Question

1+sec^2 x sin^2 x=sec^2 x =1+(1/cos^2 x) sin^2 x = 1/cos^2x =cos^2 x/cos^2 x + sin^2 x/cos^2 x = 1/cos^2 x = (cos^2x+sin^2 x)/cos^2 x = 1/cos^2 x

Answer 1

what was your problem to begin with ?

Answer 2

1 + sec2x sin2x = sec2x

Answer 3

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

Answer 4

1 + sec²x sin²x = sec²x
1 + (1/cos²x) sin²x = sec²x
1 + (sin²x/cos²x) = sec²x
1 + tan²x = sec²x

verified.

Answer 5

Or if you don't consider 1 + tan²x = sec²x to be an identity, then

1 + tan²x = sec²x
(cos²x/cos²x) + (sin²x/cos²x) = sec²x
(cos²x+sin²x)/cos²x = sec²x and knowing that `cos²x+sin²x=1`,
1/cos²x = sec²x
sec²x=sec²x

Answer 6

Will see what Lars is going to say about this :)

Answer 7

\[ \large \begin{align} 1 + \sec^2x\sin^2x &= \sec^2x \cr &= 1 +\tan^2x \cr &=1+ {\sin^2x \over \cos^2x} \cr &= 1 + {1 \over \cos^2x}\sin^2x \cr
&=1 + \sec^2x\sin^2x\end{align} \]

Answer 8

Yes, the same thing, but working the other side.

Answer 9

Just seems a bit wild though, to start verifying with making the simpler side more complicated rather than simplifying the hard side, but I like this approach to .

Answer 10

What is important in demonstrations like this is to work only one side. Otherwise you end up with 1=1 or 0=0 which is not a very good demonstration unless you turn it upside down and start with 1=1.

Answer 11

The reason I started on the right side was that this Pythagorean identity has been tattooed on my brain (and it is also one that exercise authors think is obscure, so they use it a lot).

Answer 12

Okay... I see your approach just the same way. It's probably just that I am thinking in a standard way and you are very creative!

Answer 13

Thank you guys... I'm still in the process of understanding each step.

Answer 14

Good luck:)

Answer 15

Thanks. :D

Answer 16

Ok, 1+tan^2 = sec^2 is a Pythagorean Identity.
Here is a geometric interpretation. See the pink tangent and the secant is cos + versine + exsecant on the horizontal line.

Answer 17

http://larswiki.larseighner.com/uploads/Main/trig_tangent.svg

Answer 18

The rest of the lines are basic trig identities.

Answer 19

Back soon. Have to reboot. Something (I suspect Opera) is leaking memory like a sieve.