[2 Mark Question Answer]

Question 12 Marks

Evaluate$: 14 \sin 30^\circ + 6 \cos 60^\circ - 5 \tan 45^\circ .$

Answer

$ 14 \sin 30^\circ + 6 \cos 60^\circ - 5 \tan 45^\circ$
$=14\left(\frac{1}{2}\right)+6\left(\frac{1}{2}\right)-5(1)$
$= 7 + 3 - 5$
$= 5.$

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Question 22 Marks

Evaluate $:\operatorname{cosec} (65^\circ + A) - \sec (25^\circ - A)$

Answer

$\operatorname{cosec} (65^\circ + A) - \sec (25^\circ - A)$
$=\operatorname{cosec} [90^\circ - (25^\circ - A)] - \sec (25^\circ - A)$
$= \sec (25^\circ - A) - \sec (25^\circ - A)$
$= 0.$

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Question 32 Marks

Evaluate$: \tan (55^\circ - A) - \cot (35^\circ + A)$

Answer

$\tan (55^\circ - A) - \cot (35^\circ + A)$
$= \tan [90^\circ - (35^\circ + A)] - \cot (35^\circ + A)$
$= \cot (35^\circ + A) - \cot (35^\circ + A)$
$= 0.$

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Question 42 Marks

Evaluate $: \frac{\sin 80^{\circ}}{\cos 10^{\circ}}+\sin 59^{\circ} \sec 31^{\circ}$

Answer

$\frac{\sin 80^{\circ}}{\cos 10^{\circ}}+\sin 59^{\circ} \sec 31^{\circ} $
$ =\frac{\sin \left(90^{\circ}-10^{\circ}\right)}{\cos 10^{\circ}}+\sin \left(90^{\circ}-31^{\circ}\right) \sec 31^{\circ} $
$=\frac{\cos 10^{\circ}}{\cos 10^{\circ}}+\frac{\cos 31^{\circ}}{\cos 31^{\circ}}$
$= 1 + 1$
$= 2$

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Question 52 Marks

Evaluate $: 3 \cos 80^\circ \operatorname{cosec} 10^\circ + 2 \sin 59^\circ \sec 31^\circ .$

Answer

$ 3 \cos 80^\circ \operatorname{cosec} 10^\circ + 2 \sin 59^\circ \sec 31^\circ$
$= 3 \cos (90^\circ - 10^\circ ) \operatorname{cosec} 10^\circ + 2\sin (90^\circ - 31) \sec 31^\circ$
$= 3 \sin 10^\circ \operatorname{cosec} 10^\circ + 2 \cos 31^\circ \sec 31^\circ$
$= 3 \times 1 + 2\times1$
$= 3 + 2$
$= 5$

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Question 62 Marks

Evaluate $: 3 \frac{\sin 72^{\circ}}{\cos 18^{\circ}}-\frac{\sec 32^{\circ}}{\operatorname{cosec} 58^{\circ}}$.

Answer

$3 \frac{\sin 72^{\circ}}{\cos 18^{\circ}}-\frac{\sec 32^{\circ}}{\operatorname{cosec} 58^{\circ}} $
$ =3 \frac{\sin \left(90^{\circ}-18^{\circ}\right)}{\cos 18^{\circ}}-\frac{\sec \left(90^{\circ}-58^{\circ}\right)}{\operatorname{cosec} 58^{\circ}} $
$=3 \frac{\cos 18^{\circ}}{\cos 18^{\circ}}-\frac{\operatorname{cosec} 58^{\circ}}{\operatorname{cosec} 58^{\circ}}$
$= 3 - 1$
$= 2$

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Question 72 Marks

For triangle $\text{ABC},$ show that $:\sin \left(\frac{ A + B }{2}\right)=\cos \frac{ C }{2}$

Answer

We know that for a triangle $\triangle ABC$
$\angle A + \angle B + \angle C = 180^\circ $
$\frac{\angle B +\angle A }{2}=90^{\circ}-\frac{\angle C }{2}$
$\sin \left(\frac{ A + B }{2}\right)=\sin \left(90^{\circ}-\frac{ C }{2}\right)$
$\sin \left(\frac{ A + B }{2}\right)=\cos \left(\frac{ C }{2}\right)$

