[3 marks sum]

Question 13 Marks

In the case, given below, find the value of angle $A,$ where$ 0^\circ \leq A \leq 90^\circ .\cos(90^\circ - A) · sec 77^\circ = 1$

Answer

$\cos (90^\circ - A) · \sec 77^\circ = 1$
$\Rightarrow \cos(90^\circ - A) = \frac{1}{\sec 77^{\circ}}$
$\Rightarrow \cos(90^\circ - A) = \cos 77^\circ $
$\Rightarrow 90^\circ - A = 77^\circ $
$\Rightarrow - A = 77^\circ - 90^\circ $
$\Rightarrow - A = - 13^\circ $
$\Rightarrow A = 13^\circ $

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Question 23 Marks

In the case, given below, find the value of angle $A,$ where $0^\circ \leq A \leq 90^\circ.\sin (90^\circ - 3A).\operatorname{cosec} 42^\circ = 1.$

Answer

$\sin (90^\circ - 3A). \operatorname{cosec} 42^\circ = 1$
$\Rightarrow \sin \left(90^{\circ}-3 A \right)=\frac{1}{\operatorname{cosec} 42^{\circ}}$
$\Rightarrow \cos 3 A =\frac{1}{\operatorname{cosec}\left(90^{\circ}-48^{\circ}\right)}$
$\Rightarrow \cos 3 A =\frac{1}{\sec 48^{\circ}}$
$\Rightarrow \cos 3A = \cos 48^\circ$
$\Rightarrow 3A = 48^\circ$
$\Rightarrow A = 16^\circ.$

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Question 33 Marks

A $\triangle ABC$ is right$-$angled at $B;$ find the value of $\frac{\sec A \cdot \sin C-\tan A \cdot \tan C}{\sin B}$.

Answer

Since $\triangle ABC$ is a right angled triangle, right angled at $B,$
$A + C = 90^\circ$
$\therefore \frac{\sec A \cdot \sin C-\tan A \cdot \tan C}{\sin B}$
$=\frac{\sec A\left(90^{\circ}-C\right) \sin C-\tan \left(90^{\circ}-C\right) \tan C}{\sin 90^{\circ}}$
$=\frac{\operatorname{cosec} C \sin C-\cot C \tan C}{1}$
$=\frac{1}{\sin C} \times \sin C-\frac{1}{\tan C} \times \tan C$
$= 1 - 1$
$= 0$

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Question 43 Marks

Evaluate: $\frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 59^{\circ}}{\sin 31^{\circ}}-8 \sin ^2 30^{\circ}$

Answer

$\frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 59^{\circ}}{\sin 31^{\circ}}-8 \sin ^2 30^{\circ}$
$ =\frac{\cos \left(90^{\circ}-20^{\circ}\right)}{\sin 20^{\circ}}+\frac{\cos \left(90^{\circ}-31^{\circ}\right)}{\sin 31^{\circ}}-8\left(\frac{1}{2}\right)^2 $
$=\frac{\sin 20^{\circ}}{\sin 20^{\circ}}+\frac{\sin 31^{\circ}}{\sin 31^{\circ}}-2$
$= 1 + 1 - 2$
$= 0$

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Question 53 Marks

Evaluate:$\frac{\cot ^2 41^{\circ}}{\tan ^2 49^{\circ}}-2 \frac{\sin ^2 75^{\circ}}{\cos ^2 15^{\circ}}$

Answer

$\frac{\cot ^2 41^{\circ}}{\tan ^2 49^{\circ}}-2 \frac{\sin ^2 75^{\circ}}{\cos ^2 15^{\circ}}$
$ =\frac{\left[\cot \left(90^{\circ}-49^{\circ}\right)\right]^2}{\tan ^2 49^{\circ}}-2 \frac{\left[\sin \left(90^{\circ}-15^{\circ}\right)\right]^2}{\cos ^2 15^{\circ}} $
$=\frac{\tan ^2 49^{\circ}}{\tan ^2 59^{\circ}}-2 \frac{\cos ^2 15^{\circ}}{\cos ^2 15^{\circ}}$
$= 1 - 2$
$= -1$

