Question 1: In a right angles triangle ABC , right angled at B , if \sin A = \frac{3}{5} , find all the six trigonometric ratios of \angle C .

Answer:2018-10-26_20-52-02

Given \sin A = \frac{3}{5}

\Rightarrow AB^2 + BC^2 = AC^2

\Rightarrow AB^2 = AC^2 - BC^2

\Rightarrow AB^2 = 5^2 - 3^2 = 16

\Rightarrow AB = 4

Therefore,

\sin C = \frac{Perpendicular}{Hypotenuse} = \frac{4}{5} ;      cosec \ C = \frac{Hypotenuse}{Perpendicular} = \frac{5}{4}

\cos C = \frac{Base}{Hypotenuse} = \frac{3}{5} ;      \sec C = \frac{Hypotenuse}{Base} = \frac{5}{3}

\tan C = \frac{Perpendicular}{Base} = \frac{4}{3} ;      \cot C = \frac{Base}{Perpendicular} = \frac{3}{4} 

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Question 2: If  \sin A = \frac{a^2 - b^2}{a^2 + b^2} , find the value of the other five trigonometric ratios.2018-10-26_21-00-45.jpg

Answer:

Given \sin A = \frac{a^2 - b^2}{a^2 + b^2}

By Pythagoras theorem, we have

AB^2 = AC^2 = BC^2 = (a^2 + b^2)^2 - (a^2 - b^2)^2 = 4a^2b^2

\Rightarrow AB = 2ab

\cos A = \frac{2ab}{a^2 + b^2} ;      \sec A = \frac{a^2 + b^2}{2ab}

\tan A = \frac{a^2 - b^2}{2ab} ;      \cot A = \frac{2ab}{a^2 - b^2}

\sin A = \frac{a^2 - b^2}{a^2 + b^2} ;      cosec \ A = \frac{a^2 + b^2}{a^2 - b^2}

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Question 3: If cosec \ A = 2 , find the value of \frac{1}{\tan A} + \frac{\sin A}{1+ \cos A} 2018-10-26_20-58-40.jpg

Answer:

Given cosec \ A = 2 , therefore by Pythagoras theorem, AB = \sqrt{3}

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

Hence \frac{1}{\tan A} + \frac{\sin A}{1+ \cos A} = \sqrt{3} + \frac{1}{2} (\frac{2}{2+\sqrt{3}}) 

= \sqrt{3} + \frac{1}{2+\sqrt{3}} = \frac{2\sqrt{3}+ 3 + 1}{2 + \sqrt{3}} = \frac{2(2+\sqrt{3})}{(2+\sqrt{3})} = 2

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Question 4: If \tan A = \sqrt{2}-1 , show that \sin A \cos A = \frac{\sqrt{2}}{4} 

Answer:2018-10-26_21-02-55.jpg

Given \tan A = \sqrt{2}-1

By Pythagoras theorem, we have

AC = \sqrt{(\sqrt{2}-1)^2 + 1^2} = \sqrt{4 - 2\sqrt{2}}

\therefore \sin A = \frac{\sqrt{2}-1}{\sqrt{4 - 2\sqrt{2}}} and \cos A = \frac{1}{\sqrt{4 - 2\sqrt{2}}} 

Hence, \sin A  \cos A =  \frac{\sqrt{2}-1}{\sqrt{4 - 2\sqrt{2}}} \times \frac{1}{\sqrt{4 - 2\sqrt{2}}} = \frac{\sqrt{2}-1}{(4 - 2\sqrt{2})} 

= \frac{(\sqrt{2}-1)}{2\sqrt{2}(\sqrt{2}-1)} = \frac{1}{2\sqrt{2}}

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Question 5: In \triangle ABC , right angles at C , if  \tan A = \frac{1}{\sqrt{3}} and \tan B = \sqrt{3} , show that \sin A \cos B + \cos A  \sin B = 1 2018-10-26_20-58-40

Answer:

Given \tan A = \frac{1}{\sqrt{3}}

By Pythagoras theorem, we have

AB = \sqrt{ (\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2

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

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

\therefore \sin A \cos B + \cos 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.

