Question 1: Evaluate each of the following:

Answers:

i) cosec \ 30^o + \cot 45^o = 2 + 1 = 3

ii) \cos 30^o \cos 45^o - \sin 30^o \sin 45^o = \frac{\sqrt{3}}{2} . \frac{1}{\sqrt{2}} - \frac{1}{2} . \frac{1}{\sqrt{2}} = \frac{\sqrt{3} -1}{2 \sqrt{2}}

iii) \tan 30^o \sec 45^o + \tan 60^o \sec 30^o = \frac{1}{\sqrt{3}} . \sqrt{2} + \sqrt{3} . \frac{2}{\sqrt{3}} = \frac{\sqrt{2}+ 2 \sqrt{3}}{\sqrt{3}}

iv) \sin 30^o \cos 45^o + \cos 30^o \sin 45^o = \frac{1}{2} . \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} . \frac{1}{\sqrt{2}} = \frac{\sqrt{3}+1}{2 \sqrt{2}}

v) \frac{\sin^2 45^o + \cos^2 45^o}{\tan^2 60^o} = \frac{(\frac{1}{\sqrt{2}})^2 +(\frac{1}{\sqrt{2}})^2 }{(\sqrt{3})^2} = \frac{\frac{1}{2}+ \frac{1}{2}}{3} = \frac{1}{3}

vi) \frac{\sin 30^o - \sin 90^o + 2 \cos 0^o}{\tan 30^o \tan 60^o} = \frac{\frac{1}{2} - 1 + 2 \times 1}{\frac{1}{\sqrt{3}} \times \sqrt{3}} = \frac{3}{2}

vii) \frac{\sin 60^o}{\cos^2 45^o} - \cot 30^o + 15 \cos 90^o = \frac{\frac{\sqrt{3}}{2}}{(\frac{1}{\sqrt{2}})^2} - \sqrt{3} + 15 \times 0 = \sqrt{3} -\sqrt{3} = 0

viii) \frac{5 \sin^2 30^o + \cos^2 45^o - 4 \tan^2 30^o}{2 \sin 30^o \cos 30^o + \tan 45^o}

= \frac{5 \Big(\frac{1}{2}\Big)^2 + \Big(\frac{1}{\sqrt{2}}\Big)^2 - 4 \Big(\frac{1}{\sqrt{3}}\Big)^2 }{2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} + 1} = \frac{\frac{5}{4} + \frac{1}{2} - \frac{4}{3}}{\frac{\sqrt{3}}{2} +1}  = \frac{\frac{5}{12}}{\Big(\frac{\sqrt{3}+2}{2}\Big)}  = \frac{5}{6} (2 - \sqrt{3})

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Question 2: Find the value of \theta in each of the following when 0^o < \theta < 90^o :

Answers:

i) 2 \sin 2 \theta = \sqrt{3}

\Rightarrow  \sin 2 \theta = \frac{\sqrt{3}}{2} 

\Rightarrow  \sin 2 \theta = \sin 60^o

\Rightarrow  2 \theta = 60^o \Rightarrow  \theta = 30^o

ii) 2 \cos 3\theta = 1

\Rightarrow  \cos 3 \theta = \frac{1}{2} 

\Rightarrow  \cos 3 \theta = \cos 60^o

\Rightarrow  3 \theta = 60^o \Rightarrow  \theta = 20^o

iii) \sqrt{3} \tan 2 \theta - 3 = 0

\Rightarrow  \tan 2 \theta = \frac{3}{\sqrt{3}} = \sqrt{3}

\Rightarrow \tan 2 \theta = \tan 60^o

\Rightarrow  2 \theta = 60^o  \Rightarrow  \theta = 30^o

iv) 2 \cos \theta = 1

\Rightarrow  \cos \theta = \frac{1}{2}

\Rightarrow  \cos \theta = \cos 60^o

\Rightarrow  \theta = 60^o

v) 2 \cos^2 \theta = \frac{1}{2}

\Rightarrow  \cos^2 \theta = \frac{1}{4} \Rightarrow  \cos \theta = \frac{1}{2}

\Rightarrow  \cos \theta = \cos 60^o \Rightarrow  \theta = 60^o

vi) 2 \sin^2 \theta = \frac{1}{2}

\Rightarrow  \sin^2 \theta = \frac{1}{4} \Rightarrow  \sin \theta = \frac{1}{2}

\Rightarrow  \sin \theta = \sin 30^o \Rightarrow  \theta = 30^o

vii) 3 \tan^2 \theta - 1 = 0

\Rightarrow  \tan^2 \theta = \frac{1}{3} \Rightarrow  \tan \theta = \frac{1}{\sqrt{3}}

\Rightarrow  \tan \theta = \tan 30^o \Rightarrow  \theta = 30^o

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Question 3: If \theta   is an acute angle and \tan \theta  + \cot \theta  = 2 , find the value of \tan^7 \theta  + \cot^7 \theta .

