Question 1: Evaluate  each of the following:

Answer:

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

iv) \sin 30^o \sin 45^o + \cos 30^o \cos 45^o = \frac{1}{2} \times \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \times \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 \times (\frac{1}{2} )^2 + ( \frac{1}{\sqrt{2}} )^2 - 4 ( \frac{1}{\sqrt{3}} )^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{5}{6} (2 - \sqrt{3})

\\

Question 2: Find the value of \theta in each of the following:

Answer:

i) 2 \sin \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 = \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 3 \theta = \tan 60^o

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

\\

Question 3: Evaluate / Prove the following:

Answer:

i) \frac{\cos 37^o}{\sin 53^o} = \frac{\cos (90^o-53^o)}{\sin 53^o} = \frac{\sin 53^o}{\sin 53^o} = 1

ii) \frac{\tan 54^o}{\cot 36^o} = \frac{\tan (90^o-36^o)}{\cot 36^o} = \frac{\cot 36^o}{\cot 36^o} = 1

iii) \sin 39^o - \cos 51^o = \sin 39^o - \cos (90^o-39^o) = \sin 39^o - \sin 39^o = 0

iv) \cot 34^o - \tan 56^o = \cot 34^o - \tan (90^o-34^o) = \cot 34^o - \cot 34^o = 0

v) \frac{\cos 80^o}{\sin 10^o} + \cos 59^o \ cosec \ 31^o = \frac{\cos (90^o-10^o)}{\sin 10^o} + \cos 59^o \ cosec \ (90^o-59^o)

= \frac{\sin 10^o}{\sin 10^o} + \cos 59^o \sec 59^o = 1 + 1 = 2

vi) \sec 50^o \sin 40^o + \cos 40^o \ cosec \ 50^o

= \sec 50^o \sin (90^o - 50^o) + \cos 40^o \ cosec \ (90^o-40^o)

= \sec 50^o \cos 50^o + \cos 40^o \sec 40^o = 1 + 1 = 2

vii) \Big( \frac{\sin 35^o}{\cos 55^o} \Big)^2 + \Big( \frac{\cos 55^o}{\sin 35^o} \Big)^2 - 2 \cos 60^o

= \Big( \frac{\sin (90^o-55^o)}{\cos 55^o} \Big)^2 + \Big( \frac{\cos (90^o-35^o)}{\sin 35^o} \Big)^2 - 2 \cos 60^o

= \Big( \frac{\cos 55^o}{\cos 55^o} \Big)^2 + \Big( \frac{\sin 35^o}{\sin 35^o} \Big)^2 - 2 \times \frac{1}{2}

= 1 + 1 - 1 = 1

vii)  \cos (40^o - \theta ) - \sin ( 50^o + \theta) + \frac{\cos^2 40^o + \cos^2 50^o}{\sin^2 40^o + sin^2 50^o}

= \sin (90 - (40^o - \theta) ) - \sin ( 50^o + \theta) + \frac{\cos^2 40^o + \cos^2 (90-40^o)}{\sin^2 40^o + sin^2 (90-40^o)}

= \sin (50^o + \theta ) - \sin ( 50^o + \theta) + \frac{\cos^2 40^o + \sin^2 40^o}{\sin^2 40^o + \cos^2 40^o}

= 0 + 1 = 1

viii) \cot 12^o \cot 38^o \cot 52^o \cot 60^o \cot 78^o

= \cot (90^o-78^o) \cot (90^o-52^o) \cot 52^o \cot 60^o \cot 78^o

= \tan 78^o \tan 52^o \cot 52^o \cot 60^o \cot 78^o

= \cot 60^o = \frac{1}{\sqrt{3}}

ix)  \tan 5^o \tan 25^o \tan 30^o \tan 65^o \tan 85^o

= \tan (90^o-85^o) \tan (90^o-65^o) \tan 30^o \tan 65^o \tan 85^o

= \cot 85^o \cot 65^o \tan 30^o \tan 65^o \tan 85^o

= \tan 30^o = \frac{1}{\sqrt{3}}

\\

Question 4: Express each of the following in terms of trigonometric ratios of angles between 0 and 45^o .

