Question

which of the following is an identity?

Answer 1

\[(tanx+cotx)(sinx cosx)= 1\]

Answer 2

\[cosx(2sinx + 1)= 0\]

Answer 3

\[\sin^2 x= 4 - 2\cos^2 x\]

Answer 4

No, none of the first two are identity, its a trigonometric equation because it will be true for just a particular value of x.

Answer 5

\[secxtanx - cosxcotx\]

Answer 6

[tan(x) + cot(x)] [sin(x)cos(x)]
[sin(x)/cos(x) + cos(x)/sin(x)] [sin(x)cos(x)]
[sin^2(x)/(sin(x)cos(x)) + cos^2(x)/(sin(x)cos(x))] [sin(x)cos(x)]
[(sin^2(x) + cos^2(x))/(sin(x)cos(x))] [sin(x)cos(x)]
[1/(sin(x)cos(x))] [sin(x)cos(x)]
1

Answer 7

so it would be the second one?

Answer 8

Even the third one is not an identity,
because sin^2x =1 - cos^2x.

Answer 9

im confused

Answer 10

Do you know what is pythagorean theorem? Just divide it by hypotenuse square, and you'll obtain the identity sin^2x =1 - cos^2x.

Answer 11

Now since sin^2x =1 - cos^2x., you may guess why sin^2 x= 4 - 2\cos^2 x.

Answer 12

?

Answer 13

so it woud be the 4th one?

Answer 14

@Hitarth

Answer 15

no first one.

Answer 16

oh ok. i was confused. lol

Answer 17

OH so sorry, I thought that there is a divide of tanx + cotx by sinxcosx, but it was multiply in first one. That is right. OK.

Answer 18

thanks!

Answer 19

anyone wanna help with another Q?