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Sir, i'm so much curious to know the relations between trigonometric identities and how to prove them. For instance, prove that (1)1+tanx/1+cotx=tanx.
2) u=1+sinØ/cosØ show that
b)tanØ=u^2-1/2u. The truth is that, i don't know where to start! I don't how to apply trigonometric identities! Please sir help!

(1) It is known that tanx = 1/ctnx and tanx = sinx/cosx, which makes ctnx = cosx/sinx.

That makes the fraction (1+tanx)/(1+cotx) be (1 + sinx/cosx)/(1 + cosx/sinx).
If we multiply by (sinx*cosx)/(sinx*cosx), that is the same as multiplying each term by
sinx*cosx.  This gives (sinx*cosx + sin²x)/(sinx*cosx + cos²x).
That factors to sinx(cosx+sinx)/[cosx(sinx+cosx)].

Since cosx+sinx = sinx+cosx, those terms, the 1st in the numerator and
the 2nd in the denominator, cancel.

That leaves sinx/cosx.  As stated at the start, that is the same as tanx.

(2a) It is known that 1/u = cosØ/(1+sinØ).
Multiply this by (1-sinØ)/(1-sinØ) and get cosØ(1-sinØ)/(1-sin²Ø).

Since sin²Ø + cos²Ø = 1, it can be seen that by subtracting sin²Ø from both sides gives
cos²Ø = 1 - sin²Ø. Using this in the denominator shows cosØ(1-sinØ)/(1-sin²Ø)
is the same as cosØ(1-sinØ)/cos²Ø.  A cosØ/cosØ can be cancelled, leaving (1-sinØ)/cosØ.

(2b) If we have tanØ = (u2-1)/2u, split the fraction (u2-1)/2u into two terms.
That gives tanØ = u/2 - 1/(2u).

Since u = (1 + sinØ)/cosØ, that can be split apart to, giving u = 1/cosØ + sinØ/cosØ.
That is the same as u = secØ + tanØ.

Putting this into tanØ = u/2 - 1/(2u) gives tanØ = (secØ+tanØ)/2 - 1/[2(secØ+tanØ)].

Multiplying the 1st fraction by (secØ+tanØ)/(secØ+tanØ) and bringing down the other gives
(secØ+tanØ)2/[2(secØ+tanØ)] - 1/[2(secØ+tanØ)].

They have a common denominator, so combining them gives [(secØ+tanØ)² - 1]/[2(secØ+tanØ)].
Squaring the term in the top gives [sec²Ø + 2secØtanØ + tan²Ø - 1]/[2(secØ+tanØ)].

A note on the side is:
Remembering the identity that sin²Ø + cos²Ø = 1, divide both sides of sin²Ø + cos²Ø = 1
by cos²Ø. That gives sin²Ø/cos²Ø + 1 = 1/cos²Ø, which is the same as tan²Ø + 1 = sec²Ø.
Subtracting 1 from both sides gives tan²Ø = sec²Ø - 1.

Back to the problem at hand:
Noting that the 1st and last terms in the numerator are sec²Ø - 1, that is the same as tan²Ø.
Combining that with the other tan²Ø gives (2secØtanØ + 2tan²Ø)/[2(secØ+tanØ)].

Factoring 2tanØ out of each term on the top gives 2tanØ(secØ+tanØ)/[2(secØ+tan)].
Cancelling the 2/2 and the (secØ+tanØ)/(secØ+tanØ) gives tanØ,
and that's what we were looking for.  

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Scott A Wilson


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