1. maheshmeghwal9

2. maheshmeghwal9

Why tanx≠−2 ?????

3. maheshmeghwal9

4. maheshmeghwal9

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

5. Mertsj

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

6. maheshmeghwal9

What is this?

7. Mertsj

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

8. maheshmeghwal9

but answer is given "2" in the question.

9. FoolForMath

$$\frac 12$$ is the correct answer.

10. maheshmeghwal9

k! thanx to everyone:)

11. maheshmeghwal9

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

12. FoolForMath

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

13. shivam_bhalla

This is how @Mertsj got that and it is right $\large \tan x/2 = \frac{\sin x/2 }{\cos x/2} = {\sqrt{1-cosx \over 2} \over \sqrt{1+cosx \over 2}}$ $\large \tan x/2 = {\sqrt{1-cosx } \over \sqrt{1+cosx}} * \frac{\sqrt{1+cosx}}{\sqrt{1+cosx}}$ $\large \tan x/2= {sinx \over 1+\cos x }$

14. maheshmeghwal9

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

15. maheshmeghwal9

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

16. maheshmeghwal9

First u see full cmments

17. KingGeorge

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.

18. maheshmeghwal9

I didn't understand ur 1st line:\

19. maheshmeghwal9

Plz help:)

20. KingGeorge

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.

21. maheshmeghwal9

k! thanx I got it:)