Answers:

Given \tan \theta  + \cot \theta  = 2

\tan \theta  + \frac{1}{ \tan \theta } = 2

\tan^2 \theta  - 2 \tan \theta  + 1 = 0

\Rightarrow (\tan \theta  -1)^2 = 0

\Rightarrow \tan \theta  -1 = 0

\Rightarrow \tan \theta  = 1 

Therefore \cot \theta  = 1

Hence \tan^7 \theta  + \cot^7 \theta = 1^7 + 1^7 = 2

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Question 4: if x = 30^o , verify that

Answers:

i) \sin 3x = 3 \sin x - 4 \sin^3 x

Given x = 30^o

Therefore \sin x = \frac{1}{2} ;  \sin 3x = \sin 90^o = 1

Therefore LHS = 3 \sin x - 4 \sin^3 x

= 3 \Big(\frac{1}{2}\Big) - 4\Big(\frac{1}{2}\Big)^3

= \frac{3}{2} - \frac{1}{2} = 1

Therefore RHS = LHS. Hence proved.

ii) \cos 3x = 4 \cos^3 x - 3 \cos x

Given x = 30^o

Therefore \cos x = \frac{\sqrt{3}}{2} ;  \cos 3x = \cos 90^o = 0

RHS = 4 \Big( \frac{\sqrt{3}}{2} \Big)^3 - 3 \Big( \frac{\sqrt{3}}{2} \Big) = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2}

LHS = \cos 90^o = 0

Therefore  LHS = RHS.  Hence proved.

iii) \tan 2x = \frac{2 \tan x}{1 - \tan^2 x}

LHS = \tan 2x = \tan 60^o = \sqrt{3}

RHS = \frac{2 \tan x}{1 - \tan^2 x} = \frac{2 \tan 30^o}{1 - \tan^2 30^o}

= \frac{2. \frac{1}{\sqrt{3}}}{1- \frac{1}{3}} = \sqrt{3}

iv) \sin x = \sqrt{ \frac{1- \cos 2x}{2} }

LHS = \sin x = \sin 30^o = \frac{1}{2}

RHS = \sqrt{ \frac{1- \cos 2x}{2} } = \sqrt{ \frac{1- \cos 60^o}{2} } = \sqrt{ \frac{1- \frac{1}{2} }{2} } = \frac{1}{2}

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Question 5: Find an acute angle \theta when \frac{\cos \theta - \sin \theta}{ \cos \theta + \sin \theta} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}}

Answers:

Given \frac{\cos \theta - \sin \theta}{ \cos \theta + \sin \theta} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}}

Applying componendo and dividendo we get

\frac{(\cos \theta - \sin \theta) + (\cos \theta + \sin \theta)}{(\cos \theta - \sin \theta) - (\cos \theta + \sin \theta)} = \frac{(1 - \sqrt{3}) + (1 + \sqrt{3})}{(1 - \sqrt{3}) - (1 + \sqrt{3})}

\frac{2 \cos \theta}{-2 \sin \theta} = \frac{2}{2\sqrt{3}} 

\Rightarrow \cot \theta = \frac{1}{\sqrt{3}} 

\Rightarrow \tan \theta = \sqrt{3} = \tan 60^o

\Rightarrow \theta = 60^o

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Question 6: If  \sin (A+B) = 1 and \cos (A-B) = \frac{\sqrt{3}}{2} , 0^o < A + B \leq 90^o, A > B , then find A, B .

