# Trig identities.

• May 26th 2010, 11:23 AM
spoc21
Trig identities.
Hi,
Im having a lot of trouble solving the following trig identities:

a) sec˛ x-2 sec x cos x + cos˛ = tan˛ x - sin˛ x
b) $\displaystyle tan x + 1/ tanx = 1/sin x cos x$

My working:
For a):

Trigonometric identity: secx cosx=1
Therefore:
sec˛x-2 sec x cos x + cos˛
=sec˛x + cos˛-2
Identity: sec˛x = 1/cos˛x
Use the Identity:
sec˛x + cos˛- 2
1/cos^2x +cos˛x-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 :)
• May 26th 2010, 11:57 AM
harish21
Quote:

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

a) sec˛ x-2 sec x cos x + cos˛ = tan˛ x - sin˛ x
b) $\displaystyle tan x + 1/ tanx = 1/sin x cos x$

My working:
For a):

Trigonometric identity: secx cosx=1
Therefore:
sec˛x-2 sec x cos x + cos˛
=sec˛x + cos˛-2
Identity: sec˛x = 1/cos˛x
Use the Identity:
sec˛x + cos˛- 2
1/cos^2x +cos˛x-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)$\displaystyle tanx+\frac{1}{tanx} = \frac{sinx}{cosx} +\frac{1}{\frac{sinx}{cosx}} = \frac{sinx}{cosx}+\frac{cosx}{sinx}$

$\displaystyle = \frac{sin^2x+cos^2x}{sinx.cosx} = \frac{1}{sinx.cosx}$
• May 26th 2010, 12:04 PM
harish21
(a) $\displaystyle sec^2x-2secx.cosx+cos^2x$

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

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

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

$\displaystyle =tan^2x-sin^2x$
• May 26th 2010, 01:09 PM
spoc21
Quote:

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

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

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

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

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

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