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Question 82 Marks

Express the following in term of angles between $0^\circ $ and $45^\circ :\cos74^\circ + \sec 67^\circ $

Answer

$\cos74^\circ + \sec 67^\circ $
$= \cos(90 - 16)^\circ + \sec(90 - 23)^\circ $
$= \sin 16^\circ +\operatorname{cosec} 23^\circ $

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Question 92 Marks

Express the following in term of angles between $0^\circ $ and $45^\circ : \operatorname{cosec} 68^\circ + \cot 72^\circ $

Answer

$\operatorname{cosec} 68^\circ + \cot 72^\circ $
$= \operatorname{cosec}(90 - 22)^\circ + \cot(90 - 18)^\circ $
$(\because \operatorname{cosec}(90 - \theta ) = \sec \theta$ and $\cot(90 - \theta ) = \tan\theta )$
$=\sec 22^\circ + \tan 18^\circ $

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Question 102 Marks

Express the following in term of angles between $0^\circ $ and $45^\circ :\sin 59^\circ + \tan 63^\circ $

Answer

$\sin 59^\circ + \tan 63^\circ $
$= \sin(90 - 31)^\circ + \tan(90 - 27)^\circ $
$= \cos 31^\circ + \cot 27^\circ .$

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Question 112 Marks

Show that $: \tan 10^\circ \tan 15^\circ \tan 75^\circ \tan 80^\circ = 1.$

Answer

$\text{L.H.S.}$
$=\tan 10^\circ \tan 15^\circ \tan 75^\circ \tan 80^\circ $
$= \tan (90^\circ - 80^\circ ) \tan (90^\circ - 75^\circ ) \tan 75^\circ \tan 80^\circ $
$= \cot 80^\circ \cot 75 ^\circ \tan 75^\circ \tan 80^\circ $
$= (\cot 80^\circ \tan 80^\circ )(\cot 75^\circ \tan 75^\circ )$
$= (1)(1)$
$= 1$
$=\text{R.H.S.}$

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Question 122 Marks

Evaluate $: \cos^2 25^\circ - \sin^2 65^\circ - \tan^2 45^\circ$

Answer

$\cos^2 25^\circ - \sin^2 65^\circ - \tan^2 45^\circ$
$= [\cos(90^\circ - 65^\circ)]^2 - \sin^2 65^\circ - (\tan 45^\circ)^2$
$= \sin^2 65^\circ - \sin^2 65^\circ - (1)^2$
$= 0 - 1$
$= - 1.$

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Question 132 Marks

Evaluate $: \frac{2 \tan 53^{\circ}}{\cot 37^{\circ}}-\frac{\cot 80^{\circ}}{\tan 10^{\circ}}$

Answer

$\frac{2 \tan 53^{\circ}}{\cot 37^{\circ}}-\frac{\cot 80^{\circ}}{\tan 10^{\circ}}$
$=\frac{2 \tan \left(90^{\circ}-37^{\circ}\right)}{\cot 37^{\circ}}-\frac{\cot \left(90^{\circ}-10^{\circ}\right)}{\tan 10^{\circ}}$
$=\frac{2 \cot 37^{\circ}}{\cot 37^{\circ}}-\frac{\tan 10^{\circ}}{\tan 10^{\circ}}$
$= 2 - 1$
$= 1$

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Question 142 Marks

Evaluate $: \sin^2 35^\circ - \cos^2 55^\circ$

Answer

$\sin^2 35^\circ - \cos^2 55^\circ$
$= \sin^2 35^\circ - [\cos(90^\circ - 35^\circ)]^2$
$= \sin^235^\circ - \sin^235^\circ$
$= 0$

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Question 152 Marks

Evaluate $: \sin (90^\circ - A) \sin A - \cos (90^\circ - A) \cos A$

Answer

$ \sin (90^\circ - A) \sin A - \cos (90^\circ - A) \cos A$
$= \cos A \sin A - \sin A \cos A$
$= 0 $

