100. Айниятро исбот кунед: а) \(\frac{\sin2\alpha}{1+\cos2\alpha}= g\alpha\); б) \(\frac{\sin\alpha+\sin5\alpha}{\cos\alpha+\cos5\alpha}= g3\alpha\); в) \((\sin\alpha-\cos\alpha)^2=1-\sin2\alpha\); г) \(\sin2\alpha- g\alpha=\cos2\alpha\cdot g\alpha\); д) \((\sin\alpha+\cos\alpha)^2-\sin2\alpha=1\); е) \(4\sin\alpha\cdot\cos\alpha\cdot\cos2\alpha=\sin4\alpha\).

а)

\[\frac{\sin2\alpha}{1+\cos2\alpha}=\frac{2\sin\alpha\cos\alpha}{2\cos^2\alpha}=\frac{\sin\alpha}{\cos\alpha}= g\alpha.\]

б) Аз формулаҳои ҷамъ ба ҳосил:

\[\frac{\sin\alpha+\sin5\alpha}{\cos\alpha+\cos5\alpha}=\frac{2\sin3\alpha\cos2\alpha}{2\cos3\alpha\cos2\alpha}=\frac{\sin3\alpha}{\cos3\alpha}= g3\alpha.\]

в)

\[(\sin\alpha-\cos\alpha)^2=\sin^2\alpha+\cos^2\alpha-2\sin\alpha\cos\alpha=1-\sin2\alpha.\]

г)

\[\sin2\alpha- g\alpha=2\sin\alpha\cos\alpha-\frac{\sin\alpha}{\cos\alpha}=\frac{\sin\alpha(2\cos^2\alpha-1)}{\cos\alpha}.\]

Азбаски \(2\cos^2\alpha-1=\cos2\alpha\), пас:

\[\sin2\alpha- g\alpha=\cos2\alpha\cdot g\alpha.\]

д)

\[(\sin\alpha+\cos\alpha)^2-\sin2\alpha=\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha-\sin2\alpha=1.\]

е)

\[4\sin\alpha\cos\alpha\cos2\alpha=2\sin2\alpha\cos2\alpha=\sin4\alpha.\]

Ҳамаи айниятҳо исбот шуданд.