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Trigonometry: if a = sinx+cosx and b=tanx+cotx, then a(b^2-1)-2 = ?

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Homework Statement
if a = sinx+cosx and b=tanx+cotx, then the value of a(b^2-1)-2 is
Homework Equations
sin^2(x)+cos^2(x)
b^2-1= tan^2(x) + cot^2(x) + 2 -1
b^2-1= sin^2(x)/cos^2(x) +cos^2(x)/sin^2(x) -1
b^2-1=[sin^4(x) +cos^4(x)]/sin^2(x)cos^2(x) -1
b^2-1=[1-sin^2(x)cos^2(x)]/sin^2(x)cos^2(x) -1
a(b^2-1)=sinx+cosx {[1-sin^2(x)cos^2(x)]/sin^2(x)cos^2(x) -1 }
I am not able to go any further than this step to reach the answer
 
Last edited by a moderator:

tnich

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Homework Statement: if a = sinx+cosx and b=tanx+cotx, then the value of a(b^2-1)-2 is
Homework Equations: sin^2(x)+cos^2(x)

b^2-1= tan^2(x) + cot^2(x) + 2 -1
b^2-1= sin^2(x)/cos^2(x) +cos^2(x)/sin^2(x) -1
b^2-1=[sin^4(x) +cos^4(x)]/sin^2(x)cos^2(x) -1
b^2-1=[1-sin^2(x)cos^2(x)]/sin^2(x)cos^2(x) -1
a(b^2-1)=sinx+cosx {[1-sin^2(x)cos^2(x)]/sin^2(x)cos^2(x) -1 }
I am not able to go any further than this step to reach the answer
I am not sure what answer you are looking for, but this might help. What do get when you write ##b## in the form ##b = \frac {? + ?} {sin(x)cos(x)}##?
 
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I'll get sin^2(x)+cos^2(x) which is 1
 
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the answer should be 0
But I don't know how to further proceed please give me a hint
 

SammyS

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the answer should be 0
But I don't know how to further proceed please give me a hint
Did you make an error when you copied or typed the the problem?

If you interchange the definitions for ##a## and ##b## you will get the desired result.
 
Last edited:

PeroK

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the answer should be 0
But I don't know how to further proceed please give me a hint
Did you not try some values of ##x## to see whether you get 0? If the expression is identically 0, then that means it's 0 for every value of ##x##.

Why not try at least one value of ##x##?
 
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Thank you, I think the question is wrongly printed
 

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