# Mathematics - Trigonometric Identities

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Let y represent theta

Prove:

1 + 1/tan^2y = 1/sin^2y

LS:

= 1 + 1/tan^2y
= (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y)
= (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y)
= (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y)
= (sin^2y + cos^2y) + (sin^2y + cos^2y)(sin^2y) / (sin^2y)(cos^2y/sin^2y)

... And now I confuse myself

Where did I go wrong? And please direct me on how to fix it?

• Mathematics - Trigonometric Identities -

you confused me too

I will start again with
LS = 1 + cot^2 y
= 1 + cos^2 y/sin^2 y , now find a common denominator
= (sin^2 y + cos^2 y) / sin^2 y, but sin^2 y + cos^2 y=1
= 1/sin^2 y
= RS

• Mathematics - Trigonometric Identities -

I'm not suppose to use the inverse identities yet ...

• Mathematics - Trigonometric Identities -

ok, then in

LS = 1 + 1/tan^2y

= 1 + 1/(sin^2 / cos^2 y)
= 1 + (cos^2 y)/(sin^2 y)
= ... my second line

surely you are going to use tanx = sinx/cosx !!!

we are not using "inverse identities" here

• Mathematics - Trigonometric Identities -

I meant the reciprocals of SOH CAH TOA

And ... I just got it

LS:

= 1 + 1/tan^2y
= 1 + 1/(sin^2y/cos^2y)
= 1 + 1(cos^2y/sin^2y)
= 1 + cos^2y / sin^2y
= 1 + 1-sin^2y / sin^2y
= 1(sin^2y) / sin^2y + 1 - sin^2y / sin^2y
= sin^2y + 1 - sin^2y / sin^2y
= 1/sin^2y

• Mathematics - Trigonometric Identities -

yes
correct

• Mathematics - Trigonometric Identities -

should have taken a closer look at your solution.
the last few lines make no sense with "no brackets" being used

why don't you follow the steps of my original solution, it is so straightforward.

• Mathematics - Trigonometric Identities -

cscx-cotx forms an identity with?

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