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Question 63 Marks

Evaluate: $2 \frac{\tan 57^{\circ}}{\cot 33^{\circ}}-\frac{\cot 70^{\circ}}{\tan 20^{\circ}}-\sqrt{2} \cos 45^{\circ}$

Answer

$2 \frac{\tan 57^{\circ}}{\cot 33^{\circ}}-\frac{\cot 70^{\circ}}{\tan 20^{\circ}}-\sqrt{2} \cos 45^{\circ}$
$ =2 \frac{\tan \left(90^{\circ}-33^{\circ}\right)}{\cot 33^{\circ}}-\frac{\cot \left(90^{\circ}-20^{\circ}\right)}{\tan 20^{\circ}}-\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)$
$ =2 \frac{\cot 33^{\circ}}{\cot 33^{\circ}}-\frac{\tan 20^{\circ}}{\tan 20^{\circ}}-1$
$= 2 - 1 - 1$
$= 0.$

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Question 73 Marks

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

Answer

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

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Question 83 Marks

Show that $: \sin 42^\circ \sec 48^\circ + \cos 42^\circ \operatorname{cosec} 48^\circ = 2.$

Answer

$\text{L.H.S.}$
$= \sin 42^\circ \sec 48^\circ + \cos 42^\circ \operatorname{cosec} 48^\circ$
$=\sin \left(90^{\circ}-48^{\circ}\right) \times \frac{1}{\cos 48^{\circ}}+\cos \left(90^{\circ}-48^{\circ}\right) \times \frac{1}{\sin 48^{\circ}}$
$=\cos 48^{\circ} \times \frac{1}{\cos 48^{\circ}}+\sin 48^{\circ} \times \frac{1}{\sin 48^{\circ}}$
$= 1 + 1$
$= 2$
$=\text{ R.H.S.}$

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Question 93 Marks

Evaluate: $\left(\frac{\sin 77^{\circ}}{\cos 13^{\circ}}\right)^2+\left(\frac{\cos 77^{\circ}}{\sin 13^{\circ}}\right)^2-2 \cos ^2 45^{\circ}$

Answer

$\left(\frac{\sin 77^{\circ}}{\cos 13^{\circ}}\right)^2+\left(\frac{\cos 77^{\circ}}{\sin 13^{\circ}}\right)^2-2 \cos ^2 45^{\circ}$
$=\left(\frac{\sin \left(90^{\circ}-13^{\circ}\right)}{\cos 13^{\circ}}\right)^2+\left(\frac{\cos \left(90^{\circ}-13^{\circ}\right)}{\sin 13^{\circ}}\right)^2-2\left(\cos 45^{\circ}\right)^2$
$=\left(\frac{\cos 13^{\circ}}{\cos 13^{\circ}}\right)^2+\left(\frac{\sin 13^{\circ}}{\sin 13^{\circ}}\right)^2-2\left(\frac{1}{\sqrt{2}}\right)^2$
$=(1)^2+(1)^2-2 \times \frac{1}{2} $
$ =1+1-1 $
$=1$

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Question 103 Marks

Evaluate: $\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-2$

Answer

$\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-2 $
$ =\frac{\cot \left(90^{\circ}-36^{\circ}\right)}{\tan 36^{\circ}}+\frac{\tan \left(90^{\circ}-70^{\circ}\right)}{\cot 70^{\circ}}-2 $
$ =\frac{\tan 36^{\circ}}{\tan 36^{\circ}}+\frac{\cot 70^{\circ}}{\cot 70^{\circ}}-2$
$= 1 + 1 - 2$
$= 2 - 2$
$= 0$

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MATHEMATICS [3 marks sum]

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