Hi,

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

[FONT="][/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="] [/FONT]x cos[FONT="] [/FONT]x=1

Therefore:

sec²x-2 sec x cos x + cos²

=sec²x + cos²-2

Identity: sec²[FONT="] [/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="] [/FONT]x+1/tan[FONT="] [/FONT]x =1/(sin[FONT="] [/FONT]x cosx )

L.H.S:

Tan[FONT="] [/FONT]x+1/tan[FONT="] [/FONT]x

Identity: 1/tan[FONT="] [/FONT]x = cot[FONT="] [/FONT]x :

=tan[FONT="] [/FONT]x+cot[FONT="] [/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

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

[FONT="][/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="] [/FONT]x cos[FONT="] [/FONT]x=1

Therefore:

sec²x-2 sec x cos x + cos²

=sec²x + cos²-2

Identity: sec²[FONT="] [/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="] [/FONT]x+1/tan[FONT="] [/FONT]x =1/(sin[FONT="] [/FONT]x cosx )

L.H.S:

Tan[FONT="] [/FONT]x+1/tan[FONT="] [/FONT]x

Identity: 1/tan[FONT="] [/FONT]x = cot[FONT="] [/FONT]x :

=tan[FONT="] [/FONT]x+cot[FONT="] [/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

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