Answers:

Given \sin (A+B) = 1

\Rightarrow \sin (A+B) = \sin 90^o

\Rightarrow  A + B = 90^o … … … … … i)

Similarly, \cos (A-B) = \frac{\sqrt{3}}{2} 

\Rightarrow \cos (A-B) = \cos 30^o

\Rightarrow A - B = 30^o   … … … … … ii)

Solving i) and ii) we get

A = 60^o and B = 30^o

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Question 7: ABC is a right triangle, right angled at C . If A = 30^o and AB = 40 units, find remaining two sides and \angle B in \triangle ABC .p5

Answers:

Given \angle A = 30^o and \angle C = 90^o

Since  \angle A + \angle B + \angle C = 180^o

\Rightarrow \angle B = 60^o

\frac{BC}{AB} = \sin 30^o

\Rightarrow BC = 40 \times \frac{1}{2} = 20 units

Similarly, \frac{AC}{40} = \cos 30^o

\Rightarrow AC = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} units

Hence AC = 20\sqrt{3} units, BC = 20 units and \angle B = 60^o  

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Question 8: A rhombus of side 20 cm has two angles of 60^o each. Find the length of the diagonals.p6

Answers:

Given ABCD is a rhombus. 

\Rightarrow AB = BC = CD = DA = 20 cm

Property of rhombus: Diagonals are perpendicular bisectors and AC and BD are bisectors of \angle A and \angle B

\cos \angle BAO = \frac{OA}{AB} 

\cos 30^o = \frac{OA}{20} \Rightarrow OA = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}

Similarly, \sin \angle BAO = \frac{OB}{AB} 

\sin 30^o = \frac{OB}{20} \Rightarrow OB = \frac{20}{2} = 10

Therefore AC = 2 \times OA = 20\sqrt{3} cm and DB = 2 \times OB = 20 cm

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Question 9: An equilateral triangle is inscribed in a circle of radius $latex 6 cm. Find the side.p4

Answers:

Given OA = OB = OC = 6 cm

OD \perp BC , then D is the mid point of BC. OB and OC are bisectors of \angle B and \angle C .

therefore \angle OBD = 30^o

In \triangle OBD, \angle D = 90^o we have

\angle OBD = 30^o and OB = 6 cm

Therefore \cos \angle OBD = \frac{BD}{OB}

\Rightarrow \cos 30^o = \frac{BD}{OB}

\Rightarrow \cos 30^o = \frac{BD}{6}

\Rightarrow BD = 6 \times \frac{\sqrt{3}}{2}  = 3\sqrt{3} cm

\Rightarrow BC = 2 BD = 2 \times 3\sqrt{3}  = 6\sqrt{3} cm

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Question 10: If each of \alpha, \beta and \gamma are positive acute angle, such that \sin (\alpha + \beta - \gamma) = \frac{1}{2} ,  \cos (\beta + \gamma - \alpha) = \frac{1}{2} and \tan (\gamma + \alpha - \beta) = 1 , find the value of \alpha, \beta and \gamma

Answers:

Given \sin (\alpha + \beta - \gamma) = \frac{1}{2}

\Rightarrow \sin (\alpha + \beta - \gamma) = \sin 30^o

\Rightarrow  \alpha + \beta - \gamma = 30^o … … … … … i)

Similarly, \cos (\beta + \gamma - \alpha) = \frac{1}{2}

\Rightarrow   \cos (\beta + \gamma - \alpha) = \cos 60^o

\Rightarrow  \beta + \gamma - \alpha = 60^o … … … … … ii)

and \tan (\gamma + \alpha - \beta) = 1

\Rightarrow \tan (\gamma + \alpha - \beta) = \tan 45^o

\Rightarrow \gamma + \alpha - \beta = 45^o   … … … … … iii)

Adding i) , ii) and iii) we get 

\alpha + \beta + \gamma = 135^o … … … … … iv)

Now  iv) – i) \Rightarrow 2 \gamma = 105^o \Rightarrow \gamma = 52\frac{1}{2}^o

          iv) – ii)  \Rightarrow 2 \alpha = 75^o \Rightarrow \alpha = 37 \frac{1}{2}^o

         iv) – i) \Rightarrow 2 \beta = 90^o \Rightarrow \beta = 45^o

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Question 11: In an acute angles triangle ABC , if \tan (A + B - C) = 1 and \sec (B + C - A) = 2 , find the value of A, B and C .

