\displaystyle \textbf{Question 1: }\text{In each of the following, one of the six trigonometric ratios is given. }
\displaystyle \text{Find the values of the other trigonometric ratios.}
\displaystyle \text{(i) }\sin A=\frac{2}{3}\qquad\text{(ii) }\cos A=\frac{4}{5}\qquad\text{(iii) }\tan\theta=11\qquad\text{(iv) }\cot\theta=\frac{12}{5}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given }\sin A=\frac{2}{3}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}
\displaystyle \text{Let perpendicular}=2,\quad\text{hypotenuse}=3.
\displaystyle \text{By Pythagoras theorem, base}=\sqrt{3^2-2^2}=\sqrt5.
\displaystyle \therefore \cos A=\frac{\sqrt5}{3},\quad\tan A=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}
\displaystyle \mathrm{cosec}\,A=\frac{3}{2},\quad\sec A=\frac{3}{\sqrt5}=\frac{3\sqrt5}{5},\quad\cot A=\frac{\sqrt5}{2}

\displaystyle \text{(ii) Given }\cos A=\frac{4}{5}=\frac{\text{Base}}{\text{Hypotenuse}}
\displaystyle \text{Let base}=4,\quad\text{hypotenuse}=5.
\displaystyle \text{By Pythagoras theorem, perpendicular}=\sqrt{5^2-4^2}=3.
\displaystyle \therefore \sin A=\frac35,\quad\tan A=\frac34,\quad\mathrm{cosec}\,A=\frac53
\displaystyle \sec A=\frac54,\quad\cot A=\frac43

\displaystyle \text{(iii) Given }\tan\theta=11=\frac{11}{1}=\frac{\text{Perpendicular}}{\text{Base}}
\displaystyle \text{Let perpendicular}=11,\quad\text{base}=1.
\displaystyle \text{By Pythagoras theorem, hypotenuse}=\sqrt{11^2+1^2}=\sqrt{122}.
\displaystyle \therefore \sin\theta=\frac{11}{\sqrt{122}},\quad\cos\theta=\frac{1}{\sqrt{122}},\quad\mathrm{cosec}\,\theta=\frac{\sqrt{122}}{11}
\displaystyle \sec\theta=\sqrt{122},\quad\cot\theta=\frac1{11}

\displaystyle \text{(iv) Given }\cot\theta=\frac{12}{5}=\frac{\text{Base}}{\text{Perpendicular}}
\displaystyle \text{Let base}=12,\quad\text{perpendicular}=5.
\displaystyle \text{By Pythagoras theorem, hypotenuse}=\sqrt{12^2+5^2}=13.
\displaystyle \therefore \sin\theta=\frac5{13},\quad\cos\theta=\frac{12}{13},\quad\tan\theta=\frac5{12}
\displaystyle \mathrm{cosec}\,\theta=\frac{13}{5},\quad\sec\theta=\frac{13}{12}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In a }\triangle ABC,\text{ right angled at }B,\ AB=24\text{ cm},\ BC=7\text{ cm. Determine}
\displaystyle \text{(i) }\sin A,\cos A\qquad\text{(ii) }\sin C,\cos C.
\displaystyle \text{Answer:}
\displaystyle \text{Since }\triangle ABC\text{ is right angled at }B,
\displaystyle AC=\sqrt{AB^2+BC^2}=\sqrt{24^2+7^2}
\displaystyle =\sqrt{576+49}=\sqrt{625}=25\text{ cm}.
\displaystyle \text{(i) For angle }A,\quad\text{perpendicular}=BC=7,\quad\text{base}=AB=24.
\displaystyle \therefore \sin A=\frac{BC}{AC}=\frac7{25},\qquad\cos A=\frac{AB}{AC}=\frac{24}{25}

