Solution.
1.
sec(−θ)=secθ,
because secθ=cosθ1 and cos(−θ)=cosθ.
2.
1−sin2xcot2x=1−sin2xsin2xcos2x==1−cos2x=sin2x.
As for me, this way is easier, because it is a way of proving by definition.
3.
cosx−cosxsin2x=cosx−cosx(1−cos2x)==cosx−cosx+cos3x=cos3x.
4.
I agree with John, because this equation has no solution.
sinx+cosx=2,
2sin(x+4π)=2,
sin(x+4π)=22,
and 22>1.
This equation has no real solutions.
Verify identity:
1.
cot2θ+cot2θ1=cot2θ+tan2θ,sec2θ=cos2θ1=1+tan2θ.
But 1=cot2θ. This equality is not an identity.
cot2θ+cot2θ1=sec2θ.
2.
(csc2θ−1)sin2θ=(sin2θ1−1)sin2θ==sin2θ1−sin2θsin2θ=1−sin2θ=cos2θ.
3.
1−secαcosα=0, and tanαcotα−1=0.
So, 1−secαcosα=tanαcotα−1.
4.
tanA+secAcscAcotA=tanA+sinAcosA⋅cosAsinA==tanA+cos2A=cosAsinA+cos3A=1.
This equality is not an identity.
5.
sec4θ−tan4θ=(sec2θ−tan2θ)(sec2θ+tan2θ)==cos2θ1−sin2θ⋅cos2θ1+sin2θ=cos2θcos2θ+sin2θ+sin2θ=1+2tan2θ.
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