Answers:

Given \tan (A + B - C) = 1

\Rightarrow \tan (A + B - C) = \tan 45^o

\Rightarrow A + B - C = 45^o … … … … … i)

Also \sec (B + C - A) = 2

\Rightarrow \sec B +C - A = \sec 60^o

\Rightarrow B + C - A = 60^o   … … … … … ii)

Adding i) and ii) we get 2 B = 105^o \Rightarrow B = 52\frac{1}{2}^o

Substituting in ii) we get

52\frac{1}{2}^o + C - A = 60^o \Rightarrow C - A = 7\frac{1}{2}^o    … … … … … iii)

We know A + B + C = 180^o

\Rightarrow A + 52\frac{1}{2}^o + C = 180^o

\Rightarrow A + C = 127\frac{1}{2}^o   … … … … … iv)

Adding iii) and iv) we get

2C = 135^o \Rightarrow C = 67\frac{1}{2}^o

Therefore A = 127\frac{1}{2}^o - 67\frac{1}{2}^o = 60^o

Hence A = 60^o, B = 52\frac{1}{2}^o and C = 67\frac{1}{2}^o

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Question 12: Evaluate

\sin^2 30^o \cos 45^o + 4 \tan^2 30^o + \frac{1}{2} \sin^2 90^o - 2 \cos^2 90^o + \frac{1}{24} \cos^2 0^o

Answers:

\sin^2 30^o \cos 45^o + 4 \tan^2 30^o + \frac{1}{2} \sin^2 90^o - 2 \cos^2 90^o + \frac{1}{24} \cos^2 0^o

= \Big( \frac{1}{2} \Big)^2 \times \Big( \frac{1}{\sqrt{2}} \Big)^2 + 4 \times \Big( \frac{1}{\sqrt{3}} \Big)^2 + \frac{1}{2} (1) - 2 (0)^2 + \frac{1}{24} (1)^2

= \frac{1}{4} \times \frac{1}{2} + \frac{4}{3} + \frac{1}{2} + \frac{1}{24}

= \frac{1}{8} + \frac{4}{3} + \frac{1}{2} + \frac{1}{24}  

= \frac{3+32+12+1}{24} 

= \frac{48}{24} = 2

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Question 13: Evaluate: \cot^2 30^o - 2 \cos^2 60^o - \frac{3}{4} \sec^2 45^o - 4 \sec^2 30^o

Answers:

\cot^2 30^o - 2 \cos^2 60^o - \frac{3}{4} \sec^2 45^o - 4 \sec^2 30^o

= (\sqrt{3})^2 - 2 \Big( \frac{1}{2} \Big)^2 - \frac{3}{4} (\sqrt{2})^2 - 4\Big( \frac{2}{\sqrt{3}} \Big)^2 

= 3 - \frac{1}{2} - \frac{3}{2} - \frac{16}{3}

= \frac{18-3-9-32}{6} = \frac{-26}{6} = \frac{-13}{3}

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Question 14: Find the value of x i) 2 \sin 3x = \sqrt{3}    ii) \sqrt{3} \sin x =\cos x

Answers:

i) 2 \sin 3x = \sqrt{3}

\Rightarrow \sin 3x = \frac{\sqrt{3}}{2} 

\Rightarrow \sin 3x = \sin 60^o

\Rightarrow 3x = 60^o

\Rightarrow x = 20^o

ii) \sqrt{3} \sin x = \cos x

\Rightarrow \tan x = \frac{1}{\sqrt{3}} 

\Rightarrow \tan x = \tan 30^o

\Rightarrow x = 30^o

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Question 15: If A = B = 60^o , verify

i) \cos (A - B) = \cos A \cos B + \sin A \sin B

ii) \sin (A - B) = \sin A \cos B - \cos A \sin B

iii) \tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

Answers:

Given A = B = 60^o   we have

\sin A =   \frac{\sqrt{3}}{2} ;     \cos A =   \frac{1}{2}

\sin B =   \frac{\sqrt{3}}{2} ;     \cos B =   \frac{1}{2}

\cos 0^o = 1 ;    \sin 0^o = 0

i) LHS = \cos (A - B) = \cos (60^o-60^o) = \cos 0^o = 1

    RHS = \cos A \cos B + \sin A \sin B =  \frac{1}{2} \times  \frac{1}{2} +  \frac{\sqrt{3}}{2} \times  \frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{4} = 1

    Therefore LHS = RHS. Hence proved.

ii) LHS = \sin (A - B) = \sin (60^o-60^o) = \sin 0^o = 0

     RHS =  \sin A \cos B - \cos A \sin B =  \frac{\sqrt{3}}{2} \times  \frac{1}{2} -  \frac{\sqrt{3}}{2} \times  \frac{1}{2} = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0

     Therefore LHS = RHS. Hence proved.

iii) LHS = \tan (A-B) = \tan 0^o = 0

      RHS =  \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{\sqrt{3} - \sqrt{3}}{1 + \sqrt{3} \times \sqrt{3}} = 0

     Therefore LHS = RHS. Hence proved.

