Proving the Identity: cot x + tan x = sec x csc x

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To prove the identity cot x + tan x = sec x csc x, one can express tan x as sin x/cos x and cot x as cos x/sin x. By combining these fractions, the left side simplifies to (cos^2 x + sin^2 x) / (sin x cos x), which equals 1/(sin x cos x). The right side, sec x csc x, can also be expressed as 1/(sin x cos x), confirming both sides are equal. The proof demonstrates the equivalence of the two expressions clearly.
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I need to prove algebraically this and I'm having trouble with it.

cot x + tan x = sec x csc x

Can anyone help me?
 
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What is tan x in terms of sin x and cos x? And cot x?
 
Ahh... I see.

cos^2 x + sin^2 x
------------------ = MD
sin x cos x

1
---------- = MD
sin x cos x

Then

sec x csc x= sec x csc x CQFD

I always see those problem and try to complicate things. Thanks for your help.
 

Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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