5 Marks Questions

Question 15 Marks

Prove that:
$\cos15^\circ\cos35^\circ\text{cosec }55^\circ\cos60^\circ\text{cosec }75^\circ=\frac{1}{2}$

Answer

$\text{L.H.S.}=\cos15^\circ\cos35^\circ\text{cosec }55^\circ\cos60^\circ\text{cosec }75^\circ$
$=\cos(90^\circ-75^\circ)\cos(90^\circ-55^\circ)\frac{1}{\sin55^\circ}\times\frac12\times\frac{1}{\sin75^\circ}$
$=\sin75^\circ\sin55^\circ\frac{1}{\sin55^\circ}\times\frac12\times\frac{1}{\sin75^\circ}$
$=\frac{1}{2}$
$=\text{R.H.S.}$

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Question 25 Marks

Prove that:
$\cot12^\circ\cot38^\circ\cot52^\circ\cot60^\circ\cot78^\circ=\frac{1}{\sqrt{3}}$

Answer

$\text{L.H.S.}=\cot12^\circ\cot38^\circ\cot52^\circ\cot60^\circ\cot78^\circ$
$=\tan(90^\circ-12^\circ)\tan(90^\circ-38^\circ)\times\cot52^\circ\times\frac{1}{\sqrt{3}}\times\cot78^\circ$
$=\frac{1}{\sqrt{3}}\times\tan78^\circ\times\tan52^\circ\times\cot52^\circ\times\cot78^\circ$
$=\frac{1}{\sqrt{3}}\times\tan786\circ\times\tan52^\circ\times\frac{1}{\tan52^\circ}\times\frac{1}{\tan78^\circ}$
$=\frac{1}{\sqrt{3}}$
$=\text{R.H.S.}$

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Question 35 Marks

Prove that:$\frac23\text{cosec}^258^\circ-\frac23\cot58^\circ\tan32^\circ\\-\frac52\tan13^\circ\tan37^\circ\tan45^\circ\tan53^\circ\tan77^\circ=-1.$

Answer

$\frac23\text{cosec}^258^\circ-\frac23\cot58^\circ\tan32^\circ\\-\frac52\tan13^\circ\tan37^\circ\tan45^\circ\tan53^\circ\tan77^\circ$
$=\frac23\big(\text{cosec}^258^\circ-\cot58^\circ\tan32^\circ\big)\\ \ -\frac53\tan136^\circ\tan(90^\circ-13^\circ)\tan37^\circ\tan(90^\circ-37^\circ)(\tan45^\circ)$
$=\frac23\big\{\text{cosec}^258^\circ-\cot58^\circ\tan(90^\circ-58^\circ)\big\}\\ \ -\frac53\tan13^\circ\cot13^\circ\tan37^\circ\cot37^\circ(1)$
$=\frac23\big(\text{cosec}^258^\circ-\cot58^\circ\cot58^\circ\big)\\ \ -\frac53\tan13^\circ\frac{1}{\tan13^\circ}\tan37^\circ\frac{1}{\tan37^\circ}$
$=\frac23\big(\text{cosec}^258^\circ-\cot^258^\circ\big)-\frac53$
$=\frac23-\frac53$
$=-1$
Hence Proved.

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Maths 5 Marks Questions

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