1. ## Trig identities.

Hi,
Im having a lot of trouble solving the following trig identities:

a) sec² x-2 sec x cos x + cos² = tan² x - sin² x
b) $tan x + 1/ tanx = 1/sin x cos x$

My working:
For a):

Trigonometric identity: secx cosx=1
Therefore:
sec²x-2 sec x cos x + cos²
=sec²x + cos²-2
Identity: sec²x = 1/cos²x
Use the Identity:
sec²x + cos²- 2
1/cos^2x +cos²x-2

Now I am very confused, really unsure on how to prove this identity. I would greatly appreciate any helpful tips.

For b)

tanx+1/tanx =1/(sinx cosx )
L.H.S:
Tanx+1/tanx
Identity: 1/tanx = cotx :
=tanx+cotx

Now im completely lost, the R.H.S side is turning out to be "csc x sec x" .

I would greatly appreciate any helpful tips, since I'm very confused here.

Thanks

2. Originally Posted by spoc21
Hi,
Im having a lot of trouble solving the following trig identities:

a) sec² x-2 sec x cos x + cos² = tan² x - sin² x
b) $tan x + 1/ tanx = 1/sin x cos x$

My working:
For a):

Trigonometric identity: secx cosx=1
Therefore:
sec²x-2 sec x cos x + cos²
=sec²x + cos²-2
Identity: sec²x = 1/cos²x
Use the Identity:
sec²x + cos²- 2
1/cos^2x +cos²x-2

Now I am very confused, really unsure on how to prove this identity. I would greatly appreciate any helpful tips.

For b)

tanx+1/tanx =1/(sinx cosx )
L.H.S:
Tanx+1/tanx
Identity: 1/tanx = cotx :
=tanx+cotx

Now im completely lost, the R.H.S side is turning out to be "csc x sec x" .

I would greatly appreciate any helpful tips, since I'm very confused here.

Thanks
(b) $tanx+\frac{1}{tanx} = \frac{sinx}{cosx} +\frac{1}{\frac{sinx}{cosx}} = \frac{sinx}{cosx}+\frac{cosx}{sinx}$

$= \frac{sin^2x+cos^2x}{sinx.cosx} = \frac{1}{sinx.cosx}$

3. (a) $sec^2x-2secx.cosx+cos^2x$

use the identity $sec^2x-tan^2x=1$ and $cos^2x=1-sin^2x$

$= (1+tan^2x)-2+(1-sin^2x)$

$= 1+tan^2x-2+1-sin^2x$

$=tan^2x-sin^2x$

4. Originally Posted by harish21
(a) $sec^2x-2secx.cosx+cos^2x$

use the identity $sec^2x-tan^2x=1$ and $cos^2x=1-sin^2x$

$= (1+tan^2x)-2+(1-sin^2x)$

$= 1+tan^2x-2+1-sin^2x$

$=tan^2x-sin^2x$
Thanks harish,
the $cos^2 x$ in bold is actually only $cos^2$..would this make a difference? or would the same method you described work?