\displaystyle \text{(ii) For angle }C,\quad\text{perpendicular}=AB=24,\quad\text{base}=BC=7.
\displaystyle \therefore \sin C=\frac{AB}{AC}=\frac{24}{25},\qquad\cos C=\frac{BC}{AC}=\frac7{25}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Given }15\cot A=8,\text{ find }\sin A\text{ and }\sec A.
\displaystyle \text{Answer:}
\displaystyle 15\cot A=8
\displaystyle \therefore \cot A=\frac{8}{15}=\frac{\text{Base}}{\text{Perpendicular}}
\displaystyle \text{Let base}=8,\quad\text{perpendicular}=15.
\displaystyle \text{By Pythagoras theorem, hypotenuse}=\sqrt{8^2+15^2}
\displaystyle =\sqrt{64+225}=\sqrt{289}=17.
\displaystyle \therefore \sin A=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{15}{17}
\displaystyle \text{and }\sec A=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{17}{8}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }\cot\theta=\frac{7}{8},\text{ evaluate:}
\displaystyle \text{(i) }\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}\qquad\text{(ii) }\cot^2\theta
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}
\displaystyle =\frac{1-\sin^2\theta}{1-\cos^2\theta}
\displaystyle =\frac{\cos^2\theta}{\sin^2\theta}=\cot^2\theta
\displaystyle =\left(\frac{7}{8}\right)^2=\frac{49}{64}

