Question

Inverse Trigonometric Function Question:-
Please Help:)

Answer 1

Why tanx≠−2
?????

Answer 2

Please give reason:)

Answer 3

@apoorvk & @Callisto & @shivam_bhalla Plz help:)

Answer 4

\[\tan \frac{1}{2}x=\frac{1-\cos x}{\sin x}\]

Answer 5

What is this?

Answer 6

\[\tan(\frac{1}{2}\cos^{-1} \frac{3}{5})=\frac{1-\frac{3}{5}}{\frac{4}{5}}=\frac{1}{2}\]

Answer 7

but answer is given "2" in the question.

Answer 8

\(\frac 12 \) is the correct answer.

Answer 9

k! thanx to everyone:)

Answer 10

but in the question; is the solution wrong anywhere?
Plz tell:)

Answer 11

In the componendo dividendo step, the final answer is \(\tan^2 x =\frac 1 4 \)

Answer 12

This is how @Mertsj got that and it is right

\[\large \tan x/2 = \frac{\sin x/2 }{\cos x/2} = {\sqrt{1-cosx \over 2} \over \sqrt{1+cosx \over 2}}\]
\[\large \tan x/2 = {\sqrt{1-cosx } \over \sqrt{1+cosx}} * \frac{\sqrt{1+cosx}}{\sqrt{1+cosx}}\]
\[\large \tan x/2= {sinx \over 1+\cos x }\]

Answer 13

k!k! I got that thanx a lot:)

Answer 14

I want to know why answer is not -1/2

Answer 15

First u see full cmments

Answer 16

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.

Answer 17

I didn't understand ur 1st line:\

Answer 18

Plz help:)

Answer 19

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.

Answer 20

k! thanx I got it:)