State, in terms of \(\pi\), the principal value of
\(\tan^{-1}1\),[1]
\(\sin^{-1}\left(-\dfrac{\sqrt3}2\right)\).[1]
Explain why the principal value of \(\cos^{-1}\left(-\dfrac{\sqrt2}2\right)\) cannot be \(-\dfrac\pi4\).[1]
Prove that \(\dfrac1{\sin x+1}-\dfrac1{\sin x-1}=2\sec^2x\).[3]
Hence solve the equation \(\dfrac1{\sin2x+1}-\dfrac1{\sin2x-1}=3\) for \(0<x<\pi\).[4]
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Answer:(a)(i) \(\dfrac\pi4\), (ii) \(-\dfrac\pi3\). (b) Principal arccos values lie in \([0,\pi]\). (c)(i) Proved. (ii) \(x=0.308,1.26,1.88,2.83\) radians (3 s.f.).