Trig identities.

May 2010
36
0
Hi,
Im having a lot of trouble solving the following trig identities:

[FONT=&quot][/FONT] a) sec² x-2 sec x cos x + cos² = tan² x - sin² x
b) \(\displaystyle tan x + 1/ tanx = 1/sin x cos x\)




My working:
For a):


Trigonometric identity: sec[FONT=&quot] [/FONT]x cos[FONT=&quot] [/FONT]x=1
Therefore:
sec²x-2 sec x cos x + cos²
=sec²x + cos²-2
Identity: sec²[FONT=&quot] [/FONT]x = 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)


tan[FONT=&quot] [/FONT]x+1/tan[FONT=&quot] [/FONT]x =1/(sin[FONT=&quot] [/FONT]x cosx )
L.H.S:
Tan[FONT=&quot] [/FONT]x+1/tan[FONT=&quot] [/FONT]x
Identity: 1/tan[FONT=&quot] [/FONT]x = cot[FONT=&quot] [/FONT]x :
=tan[FONT=&quot] [/FONT]x+cot[FONT=&quot] [/FONT]x

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 :)
 
Last edited by a moderator:
Feb 2010
1,036
386
Dirty South
Hi,
Im having a lot of trouble solving the following trig identities:

[FONT=&quot][/FONT] a) sec² x-2 sec x cos x + cos² = tan² x - sin² x
b) \(\displaystyle tan x + 1/ tanx = 1/sin x cos x\)




My working:
For a):


Trigonometric identity: sec[FONT=&quot] [/FONT]x cos[FONT=&quot] [/FONT]x=1
Therefore:
sec²x-2 sec x cos x + cos²
=sec²x + cos²-2
Identity: sec²[FONT=&quot] [/FONT]x = 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)


tan[FONT=&quot] [/FONT]x+1/tan[FONT=&quot] [/FONT]x =1/(sin[FONT=&quot] [/FONT]x cosx )
L.H.S:
Tan[FONT=&quot] [/FONT]x+1/tan[FONT=&quot] [/FONT]x
Identity: 1/tan[FONT=&quot] [/FONT]x = cot[FONT=&quot] [/FONT]x :
=tan[FONT=&quot] [/FONT]x+cot[FONT=&quot] [/FONT]x

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)\(\displaystyle tanx+\frac{1}{tanx} = \frac{sinx}{cosx} +\frac{1}{\frac{sinx}{cosx}} = \frac{sinx}{cosx}+\frac{cosx}{sinx}\)

\(\displaystyle = \frac{sin^2x+cos^2x}{sinx.cosx} = \frac{1}{sinx.cosx}\)
 
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Feb 2010
1,036
386
Dirty South
(a) \(\displaystyle sec^2x-2secx.cosx+cos^2x\)

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

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

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

\(\displaystyle =tan^2x-sin^2x\)
 
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May 2010
36
0
(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?