Question

(sen+cos)`2+(sen-cos)^2=2(tan^2cos^2+cot^2) help me pls

Answer 1

(sen+cos)`2+(sen-cos)^2=2(tan^2cos^2+cot^2)

Answer 2

Is `2 supposed to be ^2 ?

Answer 3

yes

Answer 4

\((\sin x + \cos x)^2+(\sin x - \cos x)^2=2(\tan^2 x \cos^2 x+\cot^2 x)\)

Are you solving an equation or proving an identity?

Answer 5

Trigonometric Identities

Answer 6

Let's try.

Answer 7

pls ty

Answer 8

\((\sin x + \cos x)^2+(\sin x - \cos x)^2=2(\tan^2 x \cos^2 x+\cot^2 x)\)
Square both quantities on the left side:
\(\sin^2 x + 2\sin x\cos x + \cos^2 x + \sin^2 x - 2 \sin x \cos x + \cos^2 x\)

Answer 9

\(\sin^2 x + \cos^2 x + \sin^2 x + \cos^2 x\)

Answer 10

\(2(\sin^2 x + \cos^2 x) \)
\(2(1)\)
\(2\)

Answer 11

The left sides simplifies to just 2.
Now let's see what we can do to the right side.

Answer 12

Did you copy the problem correctly?
Are you sure on the right side it's cot^2 x and not cos^2 x at the end?

Answer 13

is cot^2x

Answer 14

Then the identity will not work.

Answer 15

Let's work on the right side.

Answer 16

\(2(\tan^2 x \cos^2 x+\cot^2 x)\)
\(2(\dfrac{\sin^2 x}{\cos^2 x} \cos^2 x+\cot^2 x)\)
\(2(\sin^2 x+\cot^2 x)\)

Answer 17

Notice that if you had \(\cos ^2 x\) instead of \(\cot^2 x\), then \(\sin^2 x + \cos^2 x\) would add up to 1, and you'd end up with 2 on the right side, but if the right side really has \(\cot^2 x\) and not \(\cos^2 x\), then this is not an identity.

Answer 18

:/ i dont know in my exercise is cot^2x in the end

Answer 19

Ok. Then it's not an identity.
We proved that it is not an identity.