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Question 16: If \sin (A - B) = \sin A \cos B - \cos A \sin B and \cos (A-B) = \cos A \cos B + \sin A \sin B , find the value of  \sin 15^o and \cos 15^o

Answers:

i) \sin 15 = \sin (45-30) = \sin 45 \cos 30 - \cos 45 \sin 30

= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}}

ii) \cos 15 = \cos (45-30) = \cos 45 \cos 30 + \sin 45 \sin 30

=  \frac{1}{\sqrt{2}} \times  \frac{\sqrt{3}}{2} +  \frac{1}{\sqrt{2}} \times  \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}

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Question 17: In a right angled triangle, right angled at C , if  \angle B = 60^o , AB = 15 units, find the remaining angles and sides.p2.jpg

Answers:

Given \angle C = 90^o and \angle B = 60^o

Therefore \angle A = 180^o - 90^o - 60^o = 30^o

\frac{BC}{15} = \sin 30^o \Rightarrow BC = \frac{1}{2} \times 15 = 7 \frac{1}{2} units

\frac{AC}{15} = \cos 30^o \Rightarrow AC = \frac{\sqrt{3}}{2} \times 15 units

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Question 18: If \sin (A + B)  = 1 and \cos (A - B) = 1, 0^o < A + B \leq 90^o and A \geq B , find A and B .

Answers:

Given \sin (A+B) = 1

\Rightarrow \sin (A + B) = \sin 90

\Rightarrow A + B = 90^o … … … … … i)

Similarly, \cos (A - B) = 1

\Rightarrow \cos (A - B) = \cos 0^o

\Rightarrow A = B   … … … … … ii)

Solving i) and ii) A = 45^o and B = 45^o

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Question 19: If \tan (A-B) = \frac{1}{\sqrt{3}} and \tan (A + B) = \sqrt{3},  0^o < A + B \leq 90^o and A > B . Find A and B .

Answers:

Given \tan (A-B) = \frac{1}{\sqrt{3}}

\Rightarrow \tan (A-B) = \tan 30^o … … … … … i)

\Rightarrow A - B = 30^o

Also \tan (A+B) = \sqrt{3}

\Rightarrow \tan (A+B) = \tan 60^o  

\Rightarrow A + B = 60^o … … … … … ii)

Solving i) and ii) A = 45^o and B = 15^o

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Question 20: In \triangle ABC , right angled at B, \angle A = \angle C . Find values of i) \sin A \cos C + \cos A \sin C  ii) \sin A \sin B + \cos A \cos B 

Answers:p3

We have A = C = 45^o ( \ since \  \angle B = 90^o)

\Rightarrow \sin A = \sin C = \cos A = \cos C = \frac{1}{\sqrt{2}}

 

i) \sin A \cos C + \cos A \sin C =  \frac{1}{\sqrt{2}}  +  \times  \frac{1}{\sqrt{2}} +  +   \frac{1}{\sqrt{2}} +  \times  \frac{1}{\sqrt{2}} +  = \frac{1}{2} +  + \frac{1}{2} +  = 1

ii) \sin A \sin B + \cos A \cos B =  \frac{1}{\sqrt{2}} +  \times 1 +  \frac{1}{\sqrt{2}} +  \times 0 =  \frac{1}{\sqrt{2}}

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Question 21: If A and B are acute angles such that tan A = \frac{1}{2} and tan B = \frac{1}{3} and tan A = \frac{tan A + tan B}{1 - tan A tan B} . Find A + B .

Answers:

RHS = \frac{tan A + tan B}{1 - tan A tan B} = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}} = \frac{5}{5} = 1

Therefore \tan (A+B) = 1 = \tan 45^o \Rightarrow A = 45^o

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Question 22: In adjoining figure ABC is a right angled triangle at B and \triangle ABD is a right angled triangle at A . If BD \perp AC and BC = 2\sqrt{3} , find AD .p1

Answers:

\frac{2\sqrt{3}}{AB} = \tan 30^o = \frac{2\sqrt{3}}{ (\frac{1}{\sqrt{3}})} = 6

\angle CAB + 90 + \angle DBA = 180 \Rightarrow \angle DBA = 180^o - 30^o - 90^o = 60^o

Therefore \frac{AD}{AB} = \tan 60^o \Rightarrow AD = 6 \times \sqrt{3} cm

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