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

\displaystyle \textbf{Question 5: }\text{If }3\cot A=4,\text{ check whether }\frac{1-\tan^2 A}{1+\tan^2 A}=\cos^2 A-\sin^2 A\text{ or not.}
\displaystyle \text{Answer:}
\displaystyle 3\cot A=4
\displaystyle \therefore \cot A=\frac{4}{3}
\displaystyle \therefore \tan A=\frac{3}{4}
\displaystyle \text{Let perpendicular}=3,\quad\text{base}=4.
\displaystyle \text{By Pythagoras theorem, hypotenuse}=\sqrt{3^2+4^2}=5.
\displaystyle \therefore \sin A=\frac{3}{5},\qquad\cos A=\frac{4}{5}
\displaystyle \text{LHS}=\frac{1-\tan^2 A}{1+\tan^2 A}
\displaystyle =\frac{1-\left(\frac{3}{4}\right)^2}{1+\left(\frac{3}{4}\right)^2}=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}
\displaystyle =\frac{\frac{7}{16}}{\frac{25}{16}}=\frac{7}{25}
\displaystyle \text{RHS}=\cos^2 A-\sin^2 A
\displaystyle =\left(\frac{4}{5}\right)^2-\left(\frac{3}{5}\right)^2=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \therefore \frac{1-\tan^2 A}{1+\tan^2 A}=\cos^2 A-\sin^2 A\text{ is verified.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }\tan\theta=\frac{a}{b},\text{ find the value of }\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}.
\displaystyle \text{Answer:}
\displaystyle \frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}
\displaystyle =\frac{\frac{\cos\theta}{\cos\theta}+\frac{\sin\theta}{\cos\theta}}{\frac{\cos\theta}{\cos\theta}-\frac{\sin\theta}{\cos\theta}}
\displaystyle =\frac{1+\tan\theta}{1-\tan\theta}
\displaystyle =\frac{1+\frac{a}{b}}{1-\frac{a}{b}}=\frac{\frac{a+b}{b}}{\frac{b-a}{b}}
\displaystyle =\frac{a+b}{b-a}
\displaystyle \therefore \frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}=\frac{a+b}{b-a}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }3\cot\theta=2,\text{ find the value of }\frac{4\sin\theta-3\cos\theta}{2\sin\theta+6\cos\theta}.
\displaystyle \text{Answer:}
\displaystyle 3\cot\theta=2
\displaystyle \therefore \cot\theta=\frac{2}{3}=\frac{\text{Base}}{\text{Perpendicular}}
\displaystyle \text{Let base}=2,\quad\text{perpendicular}=3.
\displaystyle \text{By Pythagoras theorem, hypotenuse}=\sqrt{2^2+3^2}=\sqrt{13}.
\displaystyle \therefore \sin\theta=\frac{3}{\sqrt{13}},\qquad\cos\theta=\frac{2}{\sqrt{13}}
\displaystyle \frac{4\sin\theta-3\cos\theta}{2\sin\theta+6\cos\theta}
\displaystyle =\frac{4\left(\frac{3}{\sqrt{13}}\right)-3\left(\frac{2}{\sqrt{13}}\right)}{2\left(\frac{3}{\sqrt{13}}\right)+6\left(\frac{2}{\sqrt{13}}\right)}
\displaystyle =\frac{\frac{12-6}{\sqrt{13}}}{\frac{6+12}{\sqrt{13}}}=\frac{6}{18}=\frac{1}{3}
\displaystyle \therefore \text{The required value is }\frac{1}{3}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }\tan\theta=\frac{a}{b},\text{ prove that }\frac{a\sin\theta-b\cos\theta}{a\sin\theta+b\cos\theta}=\frac{a^2-b^2}{a^2+b^2}.
\displaystyle \text{Answer:}
\displaystyle \tan\theta=\frac{a}{b}=\frac{\text{Perpendicular}}{\text{Base}}
\displaystyle \text{Let perpendicular}=a,\quad\text{base}=b.
\displaystyle \text{By Pythagoras theorem, hypotenuse}=\sqrt{a^2+b^2}.
\displaystyle \therefore \sin\theta=\frac{a}{\sqrt{a^2+b^2}},\qquad\cos\theta=\frac{b}{\sqrt{a^2+b^2}}
\displaystyle \text{LHS}=\frac{a\sin\theta-b\cos\theta}{a\sin\theta+b\cos\theta}
\displaystyle =\frac{a\left(\frac{a}{\sqrt{a^2+b^2}}\right)-b\left(\frac{b}{\sqrt{a^2+b^2}}\right)}{a\left(\frac{a}{\sqrt{a^2+b^2}}\right)+b\left(\frac{b}{\sqrt{a^2+b^2}}\right)}
\displaystyle =\frac{\frac{a^2-b^2}{\sqrt{a^2+b^2}}}{\frac{a^2+b^2}{\sqrt{a^2+b^2}}}
\displaystyle =\frac{a^2-b^2}{a^2+b^2}=\text{RHS}
\displaystyle \therefore \frac{a\sin\theta-b\cos\theta}{a\sin\theta+b\cos\theta}=\frac{a^2-b^2}{a^2+b^2}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }\sec\theta=\frac{13}{5},\text{ show that }\frac{2\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta}=3.
