\displaystyle \textbf{Exercise - 22(A)}

\displaystyle \textbf{Question 1: }\text{From the following figure, find the values of:}
\displaystyle \text{(i) }\sin A\qquad\text{(ii) }\cos A\qquad\text{(iii) }\cot A
\displaystyle \text{(iv) }\sec C\qquad\text{(v) }\mathrm{cosec}\,C\qquad\text{(vi) }\tan C \displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ABC,\ AB=3,\ BC=4.
\displaystyle AC=\sqrt{AB^2+BC^2}=\sqrt{3^2+4^2}=\sqrt{25}=5.

\displaystyle \text{(i) }\sin A=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{BC}{AC}=\frac45.

\displaystyle \text{(ii) }\cos A=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{AB}{AC}=\frac35.

\displaystyle \text{(iii) }\cot A=\frac{\text{Base}}{\text{Perpendicular}}=\frac{AB}{BC}=\frac34.

\displaystyle \text{(iv) }\sec C=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{AC}{BC}=\frac54.

\displaystyle \text{(v) }\mathrm{cosec}\,C=\frac{\text{Hypotenuse}}{\text{Perpendicular}}=\frac{AC}{AB}=\frac53.

\displaystyle \text{(vi) }\tan C=\frac{\text{Perpendicular}}{\text{Base}}=\frac{AB}{BC}=\frac34.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{From the following figure, find the values of:}
\displaystyle \text{(i) }\cos B\qquad\text{(ii) }\tan C\qquad\text{(iii) }\sin^2B+\cos^2B
\displaystyle \text{(iv) }\sin B\cos C+\cos B\sin C \displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ABC,\ AB=8,\ BC=17.
\displaystyle AC=\sqrt{BC^2-AB^2}=\sqrt{17^2-8^2}
\displaystyle =\sqrt{289-64}=\sqrt{225}=15.

\displaystyle \text{(i) }\cos B=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{AB}{BC}=\frac8{17}.

\displaystyle \text{(ii) }\tan C=\frac{\text{Perpendicular}}{\text{Base}}=\frac{AB}{AC}=\frac8{15}.

\displaystyle \text{(iii) }\sin B=\frac{AC}{BC}=\frac{15}{17},\quad \cos B=\frac{AB}{BC}=\frac8{17}.
\displaystyle \therefore \sin^2B+\cos^2B=\left(\frac{15}{17}\right)^2+\left(\frac8{17}\right)^2
\displaystyle =\frac{225+64}{289}=\frac{289}{289}=1.

\displaystyle \text{(iv) }\sin B=\frac{15}{17},\quad\cos C=\frac{15}{17},
\displaystyle \cos B=\frac8{17},\quad\sin C=\frac8{17}.
\displaystyle \therefore \sin B\cos C+\cos B\sin C
\displaystyle =\frac{15}{17}\times\frac{15}{17}+\frac8{17}\times\frac8{17}
\displaystyle =\frac{225+64}{289}=\frac{289}{289}=1.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{From the following figure, find the values of:}
\displaystyle \text{(i) }\cos A\qquad\text{(ii) }\mathrm{cosec}\,A\qquad\text{(iii) }\tan^2A-\sec^2A
\displaystyle \text{(iv) }\sin C\qquad\text{(v) }\sec C\qquad\text{(vi) }\cot^2C-\frac{1}{\sin^2C} \displaystyle \text{Answer:}
\displaystyle \text{Let }BD\perp AC,\quad AD=3,\quad BD=4,\quad BC=12.
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle AB=\sqrt{AD^2+BD^2}=\sqrt{3^2+4^2}=5.
\displaystyle \text{In right-angled }\triangle BDC,
\displaystyle DC=\sqrt{BC^2-BD^2}=\sqrt{12^2-4^2}
\displaystyle =\sqrt{128}=8\sqrt2.

\displaystyle \text{(i) }\cos A=\frac{AD}{AB}=\frac35.

\displaystyle \text{(ii) }\mathrm{cosec}\,A=\frac{AB}{BD}=\frac54.

\displaystyle \text{(iii) }\tan A=\frac{BD}{AD}=\frac43,\quad \sec A=\frac{AB}{AD}=\frac53.
\displaystyle \therefore \tan^2A-\sec^2A=\left(\frac43\right)^2-\left(\frac53\right)^2
\displaystyle =\frac{16}{9}-\frac{25}{9}=-1.

\displaystyle \text{(iv) }\sin C=\frac{BD}{BC}=\frac4{12}=\frac13.

\displaystyle \text{(v) }\sec C=\frac{BC}{DC}=\frac{12}{8\sqrt2}
\displaystyle =\frac{3}{2\sqrt2}=\frac{3\sqrt2}{4}.