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Question 6: If \cot B = \frac{12}{5} , prove that \tan^2 B - \sin^2 B = \sin^4 B \sec^2 B 2018-10-26_21-05-04.jpg

Answer:

Given \cot B = \frac{12}{5}

By Pythagoras theorem, we have

AB = \sqrt{12^2 + 5^2 } = 13

\therefore \tan B = \frac{5}{12} ;    \sin B = \frac{5}{13} ;    \sec B =  \frac{13}{12}

LHS = \tan^2 B - \sin^2 B = ( \frac{5}{12} )^2- ( \frac{5}{13} )^2 = \frac{5^4}{12^2 \times 13^2}

RHS = \sin^4 B \sec^2 B = ( \frac{5}{13} )^4 \times ( \frac{13}{12} )^2 =  \frac{5^4}{12^2 \times 13^2}

Therefore LHS = RHS. Hence proved.

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Question 7: In the given figure, AD = DB and \angle B = 90^o . Determine (i) \sin \theta    ii) \cos \theta    iii) \tan \theta    iv) \sin^2 \theta + \cos^2  \theta 2018-10-26_21-27-15

Answer:

By Pythagoras theorem, we have

CB = \sqrt{b^2-a^2}

CD = \sqrt{(\frac{a}{2})^2 + (\sqrt{b^2-a^2})^2} = \frac{\sqrt{4b^2-3a^2}}{2}

Therefore:

i) \sin \theta = \frac{BD}{CD} = \frac{\frac{a}{2}}{\frac{\sqrt{4b^2-3a^2}}{2}} = \frac{a}{\sqrt{4b^2-3a^2}}

ii) \cos \theta = \frac{BC}{CD} = \frac{\sqrt{b^2-a^2}}{\frac{\sqrt{4b^2-3a^2}}{2}} =  \frac{2\sqrt{b^2-a^2}}{\sqrt{4b^2-3a^2}}

iii) \tan \theta = \frac{BD}{BC} = \frac{\frac{a}{2}}{\sqrt{b^2-a^2}} = \frac{a}{2\sqrt{b^2-a^2}}

iv) \sin^2 \theta + \cos^2  \theta = ( \frac{a}{\sqrt{4b^2-3a^2}} )^2 + ( \frac{2\sqrt{b^2-a^2}}{\sqrt{4b^2-3a^2}} )^2 = \frac{a^2}{4b^2-3a^2} + \frac{4(b^2-a^2)}{4b^2-3a^2} = 1

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Question 8: In a \triangle ABC , right angles at C and \angle A = \angle B ,  i) is \cos A = \cos B ii) is \tan A = \tan B

Answer:

Given \angle C = 90^o

\Rightarrow \angle A = \angle B = 45^o

i) \cos A = \cos B = \cos 45^o = \frac{1}{\sqrt{2}}  

ii) \tan A = \tan B = \tan 45^o = 1

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Question 9: If 5 \tan A = 4 , show that \frac{5 \sin A - 3 \cos A}{5 \sin A + 2 \cos A} = \frac{1}{6}

Answer:

LHS =  \frac{5 \sin A - 3 \cos A}{5 \sin A + 2 \cos A}  =  \frac{5 \tan A - 3}{5 \tan A + 2} =  \frac{5 .\frac{4}{5} - 3}{5 .\frac{4}{5} + 2} =  \frac{1}{6}

= RHS. Hence proved.

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Question 10: If  \tan A + \frac{1}{\tan A} = 2 , find the value of \tan^2 A + \frac{1}{\tan^2 A}

Answer:

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

Squaring on both sides,

(\tan A + \frac{1}{\tan A} )^2 = 4

\Rightarrow \tan^2 A + \frac{1}{\tan^2 A} + 2 = 4

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

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Question 11: If \cot A = \frac{7}{8} , evaluate i) \frac{(1+\sin A)(1- \sin A)}{(1+\cos A)(1-\cos A)} ii) \cot^2 A 2018-10-26_21-12-14.jpg

Answer:

i) \frac{(1+\sin A)(1- \sin A)}{(1+\cos A)(1-\cos A)}  = \frac{1-\sin^2 A}{1-\cos^2 A}  = \frac{\cos^2 A}{\sin^2 A}  = \cot^2 A = \frac{49}{64}

ii)  \cot^2 A= ( \frac{7}{8})^2 = \frac{49}{64}

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Question 12: If  3 \cot A = 4 , check if  \frac{1-\tan^2 A}{1+\tan^2 A} = \cos^2 A - \sin^2 A 2018-10-26_20-52-02

Answer:

Given \cot A = \frac{4}{3}  \Rightarrow \tan A = \frac{3}{4}

LHS =  \frac{1-\tan^2 A}{1+\tan^2 A} =  \frac{1-\frac{9}{16}}{1+\frac{9}{16}} = \frac{16-9}{16+9} = \frac{7}{25}

RHS = \cos^2 A - \sin^2 A = ( \frac{4}{5} )^2 - ( \frac{3}{5} )^2 = \frac{16-9}{25} = \frac{7}{25}

Therefore LHS = RHS. Hence proved.

