Can This Trigonometric Identity Be Verified?

AI Thread Summary
The discussion focuses on verifying the trigonometric identity 1 + sec(-∅)/sin(-∅ + tan(-∅) = -csc ∅. Participants suggest starting by multiplying both sides by (sin∅ + tan∅) to simplify the equation. There is also a recommendation to rewrite sec, csc, and tan in terms of sine and cosine for clarity. One user notes that another related problem has already been posted in a separate thread. The conversation emphasizes the importance of proper manipulation of trigonometric functions to verify identities.
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Homework Statement



verify the identity :
1+sec(-∅)/sin(-∅+tan(-∅) = -csc ∅

Homework Equations





The Attempt at a Solution


1+sec∅/-sin-tan∅ = -csc∅

I don't know where to start from, does anyone have any idea?
 
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louie3006 said:
1+sec∅/-sin-tan∅ = -csc∅

I don't know where to start from, does anyone have any idea?

Hi louie3006!

Just do the obvious … multiply both sides by sin∅ + tan∅ :smile:
 
do you mean conjugate the left side of the equation ?
 
are those minus signs before ∅?
i agree with tiny-tim, multiply both side by (sin∅ + tan∅) and you should see it.
if it helps, you can rewrite sec, csc and tan in terms of sin and cos.
 
I need to verify the identity of this problem

tan^2(x/2)=(sec x-1)/(sec x+1)
 
You have already started a new thread for this problem, which is the right thing to do.
 
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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