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Question 162 Marks

Evaluate $: \sin 42^\circ \sin 48^\circ - \cos 42^\circ \cos 48^\circ $

Answer

$\sin 42^\circ \sin 48^\circ - \cos 42^\circ \cos 48^\circ $
$= \sin(90^\circ - 48^\circ )\sin 48^\circ - \cos(90^\circ - 48^\circ )\cos 48^\circ $
$= \cos 48^\circ \sin 48^\circ - \sin 48^\circ \cos 48^\circ $
$= \cos 48^\circ \sin 48^\circ - \cos 48^\circ \sin 48^\circ $
$= 0$

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Question 172 Marks

Evaluate $: \sin 15^\circ \cos 15^\circ - \cos 75^\circ \sin 75^\circ $

Answer

$\sin 15^\circ \cos 15^\circ - \cos 75^\circ \sin 75^\circ $
$= \sin(90^\circ - 75^\circ ) \cos15^\circ - \cos 75^\circ \sin (90^\circ -15^\circ )$
$= \cos 75^\circ \cos 15^\circ - \cos 75^\circ \cos 15^\circ $
$= 0.$

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Question 182 Marks

Evaluate $: \sec^2 18^\circ - cosec^2 72^\circ$

Answer

$\sec^2 18^\circ - \operatorname{cosec}^2 72^\circ$
$= [\sec (90^\circ - 72^\circ)]^2 - \operatorname{cosec}^272^\circ$
$= \operatorname{cosec}^2 72^\circ - \operatorname{cosec}^2 72^\circ$
$= 0$

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Question 192 Marks

Evaluate $: \sin^2 40^\circ - \cos^2 50^\circ$

Answer

$\sin^2 40^\circ - \cos^2 50^\circ$
$= \sin^2(90^\circ - 50^\circ) - \cos^2 50^\circ$
$= \cos^250^\circ - \cos^250^\circ$
$= 0$

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Question 202 Marks

Evaluate $: \frac{\cos 55^{\circ}}{\sin 35^{\circ}}+\frac{\cot 35^{\circ}}{\tan 55^{\circ}}$

Answer

$\frac{\cos 55^{\circ}}{\sin 35^{\circ}}+\frac{\cot 35^{\circ}}{\tan 55^{\circ}}$
$=\frac{\cos \left(90^{\circ}-35^{\circ}\right)}{\sin 35^{\circ}}+\frac{\cot \left(90^{\circ}-55^{\circ}\right)}{\tan 55^{\circ}}$
$=\frac{\sin 35^{\circ}}{\sin 35^{\circ}}+\frac{\tan 55^{\circ}}{\tan 55^{\circ}}$
$= 1 + 1$
$= 2$

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Question 212 Marks

Evaluate $: \frac{\sec 75^{\circ}}{\operatorname{cosec} 15^{\circ}}$

Answer

$ \frac{\sec 75^{\circ}}{\operatorname{cosec} 15^{\circ}}$
$= \frac{\sec \left(90^{\circ}-15^{\circ}\right)}{\operatorname{cosec} 15^{\circ}}$
$= \frac{\operatorname{cosec} 15^{\circ}}{\operatorname{cosec} 15^{\circ}}$
$= 1$

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Question 222 Marks

Evaluate $: \frac{\tan 47^{\circ}}{\cot 43^{\circ}}$

Answer

$\frac{\tan 47^{\circ}}{\cot 43^{\circ}}$
$=\frac{\tan \left(90^{\circ}-43^{\circ}\right)}{\cot 43^{\circ}}$
$=\frac{\cot 43^{\circ}}{\cot 43^{\circ}}$
$=1$

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Question 232 Marks

Evaluate $: \frac{\cos 22^{\circ}}{\sin 68^{\circ}}$

Answer

$\frac{\cos 22^{\circ}}{\sin 68^{\circ}}$
$=\frac{\cos \left(90^{\circ}-68^{\circ}\right)}{\sin 68^{\circ}}$
$=\frac{\sin 68^{\circ}}{\sin 68^{\circ}}$
$=1$

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