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Question 13: If \tan A = \frac{a}{b} , find the value of  \frac{\cos A + \sin A}{\cos A - \sin A}

Answer:

Given \frac{\cos A + \sin A}{\cos A - \sin A}

Dividing both the numerator and denominator by \cos A we get

= \frac{1 + \tan A}{1 - \tan A} = \frac{1 + \frac{a}{b}}{1 - \frac{a}{b}} = \frac{b-a}{b+a}

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Question 14: If \cos A = \frac{12}{13} , find \sin A (1 - \tan A) = \frac{35}{156} 2018-10-26_21-14-31.jpg

Answer:

By Pythagoras theorem, BC = \sqrt{169 - 144} = 5

Therefore, \sin A = \frac{5}{13} and \tan A = \frac{5}{12}

Hence \sin A (1 - \tan A) =  \frac{5}{13} (1 -  \frac{5}{12} ) = \frac{35}{156}

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Question 15: If \sec A = \frac{5}{4} , find the value of \frac{\sin A - 2 \cos A}{\tan A - \cot A} 2018-10-26_20-52-02

Answer:

By Pythagoras theorem,

BC = \sqrt{5^2  - 4^2} = \sqrt{9} = 3

Therefore \sin A = \frac{3}{5} ;   \tan A = \frac{3}{4} ;    \cos A = \frac{4}{5} ;   \cot A = \frac{4}{3}

\Rightarrow \frac{\sin A - 2 \cos A}{\tan A - \cot A} =  \frac{\frac{3}{5} - 2 .\frac{4}{5}}{\frac{3}{4} -\frac{4}{3}}  = \frac{-5}{5} \times \frac{12}{-7} = \frac{12}{7}

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Question 16: If \sin A = \frac{3}{5} ; evaluate \frac{\cos A - \frac{1}{\tan A}}{2 \cot A} 2018-10-26_20-52-02

Answer:

By Pythagoras theorem we get

AC = \sqrt{4^2 + 3^2 } = \sqrt{25} = 5

Therefore      \cos A = \frac{4}{5} ;   \tan A = \frac{3}{4} ;    \cot A = \frac{4}{3}

Therefore      \frac{\cos A - \frac{1}{\tan A}}{2 \cot A} =  \frac{\frac{4}{5} - \frac{1}{\frac{3}{4}}}{2 . \frac{4}{3}} = \frac{(12-20) . 3}{15.(8)} = \frac{-8}{15} \times \frac{3}{8} = \frac{-1}{5}

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Question 17: If \sec A = \frac{5}{4} , verify \frac{3 \sin A - 4 \sin^3 A}{4 \cos^3 A - 3 \cos A}  = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A} 2018-10-26_20-52-02

Answer:

By Pythagoras theorem we get

BC = \sqrt{5^2 - 4^2} = \sqrt{9} = 3

Therefore      \sin A = \frac{3}{5} ;   \cos A = \frac{4}{5} ;    \tan A = \frac{3}{4}

LHS =  \frac{3 \sin A - 4 \sin^3 A}{4 \cos^3 A - 3 \cos A}

=  \frac{3 (\frac{3}{5}) - 4 (\frac{3}{5})^3}{4 (\frac{4}{5})^3 - 3 (\frac{4}{5})}  = \frac{\frac{9}{5} - \frac{108}{125}}{\frac{256}{125} - \frac{12}{15}} = \frac{225-108}{256-300} = \frac{-117}{44}

RHS = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A}

= \frac{3 . \frac{3}{4} - (\frac{3}{4})^3 }{1 - 3 (\frac{3}{4})^2}  = \frac{\frac{9}{4} - \frac{27}{64}}{1 - \frac{27}{16}} = \frac{(144-27) (16)}{(64 )(16-27)} = \frac{117 \times 16}{64 \times (-11)} = \frac{-117}{44}

Therefore LHS = RHS. Hence proved.