Answer:

i) \sin 85^o + \ cosec \  85^o = \sin (90^o-5^o) + \ cosec \ (90^o-5^o) = \cos 5^o + \sec 5^o

ii) cosec \ 69^o + \cot  69^o  = \ cosec \ (90^o-21^o) + \cot (90^o-21^o) = \sec 21^o + \tan 21^o

\\

Question 5: Prove that:

i) \tan 1^o \tan 2^o \tan 3^o \cdots \tan 89^o = 1

ii) \cos 1^o \cos 2^o \cos 3^o \cdots \cos 180^o  = 0

Answer:

i) LHS = \tan 1^o \tan 2^o \tan 3^o \cdots \tan 89^o

= \tan (90^o-89^o) \tan (90^o-88^o) \tan (90^o-87^o) \cdots \tan 87^o \tan 88^o \tan 89^o

= \cot 89^o \cot 88^o \cot 87^o \cdots  \tan 87^o \tan 88^o \tan 89^o

= 1 \times 1 \times 1 \cdots \times 1 = 1 = RHS

Hence Proved.

ii) LHS =  \cos 1^o \cos 2^o \cos 3^o \cdots \cos 180^o

= \cos 1^o \cos 2^o \cos 3^o \cdots \cos 89^o \cos 90^o \cos 91^o  \cdots \cos 179^o \cos 180^o

= \cos 1^o \cos 2^o \cos 3^o \cdots \cos 89^o (0) \cos 91^o  \cdots \cos 179^o \cos 180^o

= 0 = RHS.

Hence Proved.

\\

Question 6: If A + B = 90^o , prove that: 

\sqrt{ \frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A} } = \tan A

Answer:

Given A + B = 90^o \Rightarrow B = 90^o - A

LHS = \sqrt{ \frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A} }  

= \sqrt{ \frac{\tan A \tan (90^o - A) + \tan A \cot (90^o - A)}{\sin A \sec (90^o - A)} - \frac{\sin^2 (90^o - A)}{\cos^2 A} }  

= \sqrt{ \frac{\tan A \cot A + \tan^2 A }{\sin A \ cosec A} - \frac{\cos^2 A}{\cos^2 A} } 

= \sqrt{1 + \tan^2 A - 1} = \sqrt{\tan^2 A} = \tan A =  RHS

Hence Proved. 

\\

Question 7: If A, \ B and C are interior angles of triangle ABC, prove that

\tan \big( \frac{B+C}{2} \Big) = \cot \Big( \frac{A}{2} \Big)

Answer:

Given A + B + C = 180^o

\Rightarrow B + C = 180^o - A

\Rightarrow  \big( \frac{B+C}{2} \Big) = \Big( 90^o - \frac{A}{2} \Big)

\Rightarrow \tan \big( \frac{B+C}{2} \Big) = \tan \Big( 90^o - \frac{A}{2} \Big)

\Rightarrow \tan \big( \frac{B+C}{2} \Big) = \cot \Big( \frac{A}{2} \Big)

\\

Question 8: Find \theta if  \sin (\theta + 36^o) = \cos \theta when \theta + 36^o is an acute angle.

Answer:

Given \sin (\theta + 36^o) = \cos \theta  

\Rightarrow \cos [90^o - (\theta + 36^o) ] = \cos \theta

\Rightarrow [ (\theta + 36^o) ] = \theta

\Rightarrow 2 \theta = 54^o

\Rightarrow \theta = 27^o

\\

Question 9: If \tan 2 \theta = \cot (\theta + 6^o) , when 2 \theta and \theta+6^o are acute angles, find the value of  \theta .

Answer:

Given \tan 2 \theta = \cot (\theta + 6^o)

\Rightarrow \cot (90^o- 2 \theta) = \cot (\theta+6^o)

\Rightarrow 90^o - 2 \theta = \theta + 6^o

\Rightarrow 84^o = 3 \theta

\Rightarrow \theta = 28^o

\\

Question 10: If A, B and C are interior angles of \triangle ABC , show that

i) \sin \Big( \frac{B+C}{2} \Big) = \cos \Big( \frac{A}{2} \Big)    ii) \cos \Big( \frac{B+C}{2} \Big) = \sin \Big( \frac{A}{2}  \Big)

Answer:

i) Given A + B + C = 180^o

\Rightarrow B + C = 180^o - A

\Rightarrow \Big( \frac{B+C}{2}   \Big) = \Big( 90^o - \frac{A}{2}  \Big)

\Rightarrow \sin \Big( \frac{B+C}{2}   \Big) = \sin \Big(90^o - \frac{A}{2}  \Big) 

