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Inverse Trigonometric Function Question:- Please Help:)

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Why tanx≠−2 ?????
Please give reason:)

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Other answers:

\[\tan \frac{1}{2}x=\frac{1-\cos x}{\sin x}\]
What is this?
\[\tan(\frac{1}{2}\cos^{-1} \frac{3}{5})=\frac{1-\frac{3}{5}}{\frac{4}{5}}=\frac{1}{2}\]
but answer is given "2" in the question.
\(\frac 12 \) is the correct answer.
k! thanx to everyone:)
but in the question; is the solution wrong anywhere? Plz tell:)
In the componendo dividendo step, the final answer is \(\tan^2 x =\frac 1 4 \)
This is how @Mertsj got that and it is right \[\large \tan x/2 = \frac{\sin x/2 }{\cos x/2} = {\sqrt{1-cosx \over 2} \over \sqrt{1+cosx \over 2}}\] \[\large \tan x/2 = {\sqrt{1-cosx } \over \sqrt{1+cosx}} * \frac{\sqrt{1+cosx}}{\sqrt{1+cosx}}\] \[\large \tan x/2= {sinx \over 1+\cos x }\]
k!k! I got that thanx a lot:)
I want to know why answer is not -1/2
First u see full cmments
It seems that trying to attack it your way, and ending with a \(\tan^2\), there's no real easy way to see that it can't be -1/2. However, doing it mertsj's method, only outputs the single solution.
I didn't understand ur 1st line:\
Plz help:)
I'm going off of what foolformath said, "In the componendo dividendo step, the final answer is \(\tan^2(x)=\frac{1}{4}\)." I can't see a good reason why you need to exclude the possible answer of -1/2 here. Also, \(\sin(\cos^{-1}(3/5))\) should have two solutions. Namely, 4/5 and -4/5. So you should still get both solutions.
k! thanx I got it:)

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