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

[Crystal037]

Homework Statement

if a = sinx+cosx and b=tanx+cotx, then the value of a(b^2-1)-2 is

Relevant 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

 

[tnich]

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 $b = \frac {? + ?} {sin(x)cos(x)}$?

 

[Crystal037]

I'll get sin^2(x)+cos^2(x) which is 1

 

[Crystal037]

the answer should be 0
But I don't know how to further proceed please give me a hint

 

[SammyS]

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.

 

[PeroK]

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 $x$?

 

[Crystal037]

Thank you, I think the question is wrongly printed