\Rightarrow \sin \Big( \frac{B+C}{2}   \Big) = \cos \Big( \frac{A}{2}  \Big) 

ii) Similarly,

\Rightarrow \cos \Big( \frac{B+C}{2}  \Big) = \cos \Big( 90^o - \frac{A}{2}   \Big) 

\Rightarrow \cos \Big( \frac{B+C}{2}  \Big) = \sin \Big( \frac{A}{2}  \Big) 

\\

Question 11: Find the value of \theta for

i) \cos 2\theta = \sin 4\theta where 2\theta, 4\theta < 90^o

ii) \sin 3\theta = \cos (\theta - 6^o) where 3\theta, (\theta-6^o) < 90^o

Answer:

i) \cos 2\theta = \sin 4\theta

\Rightarrow \cos 2\theta = \cos (90^o - 4\theta)

\Rightarrow 2\theta = 90^o- 4\theta

\Rightarrow 6\theta = 90^o

\Rightarrow \theta = 15^o

ii) \sin 3\theta = \cos (\theta - 6^o)

\Rightarrow \sin 3\theta = \sin [90^o - (\theta-6)]

\Rightarrow 3\theta = 90^o - (\theta - 6^o)

\Rightarrow 3\theta = 90^o - \theta +6^o

\Rightarrow 4\theta = 96^o \Rightarrow \theta = 19^o

\\

Question 12: Prove the following:

i) \tan 20^o \tan 35^o \tan 45^o \tan 55^o \tan 70^o = 1

ii) \frac{\cos 80^o}{\sin 10^o} + \cos 59^o \ cosec \ 31^o = 2

iii) \frac{\cos (90^o - A) \sec (90^o-A) \tan A}{cosec \ (90^o - A) \sin (90^o - A) \cot (90^o- A)} + \frac{\tan (90^o-A)}{\cot A} = 2

Answer:

i) LHS = \tan 20^o \tan 35^o \tan 45^o \tan 55^o \tan 70^o

= \tan (90^o-70^o) \tan (90^o-55^o) \tan 45^o \tan 55^o \tan 70^o

= \cot 70^o \cot 55^o \tan 45^o \tan 55^o \tan 70^o

= \tan 45^o = 1 = RHS

Hence Proved.

ii) LHS = \frac{\cos 80^o}{\sin 10^o} + \cos 59^o \ cosec \ 31^o 

=  \frac{\cos (90^o-10^o)}{\sin 10^o} + \cos 59^o \ cosec \ (90^o- 59^o)

=  \frac{\sin 10^o}{\sin 10^o} + \cos 59^o \sec 59^o

= 1 + 1 = 2 = RHS

Hence Proved.  

iii) LHS =  \frac{\cos (90^o - A) \sec (90^o-A) \tan A}{cosec \ (90^o - A) \sin (90^o - A) \cot (90^o- A)} + \frac{\tan (90^o-A)}{\cot A} 

= \frac{\sin A \ cosec \ A \tan A}{\sec A \cos A \tan A} + \frac{\cot A}{ \cot A} 

= 1 + 1 = 2

\\

Question 13: What is the max value of i) \frac{1}{\sec A} and ii) \frac{1}{cosec \ A}

Answer:

i) \frac{1}{\sec A} = \cos A \Rightarrow max \ value = 1

ii) \frac{1}{cosec \ A} = \sin A \Rightarrow max \ value = 1

\\

Question 14: If \tan A = \frac{4}{5} , find the value of \frac{\cos A - \sin A}{\cos A + \sin A}

Answer:

\frac{\cos A - \sin A}{\cos A + \sin A} = \frac{1 - \tan A}{1 + \tan A} = \frac{1 - \frac{4}{5}}{1 + \frac{4}{5}} = \frac{1}{9}

\\

Question 15: If \tan A = \frac{1}{\sqrt{5}} , what is the value of \frac{cosec^2 \ A - \sec^2 A}{cosec^2 \ A + \sec^2 A}

Answer:

\frac{cosec^2 \ A - \sec^2 A}{cosec^2 \ A + \sec^2 A}

=  \frac{\frac{1}{\sin^2 A} - \frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A} + \frac{1}{\cos^2 A}}

= \frac{\cos^2 A - \sin^2 A}{\cos^2 A + \sin^2 A}

= \frac{1 - \tan^2 A}{1 + \tan^2 A}

= \frac{1 - (\frac{1}{\sqrt{5}})^2}{1 + (\frac{1}{\sqrt{5}})^2}

= \frac{4}{6} = \frac{2}{3}

\\

Advertisements