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 $\angle C$.

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.

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 $\frac{1}{\tan A}$ $+$ $\frac{\sin A}{1+ \cos A}$

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 $\sin A \cos A =$ $\frac{\sqrt{2}}{4}$

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 $\sin A \cos B + \cos A \sin B = 1$

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 $\tan^2 B - \sin^2 B = \sin^4 B \sec^2 B$

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) $\sin^2 \theta + \cos^2 \theta$

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 $\tan A = \tan B$

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 $\frac{5 \sin A - 3 \cos A}{5 \sin A + 2 \cos A}$ $=$ $\frac{1}{6}$

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 $\tan^2 A +$ $\frac{1}{\tan^2 A}$

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

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) $\cot^2 A$

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

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

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 $\sin A (1 - \tan A) =$ $\frac{35}{156}$

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 $\frac{\sin A - 2 \cos A}{\tan A - \cot A}$

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 $\frac{\cos A - \frac{1}{\tan A}}{2 \cot A}$

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 $\frac{3 \sin A - 4 \sin^3 A}{4 \cos^3 A - 3 \cos A}$ $=$ $\frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A}$

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

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 $\sin A - \cos A$

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 $\tan A$

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 $\angle A = \angle B$

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

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

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 $\angle A = \angle B$

$\tan A =$ $\frac{BC}{AC}$ 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 $4 \sin^2 A - 3 \cos^2 A + 2$

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) $\sin^2 \theta + \cos^2 \theta$

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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