\displaystyle \text{Answer:}
\displaystyle \sec\theta=\frac{13}{5}=\frac{\text{Hypotenuse}}{\text{Base}}
\displaystyle \text{Let hypotenuse}=13,\quad\text{base}=5.
\displaystyle \text{By Pythagoras theorem, perpendicular}=\sqrt{13^2-5^2}
\displaystyle =\sqrt{169-25}=\sqrt{144}=12.
\displaystyle \therefore \sin\theta=\frac{12}{13},\qquad\cos\theta=\frac{5}{13}
\displaystyle \text{LHS}=\frac{2\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta}
\displaystyle =\frac{2\left(\frac{12}{13}\right)-3\left(\frac{5}{13}\right)}{4\left(\frac{12}{13}\right)-9\left(\frac{5}{13}\right)}
\displaystyle =\frac{\frac{24-15}{13}}{\frac{48-45}{13}}=\frac{9}{3}=3=\text{RHS}
\displaystyle \therefore \frac{2\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta}=3.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }\cos\theta=\frac{12}{13},\text{ show that }\sin\theta(1-\tan\theta)=\frac{35}{156}.
\displaystyle \text{Answer:}
\displaystyle \cos\theta=\frac{12}{13}=\frac{\text{Base}}{\text{Hypotenuse}}
\displaystyle \text{Let base}=12,\quad\text{hypotenuse}=13.
\displaystyle \text{By Pythagoras theorem, perpendicular}=\sqrt{13^2-12^2}
\displaystyle =\sqrt{169-144}=\sqrt{25}=5.
\displaystyle \therefore \sin\theta=\frac{5}{13},\qquad\tan\theta=\frac{5}{12}
\displaystyle \text{LHS}=\sin\theta(1-\tan\theta)
\displaystyle =\frac{5}{13}\left(1-\frac{5}{12}\right)
\displaystyle =\frac{5}{13}\left(\frac{7}{12}\right)=\frac{35}{156}=\text{RHS}
\displaystyle \therefore \sin\theta(1-\tan\theta)=\frac{35}{156}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }\tan\theta=\frac{1}{\sqrt7},\text{ show that }\frac{\mathrm{cosec}^2\theta-\sec^2\theta}{\mathrm{cosec}^2\theta+\sec^2\theta}=\frac34.
\displaystyle \text{Answer:}
\displaystyle \tan\theta=\frac{1}{\sqrt7}=\frac{\text{Perpendicular}}{\text{Base}}
\displaystyle \text{Let perpendicular}=1,\quad\text{base}=\sqrt7.
\displaystyle \text{By Pythagoras theorem, hypotenuse}=\sqrt{1^2+(\sqrt7)^2}
\displaystyle =\sqrt8=2\sqrt2.
\displaystyle \therefore \mathrm{cosec}\,\theta=\frac{2\sqrt2}{1}=2\sqrt2,\qquad\sec\theta=\frac{2\sqrt2}{\sqrt7}
\displaystyle \therefore \mathrm{cosec}^2\theta=8,\qquad\sec^2\theta=\frac87
\displaystyle \text{LHS}=\frac{\mathrm{cosec}^2\theta-\sec^2\theta}{\mathrm{cosec}^2\theta+\sec^2\theta}
\displaystyle =\frac{8-\frac87}{8+\frac87}=\frac{\frac{48}{7}}{\frac{64}{7}}=\frac{48}{64}=\frac34=\text{RHS}
\displaystyle \therefore \frac{\mathrm{cosec}^2\theta-\sec^2\theta}{\mathrm{cosec}^2\theta+\sec^2\theta}=\frac34.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }\sec\theta=\frac54,\text{ find the value of }\frac{\sin\theta-2\cos\theta}{\tan\theta-\cot\theta}.
\displaystyle \text{Answer:}
\displaystyle \sec\theta=\frac54=\frac{\text{Hypotenuse}}{\text{Base}}
\displaystyle \text{Let hypotenuse}=5,\quad\text{base}=4.
\displaystyle \text{By Pythagoras theorem, perpendicular}=\sqrt{5^2-4^2}=3.
\displaystyle \therefore \sin\theta=\frac35,\quad\cos\theta=\frac45,\quad\tan\theta=\frac34,\quad\cot\theta=\frac43
\displaystyle \frac{\sin\theta-2\cos\theta}{\tan\theta-\cot\theta}
\displaystyle =\frac{\frac35-2\left(\frac45\right)}{\frac34-\frac43}
\displaystyle =\frac{-1}{-\frac7{12}}=\frac{12}{7}
\displaystyle \therefore \text{The required value is }\frac{12}{7}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }\cos\theta=\frac{3}{5},\text{ find the value of }\frac{\sin\theta-\frac{1}{\tan\theta}}{2\tan\theta}.
\displaystyle \text{Answer:}
\displaystyle \cos\theta=\frac{3}{5}=\frac{\text{Base}}{\text{Hypotenuse}}
\displaystyle \text{Let base}=3,\quad\text{hypotenuse}=5.
\displaystyle \text{By Pythagoras theorem, perpendicular}=\sqrt{5^2-3^2}