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Question 18: If \sin A = \frac{3}{4} , prove that \sqrt{\frac{cosec^2 \ A - \cot^2 A}{\sec^2 A -1}} = \frac{\sqrt{7}}{3}

Answer:2018-10-26_21-17-47.jpg

Given \sin A = \frac{3}{4}

By Pythagoras theorem, AB = \sqrt{4^2 - 3^2} = \sqrt{7}

Therefore \cos A = \frac{\sqrt{7}}{4} ;  \sec A =  \frac{4}{\sqrt{7}} ;  \tan A =   \frac{3}{\sqrt{7}} ;  \cot A =  \frac{\sqrt{7}}{3} ; cosec \ A = \frac{4}{3}

LHS = \sqrt{\frac{cosec^2 \ A - \cot^2 A}{\sec^2 A -1}} = \frac{\sqrt{7}}{3}

=  \sqrt{\frac{(\frac{4}{3})^2 - (\frac{\sqrt{7}}{3})^2}{(\frac{4}{\sqrt{7}})^2 -1}}  =  \sqrt{\frac{\frac{16}{9} - \frac{7}{9}}{\frac{16}{7} -1} }  =  \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3}

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Question 19: If 8  \tan A = 15 , find \sin A - \cos A 2018-10-26_21-18-58.jpg

Answer:

Given 8 \tan A = 15 \Rightarrow \tan A = \frac{15}{8}

By Pythagoras theorem, we get

AC = \sqrt{15^2 + 8^2} = \sqrt{289} = 17

Therefore \sin A = \frac{15}{17}   and \cos A = \frac{8}{17}

Hence \sin A - \cos A = \frac{15}{17} - \frac{5}{17} = \frac{7}{17}

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Question 20: If 3 \cos A - 4 \sin A = 2 \cos A + \sin A , find \tan A

Answer:

Given 3 \cos A - 4 \sin A = 2 \cos A + \sin A , find \tan A

\Rightarrow \cos A = 5 \sin A

\Rightarrow \frac{\sin A}{\cos A} = \frac{1}{5}

\Rightarrow \tan A = \frac{1}{5}

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Question 21: If \angle A and \angle B are acute angles such that \cos A = \cos B , then show that \angle A = \angle B

Answer:2018-10-26_21-20-22

\cos A = \frac{AC}{AB}

\cos B = \frac{BC}{AB}

Since \cos A = \cos B

\Rightarrow \frac{AC}{AB} = \frac{BC}{AB} 

\Rightarrow AC = BC

\therefore \angle A = \angle B (angles opposite equal sides of a triangle are equal)

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Question 22: In a triangle, \angle A and \angle B are acute angles such that \tan A = \tan B , show that \angle A = \angle B 2018-10-26_21-20-22

Answer:

\tan A = \frac{BC}{AC}  and \tan B = \frac{AC}{BC} 

Given that \tan A = \tan B

\Rightarrow \frac{BC}{AC} = \frac{AC}{BC}

\Rightarrow BC^2 = AC^2 

\Rightarrow BC = AC

\therefore \angle A = \angle B (angles opposite equal sides of a triangle are equal)

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Question 23: If A is an acute angle such that 3 \sin A = 4 \cos A , find the value of 4 \sin^2 A - 3 \cos^2 A + 2 2018-10-26_21-21-59

Answer:

Given 3 \sin A = 4 \cos A

\Rightarrow \tan A = \frac{4}{3}

Therefore AC = \sqrt{4^2+3^2} = 5

By Pythagoras theorem, \sin A = \frac{4}{5} and \cos A = \frac{3}{5}

Hence, 4 \sin^2 A - 3 \cos^2 A + 2

= 4 ( \frac{4}{5} )^2 - 3 ( \frac{3}{5} )^2 + 2  = \frac{64}{25} - \frac{18}{25} + 2   = \frac{64-18+50}{25} = \frac{96}{25}

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Question 24: In adjoining figure, \triangle ABC is right angles at B and D is the midpoint of BC . AC = 5 \ cm , BC = 4 \ cm and \angle BAD = \theta . Find i) \tan \theta   ii) \sin \theta   iii) \sin^2 \theta + \cos^2 \theta

Answer:2018-10-26_21-23-43

By Pythagoras theorem

AB = \sqrt{5^2 - 4^2} = 3

Therefore i) \tan \theta = \frac{2}{3}

ii) By Pythagoras theorem

AD = \sqrt{3^2 + 2^2} = \sqrt{13}

Therefore \sin \theta = \frac{2}{\sqrt{13}} and \cos \theta = \frac{3}{\sqrt{13}}

iii) \sin^2 \theta + \cos^2 \theta = \frac{4}{13} + \frac{9}{13} = 1

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