Trig identities.

spoc21

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:

harish21

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

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

spoc21

harish21

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

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

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

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

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

spoc21

spoc21

(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?