\displaystyle =\sqrt{25-9}=\sqrt{16}=4.
\displaystyle \therefore \sin\theta=\frac45,\qquad\tan\theta=\frac43
\displaystyle \frac{\sin\theta-\frac{1}{\tan\theta}}{2\tan\theta}
\displaystyle =\frac{\frac45-\frac34}{2\left(\frac43\right)}
\displaystyle =\frac{\frac{16-15}{20}}{\frac83}=\frac{1}{20}\times\frac38=\frac{3}{160}
\displaystyle \therefore \text{The required value is }\frac{3}{160}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In Fig. 10.35, find }\tan P\text{ and }\cot R.\text{ Is }\tan P=\cot R?
\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle PQR,\quad PR=13\text{ cm},\quad PQ=12\text{ cm}.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle PR^2=PQ^2+RQ^2
\displaystyle 13^2=12^2+RQ^2
\displaystyle RQ^2=169-144=25
\displaystyle \therefore RQ=5\text{ cm}.
\displaystyle \tan P=\frac{\text{Perpendicular}}{\text{Base}}=\frac{RQ}{PQ}=\frac{5}{12}
\displaystyle \cot R=\frac{\text{Base}}{\text{Perpendicular}}=\frac{RQ}{PQ}=\frac{5}{12}
\displaystyle \therefore \tan P=\cot R=\frac{5}{12}.
\displaystyle \therefore \text{Yes, }\tan P=\cot R.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }\angle A\text{ and }\angle B\text{ are acute angles such that }\cos A=\cos B,\text{ then show that}
\displaystyle \angle A=\angle B.
\displaystyle \text{Answer:}
\displaystyle \text{Consider two right-angled triangles having acute angles }A\text{ and }B.
\displaystyle \text{Take the hypotenuse of each triangle equal to }1.
\displaystyle \cos A=\frac{\text{Base corresponding to }A}{1},\qquad\cos B=\frac{\text{Base corresponding to }B}{1}
\displaystyle \text{Given, }\cos A=\cos B
\displaystyle \therefore \text{The corresponding bases of the two triangles are equal.}
\displaystyle \text{Also, their hypotenuses are equal.}
\displaystyle \text{By Pythagoras theorem, their corresponding perpendiculars are also equal.}
\displaystyle \therefore \text{The two right-angled triangles are congruent by RHS congruence.}
\displaystyle \therefore \angle A=\angle B.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In a }\triangle ABC,\text{ right angled at }A,\text{ if }\tan C=\sqrt3,\text{ find the value of}
\displaystyle \sin B\cos C+\cos B\sin C.\hfill\text{[CBSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \tan C=\sqrt3=\frac{AB}{AC}
\displaystyle \text{Let }AB=\sqrt3k\text{ and }AC=k.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle BC=\sqrt{AB^2+AC^2}
\displaystyle =\sqrt{3k^2+k^2}=\sqrt{4k^2}=2k.
\displaystyle \therefore \sin B=\frac{AC}{BC}=\frac12,\qquad\cos B=\frac{AB}{BC}=\frac{\sqrt3}{2}
\displaystyle \cos C=\frac{AC}{BC}=\frac12,\qquad\sin C=\frac{AB}{BC}=\frac{\sqrt3}{2}
\displaystyle \therefore \sin B\cos C+\cos B\sin C
\displaystyle =\left(\frac12\right)\left(\frac12\right)+\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt3}{2}\right)
\displaystyle =\frac14+\frac34=1
\displaystyle \therefore \text{The required value is }1.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{State whether the following statements are true or false. Justify your answer.}
\displaystyle \text{(i) The value of }\tan A\text{ is always less than }1.
\displaystyle \text{(ii) }\sec A=\frac{12}{5}\text{ for some value of angle }A.
\displaystyle \text{(iii) }\cos A\text{ is the abbreviation used for the cosecant of angle }A.
\displaystyle \text{(iv) }\cot A\text{ is the product of cot and }A.
\displaystyle \text{(v) }\sin\theta=\frac{4}{3}\text{ for some angle }\theta.
\displaystyle \text{Answer:}
\displaystyle \text{(i) False.}
\displaystyle \tan A=\frac{\text{Perpendicular}}{\text{Base}}
\displaystyle \text{If perpendicular}>\text{base, then }\tan A>1.
\displaystyle \therefore \tan A\text{ is not always less than }1.

