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

Answer:

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

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

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

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

\displaystyle \Rightarrow AB = 4

Therefore,

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

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

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

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\displaystyle \text{Question 2: If } \sin A = \frac{a^2 - b^2}{a^2 + b^2} \text{, find the value of the other five trigonometric ratios.}  

Answer:

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

By Pythagoras theorem, we have

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

\displaystyle \Rightarrow AB = 2ab

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

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

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

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\displaystyle \text{Question 3: If } \mathrm{cosec}  A = 2 \text{ ,  find the value of } \frac{1}{\tan A} + \frac{\sin A}{1+ \cos A}

Answer:

\displaystyle \text{Given } \mathrm{cosec}  A = 2 , therefore \displaystyle \text{By Pythagoras theorem,  } AB = \sqrt{3}

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

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

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

Answer:

\displaystyle \text{Given } \tan A = \sqrt{2}-1

By Pythagoras theorem, we have

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

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

\displaystyle \text{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})}  

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

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\displaystyle \text{Question 5: In } \triangle ABC \text{ , right angles at } C \text{ , if  } \tan A = \frac{1}{\sqrt{3}}  \text{ and }  \tan B = \sqrt{3} \\ \\ \text{ , show that } \sin A \cos B + \cos A \sin B = 1

Answer:

\displaystyle \text{Given } \tan A = \frac{1}{\sqrt{3}}  

By Pythagoras theorem, we have

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

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

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

\displaystyle \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

\displaystyle \therefore LHS = RHS.

Hence proved.

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\displaystyle \text{Question 6: If  } \cot B = \frac{12}{5} \text{ , prove that } \tan^2 B - \sin^2 B = \sin^4 B \sec^2 B

Answer:

\displaystyle \text{Given } \cot B = \frac{12}{5}  

By Pythagoras theorem, we have

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

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

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

\displaystyle \text{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, \displaystyle AD = DB  \text{ and }  \angle B = 90^o . Determine (\displaystyle \text{i) } \sin \theta \displaystyle \text{ii) } \cos \theta \displaystyle \text{iii) } \tan \theta \displaystyle \text{iv) } \sin^2 \theta + \cos^2 \theta

Answer:

By Pythagoras theorem, we have

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

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

Therefore:

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

\displaystyle \text{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}}  

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

\displaystyle \text{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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\displaystyle \text{Question 8: In a } \triangle ABC \text{ , right angles at }  C  \text{ and }  \angle A = \angle B ,

i) is \displaystyle \cos A = \cos B              ii) is \displaystyle \tan A = \tan B

Answer:

\displaystyle \text{Given } \angle C = 90^o

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

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

\displaystyle \text{ii) } \tan A = \tan B = \tan 45^o = 1

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

Answer:

\displaystyle \text{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}  

\displaystyle = RHS. Hence proved.

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

Answer:

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

Squaring on both sides,

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

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

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

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\displaystyle \text{Question 11: If } \cot A = \frac{7}{8} \text{ , evaluate }

\displaystyle \text{i) } \frac{(1+\sin A)(1- \sin A)}{(1+\cos A)(1-\cos A)}  \text{          ii) } \cot^2 A

Answer:

\displaystyle \text{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}  

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

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\displaystyle \text{Question 12: If } 3 \cot A = 4 \text{ , check if } \frac{1-\tan^2 A}{1+\tan^2 A} = \cos^2 A - \sin^2 A

Answer:

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

\displaystyle \text{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}  

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

Answer:

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

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

\displaystyle = \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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\displaystyle \text{Question 14: If } \cos A = \frac{12}{13} \text{ , find } \sin A (1 - \tan A) = \frac{35}{156}

Answer:

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

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

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

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\displaystyle \text{Question 15: If } \sec A = \frac{5}{4} \text{ , find the value of } \frac{\sin A - 2 \cos A}{\tan A - \cot A}

Answer:

By Pythagoras theorem,

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

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

\displaystyle \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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\displaystyle \text{Question 16: If  } \sin A = \frac{3}{5} \text{ ; evaluate} \frac{\cos A - \frac{1}{\tan A}}{2 \cot A}

Answer:

By Pythagoras theorem we get

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

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

\displaystyle \text{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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\displaystyle \text{Question 17: If } \sec A = \frac{5}{4} \text{ , 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}

Answer:

By Pythagoras theorem we get

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

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

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

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

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

Answer:

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

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

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

\displaystyle \text{LHS } = \sqrt{\frac{\mathrm{cosec} ^2 A - \cot^2 A}{\sec^2 A -1}} = \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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\displaystyle \text{Question 19: If } 8 \tan A = 15 \text{ , find } \sin A - \cos A

Answer:

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

By Pythagoras theorem, we get

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

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

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

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

Answer:

\displaystyle \text{Given } 3 \cos A - 4 \sin A = 2 \cos A + \sin A , find \displaystyle \tan A

\displaystyle \Rightarrow \cos A = 5 \sin A

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

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

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

Answer:

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

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

Since \displaystyle \cos A = \cos B

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

\displaystyle \Rightarrow AC = BC

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

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Question 22: In a triangle, \displaystyle \angle A  \text{ and }  \angle B are acute angles such that \displaystyle \tan A = \tan B , show that \displaystyle \angle A = \angle B

Answer:

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

Given that \displaystyle \tan A = \tan B

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

\displaystyle \Rightarrow BC^2 = AC^2

\displaystyle \Rightarrow BC = AC

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

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Question 23: If \displaystyle A is an acute angle such that \displaystyle 3 \sin A = 4 \cos A , find the value of \displaystyle 4 \sin^2 A - 3 \cos^2 A + 2

Answer:

\displaystyle \text{Given } 3 \sin A = 4 \cos A

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

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

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

\displaystyle \text{Hence } 4 \sin^2 A - 3 \cos^2 A + 2

\displaystyle = 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, \displaystyle \triangle ABC is right angles at \displaystyle B  \text{ and }  D is the midpoint of \displaystyle BC . \displaystyle AC = 5 \text{ cm } , \displaystyle BC = 4 \text{ cm }  \text{ and }  \angle BAD = \theta . Find \displaystyle \text{i) } \tan \theta \displaystyle \text{ii) } \sin \theta \displaystyle \text{iii) } \sin^2 \theta + \cos^2 \theta

Answer:

By Pythagoras theorem

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

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

ii) By Pythagoras theorem

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

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

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

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