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maheshmeghwal9
Group Title
Inverse Trigonometric Function Question:
Please Help:)
 2 years ago
 2 years ago
maheshmeghwal9 Group Title
Inverse Trigonometric Function Question: Please Help:)
 2 years ago
 2 years ago

This Question is Closed

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
Why tanx≠−2 ?????
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
Please give reason:)
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
@apoorvk & @Callisto & @shivam_bhalla Plz help:)
 2 years ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.2
\[\tan \frac{1}{2}x=\frac{1\cos x}{\sin x}\]
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
What is this?
 2 years ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.2
\[\tan(\frac{1}{2}\cos^{1} \frac{3}{5})=\frac{1\frac{3}{5}}{\frac{4}{5}}=\frac{1}{2}\]
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
but answer is given "2" in the question.
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
\(\frac 12 \) is the correct answer.
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
k! thanx to everyone:)
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
but in the question; is the solution wrong anywhere? Plz tell:)
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
In the componendo dividendo step, the final answer is \(\tan^2 x =\frac 1 4 \)
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
This is how @Mertsj got that and it is right \[\large \tan x/2 = \frac{\sin x/2 }{\cos x/2} = {\sqrt{1cosx \over 2} \over \sqrt{1+cosx \over 2}}\] \[\large \tan x/2 = {\sqrt{1cosx } \over \sqrt{1+cosx}} * \frac{\sqrt{1+cosx}}{\sqrt{1+cosx}}\] \[\large \tan x/2= {sinx \over 1+\cos x }\]
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
k!k! I got that thanx a lot:)
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
I want to know why answer is not 1/2
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
First u see full cmments
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
I didn't understand ur 1st line:\
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
Plz help:)
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
k! thanx I got it:)
 2 years ago
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