\displaystyle \text{(vi) }\cot C=\frac{DC}{BD}=\frac{8\sqrt2}{4}=2\sqrt2.
\displaystyle \therefore \cot^2C-\frac{1}{\sin^2C}=(2\sqrt2)^2-\frac{1}{(1/3)^2}
\displaystyle =8-9=-1.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{From the following figure, find the values of:}
\displaystyle \text{(i) }\sin B\qquad\text{(ii) }\tan C\qquad\text{(iii) }\sec^2B-\tan^2B
\displaystyle \text{(iv) }\sin^2C+\cos^2C \displaystyle \text{Answer:}
\displaystyle \text{Given }AD\perp BC,\quad AB=13,\quad BD=5,\quad DC=16.
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle AD=\sqrt{AB^2-BD^2}=\sqrt{13^2-5^2}
\displaystyle =\sqrt{169-25}=\sqrt{144}=12.
\displaystyle \text{In right-angled }\triangle ADC,
\displaystyle AC=\sqrt{AD^2+DC^2}=\sqrt{12^2+16^2}
\displaystyle =\sqrt{400}=20.

\displaystyle \text{(i) }\sin B=\frac{AD}{AB}=\frac{12}{13}.

\displaystyle \text{(ii) }\tan C=\frac{AD}{DC}=\frac{12}{16}=\frac34.

\displaystyle \text{(iii) }\sec B=\frac{AB}{BD}=\frac{13}{5},\quad \tan B=\frac{AD}{BD}=\frac{12}{5}.
\displaystyle \therefore \sec^2B-\tan^2B=\left(\frac{13}{5}\right)^2-\left(\frac{12}{5}\right)^2
\displaystyle =\frac{169-144}{25}=\frac{25}{25}=1.

\displaystyle \text{(iv) }\sin C=\frac{AD}{AC}=\frac{12}{20}=\frac35,
\displaystyle \cos C=\frac{DC}{AC}=\frac{16}{20}=\frac45.
\displaystyle \therefore \sin^2C+\cos^2C=\left(\frac35\right)^2+\left(\frac45\right)^2
\displaystyle =\frac9{25}+\frac{16}{25}=1.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Given: }\sin A=\frac35,\text{ find:}
\displaystyle \text{(i) }\tan A\qquad\text{(ii) }\cos A

\displaystyle \text{Answer:}
\displaystyle \text{Given }\sin A=\frac35=\frac{\text{Perpendicular}}{\text{Hypotenuse}}.
\displaystyle \text{Let perpendicular }=3x\text{ and hypotenuse }=5x.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Base}=\sqrt{(5x)^2-(3x)^2}=\sqrt{25x^2-9x^2}=4x.

\displaystyle \text{(i) }\tan A=\frac{\text{Perpendicular}}{\text{Base}}=\frac{3x}{4x}=\frac34.

\displaystyle \text{(ii) }\cos A=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{4x}{5x}=\frac45.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{From the following figure, find the values of:}
\displaystyle \text{(i) }\sin A\qquad\text{(ii) }\sec A\qquad\text{(iii) }\cos^2A+\sin^2A \displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ABC,\ AB=a,\ BC=a.
\displaystyle AC=\sqrt{AB^2+BC^2}=\sqrt{a^2+a^2}=\sqrt{2a^2}=a\sqrt2.

\displaystyle \text{(i) }\sin A=\frac{BC}{AC}=\frac{a}{a\sqrt2}=\frac{1}{\sqrt2}=\frac{\sqrt2}{2}.

\displaystyle \text{(ii) }\sec A=\frac{AC}{AB}=\frac{a\sqrt2}{a}=\sqrt2.

\displaystyle \text{(iii) }\cos A=\frac{AB}{AC}=\frac{a}{a\sqrt2}=\frac{1}{\sqrt2},
\displaystyle \sin A=\frac{BC}{AC}=\frac{a}{a\sqrt2}=\frac{1}{\sqrt2}.
\displaystyle \therefore \cos^2A+\sin^2A=\left(\frac{1}{\sqrt2}\right)^2+\left(\frac{1}{\sqrt2}\right)^2
\displaystyle =\frac12+\frac12=1.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Given: }\cos A=\frac{5}{13},\text{ evaluate:}
\displaystyle \text{(i) }\frac{\sin A-\cot A}{2\tan A}\qquad\text{(ii) }\cot A+\frac{1}{\cos A}
\displaystyle \text{Answer:}
\displaystyle \text{Given }\cos A=\frac{5}{13}=\frac{\text{Base}}{\text{Hypotenuse}}.
\displaystyle \text{Let base }=5x\text{ and hypotenuse }=13x.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Perpendicular}=\sqrt{(13x)^2-(5x)^2}=\sqrt{144x^2}=12x.
\displaystyle \therefore \sin A=\frac{12}{13},\quad\tan A=\frac{12}{5},\quad\cot A=\frac{5}{12}.

