Question

Prove the trigonometric identity:

Answer 1

\[(1-\sin^2x)\left( 1+{1 \over \tan^2x} \right)=1\]

Answer 2

sorry wrong one

Answer 3

typing it up again

Answer 4

( 1 + tan^2x) ??

Answer 5

\[(1-\cos^2x)\left( 1+ {1 \over \tan^2x} \right)=1\]

Answer 6

The above is the identity I am talking about

Answer 7

Show your steps please

Answer 8

1st brackets are sin^2 2nd brackets are cosec^2 or 1/sin^2

so its\[\sin^2 \times \frac{1}{\sin^2} = 1\]

Answer 9

Use the Pythagorean identities to help you simplify that:\[\sin^2 x + \cos^2 x = 1\]\[1+\cot^2 x=\csc^2 x\]Start with this substitution: \(\dfrac{1}{\tan^2 x}=\cot^2 x\)

Answer 10

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

Answer 11

How did you do that? Also, make sure you're looking at the new identity, the first one had sin2x instead of cos2x

Answer 12

Individual steps are appreciated

Answer 13

I don't get it? @experimentX

Answer 14

it all starts with

\[\sin^2 + \cos^2 = 1\]

subtract cos^2 from both sides

\[\sin^1 = 1 - \cos^2\]

start again with sin^2 + cos^2 = 1

divide through by sin^2

\[1 + \frac{\cos^2}{\sin^2} = \frac{1}{\sin^2}\]

this the the 2nd brackets since \[\frac{1}{\tan^2} = \frac{\cos^2}{\sin^2}\]

so you end up with

\[\sin^2 \times \frac{1}{\sin^2} = 1\]

Answer 15

should be sin^2 = 1 - cos^2