\displaystyle \text{(ii) True.}
\displaystyle \sec A=\frac{\text{Hypotenuse}}{\text{Base}}
\displaystyle \text{Since hypotenuse}>\text{base, a ratio of }\frac{12}{5}\text{ is possible.}
\displaystyle \therefore \sec A=\frac{12}{5}\text{ for some value of angle }A.

\displaystyle \text{(iii) False.}
\displaystyle \cos A\text{ is the abbreviation for cosine of angle }A.
\displaystyle \mathrm{cosec}\,A\text{ is the abbreviation for cosecant of angle }A.

\displaystyle \text{(iv) False.}
\displaystyle \cot A\text{ denotes the cotangent of angle }A\text{ and is not the product of cot and }A.

\displaystyle \text{(v) False.}
\displaystyle \sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}
\displaystyle \text{Since perpendicular}<\text{hypotenuse, }\sin\theta<1\text{ for an acute angle.}
\displaystyle \text{But }\frac43>1.
\displaystyle \therefore \sin\theta=\frac43\text{ is not possible.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }\sin\theta=\frac{12}{13},\text{ find the value of }\frac{\sin^2\theta-\cos^2\theta}{2\sin\theta\cos\theta}\times\frac{1}{\tan^2\theta}.
\displaystyle \text{Answer:}
\displaystyle \sin\theta=\frac{12}{13}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}
\displaystyle \text{Let perpendicular}=12,\quad\text{hypotenuse}=13.
\displaystyle \text{By Pythagoras theorem, base}=\sqrt{13^2-12^2}=5.
\displaystyle \therefore \cos\theta=\frac{5}{13},\qquad\tan\theta=\frac{12}{5}
\displaystyle \frac{\sin^2\theta-\cos^2\theta}{2\sin\theta\cos\theta}\times\frac{1}{\tan^2\theta}
\displaystyle =\frac{\left(\frac{12}{13}\right)^2-\left(\frac{5}{13}\right)^2}{2\left(\frac{12}{13}\right)\left(\frac{5}{13}\right)}\times\frac{1}{\left(\frac{12}{5}\right)^2}
\displaystyle =\frac{\frac{144-25}{169}}{\frac{120}{169}}\times\frac{25}{144}
\displaystyle =\frac{119}{120}\times\frac{25}{144}=\frac{2975}{17280}=\frac{595}{3456}
\displaystyle \therefore \text{The required value is }\frac{595}{3456}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }\sec A=\frac{5}{4},\text{ verify that }\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}.
\displaystyle \text{Answer:}
\displaystyle \sec A=\frac54=\frac{\text{Hypotenuse}}{\text{Base}}
\displaystyle \text{Let hypotenuse}=5,\quad\text{base}=4.
\displaystyle \text{By Pythagoras theorem, perpendicular}=\sqrt{5^2-4^2}=3.
\displaystyle \therefore \sin A=\frac35,\qquad\cos A=\frac45,\qquad\tan A=\frac34
\displaystyle \text{LHS}=\frac{3\sin A-4\sin^3 A}{4\cos^3 A-3\cos A}
\displaystyle =\frac{3\left(\frac35\right)-4\left(\frac35\right)^3}{4\left(\frac45\right)^3-3\left(\frac45\right)}
\displaystyle =\frac{\frac95-\frac{108}{125}}{\frac{256}{125}-\frac{12}{5}}
\displaystyle =\frac{\frac{117}{125}}{-\frac{44}{125}}=-\frac{117}{44}
\displaystyle \text{RHS}=\frac{3\tan A-\tan^3 A}{1-3\tan^2 A}
\displaystyle =\frac{3\left(\frac34\right)-\left(\frac34\right)^3}{1-3\left(\frac34\right)^2}
\displaystyle =\frac{\frac94-\frac{27}{64}}{1-\frac{27}{16}}
\displaystyle =\frac{\frac{117}{64}}{-\frac{11}{16}}=-\frac{117}{44}
\displaystyle \therefore \text{LHS}=\text{RHS}=-\frac{117}{44}.
\displaystyle \text{Hence verified.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }\cot\theta=\frac{3}{4},\text{ prove that }\sqrt{\frac{\sec\theta-\mathrm{cosec}\,\theta}{\sec\theta+\mathrm{cosec}\,\theta}}=\frac{1}{\sqrt7}.
\displaystyle \text{Answer:}
\displaystyle \cot\theta=\frac{3}{4}=\frac{\text{Base}}{\text{Perpendicular}}
\displaystyle \text{Let base}=3,\quad\text{perpendicular}=4.
\displaystyle \text{By Pythagoras theorem, hypotenuse}=\sqrt{3^2+4^2}=5.
\displaystyle \therefore \sec\theta=\frac{5}{3},\qquad\mathrm{cosec}\,\theta=\frac{5}{4}
\displaystyle \text{LHS}=\sqrt{\frac{\sec\theta-\mathrm{cosec}\,\theta}{\sec\theta+\mathrm{cosec}\,\theta}}
\displaystyle =\sqrt{\frac{\frac53-\frac54}{\frac53+\frac54}}
\displaystyle =\sqrt{\frac{\frac{5}{12}}{\frac{35}{12}}}
\displaystyle =\sqrt{\frac17}=\frac{1}{\sqrt7}=\text{RHS}
\displaystyle \therefore \sqrt{\frac{\sec\theta-\mathrm{cosec}\,\theta}{\sec\theta+\mathrm{cosec}\,\theta}}=\frac{1}{\sqrt7}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }3\cos\theta-4\sin\theta=2\cos\theta+\sin\theta,\text{ find }\tan\theta.
\displaystyle \text{Answer:}
\displaystyle 3\cos\theta-4\sin\theta=2\cos\theta+\sin\theta
\displaystyle 3\cos\theta-2\cos\theta=4\sin\theta+\sin\theta
\displaystyle \cos\theta=5\sin\theta
\displaystyle \frac{\sin\theta}{\cos\theta}=\frac{1}{5}
\displaystyle \therefore \tan\theta=\frac{1}{5}.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If }\sin A=y,\text{ then express }\cos A\text{ and }\tan A\text{ in terms of }y.
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \sin A=y
\displaystyle \sin^2 A+\cos^2 A=1
\displaystyle y^2+\cos^2 A=1
\displaystyle \cos^2 A=1-y^2
\displaystyle \therefore \cos A=\sqrt{1-y^2}
\displaystyle \tan A=\frac{\sin A}{\cos A}
\displaystyle \therefore \tan A=\frac{y}{\sqrt{1-y^2}}.
\displaystyle \\

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