = \dfrac{\pi }{3}
So, the equation holds only for the values A=π3,π4A = \dfrac{\pi }{3},\dfrac{\pi }{4}
Therefore, the magnitude of angle AA is π3\dfrac{\pi }{3} or π4\dfrac{\pi }{4} , if tan⁡A−2cos⁡Atan⁡A+2cos⁡A−1=0\tan A - 2\cos A\tan A + 2\cos A - 1 = 0
So, the correct answer is “ π3\dfrac{\pi }{3} or π4\dfrac{\pi }{4} ”.

Note : Since, we are asked about the magnitude of the angle so we need not find the general solution. If we were asked to find the general solution we would have found it specifically. The general solution of a trigonometric function is the set of all solutions of that function in the real number system, while the principal solution is the special case of general solution where the solution lies between [0,2π][0,2\pi ]