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 $\displaystyle \angle C$.

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

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

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

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

$\displaystyle \Rightarrow AB = 4$

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

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

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

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

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

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

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

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

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

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 ($\displaystyle \text{i) } \sin \theta$ $\displaystyle \text{ii) } \cos \theta$ $\displaystyle \text{iii) } \tan \theta$ $\displaystyle \text{iv) } \sin^2 \theta + \cos^2 \theta$

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 $\displaystyle \tan A = \tan B$

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

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

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

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

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

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

$\displaystyle =$ RHS. Hence proved.

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

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

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

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

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

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

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

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

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

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

Therefore LHS = RHS. Hence proved.

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

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

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

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

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

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

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

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

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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$\displaystyle \text{Question 17: If } \sec A = \frac{5}{4} \text{ , verify} \frac{3 \sin A - 4 \sin^3 A}{4 \cos^3 A - 3 \cos A} = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A}$

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

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

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

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

$\displaystyle \text{LHS } = \sqrt{\frac{\mathrm{cosec} ^2 A - \cot^2 A}{\sec^2 A -1}} = \sqrt{\frac{(\frac{4}{3})^2 - (\frac{\sqrt{7}}{3})^2}{(\frac{4}{\sqrt{7}})^2 -1}} = \sqrt{\frac{\frac{16}{9} - \frac{7}{9}}{\frac{16}{7} -1} } = \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3}$

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$\displaystyle \text{Question 19: If } 8 \tan A = 15 \text{ , find } \sin A - \cos A$

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

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 $\displaystyle \tan A$

$\displaystyle \text{Given } 3 \cos A - 4 \sin A = 2 \cos A + \sin A$, 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 $\displaystyle \angle A = \angle B$

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

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

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 $\displaystyle \angle A = \angle B$

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

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 $\displaystyle 4 \sin^2 A - 3 \cos^2 A + 2$

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

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

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

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

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

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

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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 $\displaystyle \text{i) } \tan \theta$ $\displaystyle \text{ii) } \sin \theta$ $\displaystyle \text{iii) } \sin^2 \theta + \cos^2 \theta$

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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