\displaystyle \text{(i) }\frac{\sin A-\cot A}{2\tan A}=\frac{\frac{12}{13}-\frac{5}{12}}{2\times\frac{12}{5}}
\displaystyle =\frac{\frac{144-65}{156}}{\frac{24}{5}}=\frac{79}{156}\times\frac{5}{24}=\frac{395}{3744}.

\displaystyle \text{(ii) }\cot A+\frac{1}{\cos A}=\frac{5}{12}+\frac{13}{5}
\displaystyle =\frac{25+156}{60}=\frac{181}{60}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Given: }\sec A=\frac{29}{21},\text{ evaluate: }\sin A-\frac{1}{\tan A}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\sec A=\frac{29}{21}=\frac{\text{Hypotenuse}}{\text{Base}}.
\displaystyle \text{Let hypotenuse }=29x\text{ and base }=21x.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Perpendicular}=\sqrt{(29x)^2-(21x)^2}=\sqrt{400x^2}=20x.
\displaystyle \therefore \sin A=\frac{20}{29},\quad\tan A=\frac{20}{21}.
\displaystyle \sin A-\frac{1}{\tan A}=\frac{20}{29}-\frac{21}{20}
\displaystyle =\frac{400-609}{580}=-\frac{209}{580}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Given: }\tan A=\frac45,\text{ find: }\frac{\mathrm{cosec}\,A}{\cot A-\sec A}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\tan A=\frac45=\frac{\text{Perpendicular}}{\text{Base}}.
\displaystyle \text{Let perpendicular }=4x\text{ and base }=5x.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Hypotenuse}=\sqrt{(4x)^2+(5x)^2}=\sqrt{41}\,x.
\displaystyle \therefore \mathrm{cosec}\,A=\frac{\sqrt{41}}4,\quad\cot A=\frac54,\quad\sec A=\frac{\sqrt{41}}5.
\displaystyle \frac{\mathrm{cosec}\,A}{\cot A-\sec A}=\frac{\frac{\sqrt{41}}4}{\frac54-\frac{\sqrt{41}}5}
\displaystyle =\frac{\frac{\sqrt{41}}4}{\frac{25-4\sqrt{41}}{20}}=\frac{5\sqrt{41}}{25-4\sqrt{41}}
\displaystyle =\frac{5\sqrt{41}(25+4\sqrt{41})}{625-16(41)}
\displaystyle =-\frac{5\sqrt{41}(25+4\sqrt{41})}{31}
\displaystyle =-\frac{820+125\sqrt{41}}{31}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Given: }4\cot A=3,\text{ find:}
\displaystyle \text{(i) }\sin A\qquad\text{(ii) }\sec A\qquad\text{(iii) }\mathrm{cosec}^2A-\cot^2A
\displaystyle \text{Answer:}
\displaystyle 4\cot A=3\Rightarrow\cot A=\frac34=\frac{\text{Base}}{\text{Perpendicular}}.
\displaystyle \text{Let base }=3x\text{ and perpendicular }=4x.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Hypotenuse}=\sqrt{(3x)^2+(4x)^2}=\sqrt{25x^2}=5x.

\displaystyle \text{(i) }\sin A=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{4x}{5x}=\frac45.

\displaystyle \text{(ii) }\sec A=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{5x}{3x}=\frac53.

\displaystyle \text{(iii) }\mathrm{cosec}\,A=\frac54,\quad\cot A=\frac34.
\displaystyle \therefore \mathrm{cosec}^2A-\cot^2A=\left(\frac54\right)^2-\left(\frac34\right)^2
\displaystyle =\frac{25}{16}-\frac9{16}=\frac{16}{16}=1.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Given: }\cos A=0.6,\text{ find all other trigonometrical ratios}
\displaystyle \text{for angle }A.
\displaystyle \text{Answer:}
\displaystyle \cos A=0.6=\frac35=\frac{\text{Base}}{\text{Hypotenuse}}.
\displaystyle \text{Let base }=3x\text{ and hypotenuse }=5x.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Perpendicular}=\sqrt{(5x)^2-(3x)^2}=\sqrt{16x^2}=4x.
\displaystyle \therefore \sin A=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac45.
\displaystyle \tan A=\frac{\text{Perpendicular}}{\text{Base}}=\frac43.
\displaystyle \cot A=\frac{\text{Base}}{\text{Perpendicular}}=\frac34.
\displaystyle \sec A=\frac{\text{Hypotenuse}}{\text{Base}}=\frac53.
\displaystyle \mathrm{cosec}\,A=\frac{\text{Hypotenuse}}{\text{Perpendicular}}=\frac54.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In a right-angled triangle, it is given that }A\text{ is an acute angle}
\displaystyle \text{and }\tan A=\frac{5}{12}.\text{ Find the values of:}
\displaystyle \text{(i) }\cos A\qquad\text{(ii) }\sin A\qquad\text{(iii) }\frac{\cos A+\sin A}{\cos A-\sin A}
\displaystyle \text{Answer:}
\displaystyle \tan A=\frac{5}{12}=\frac{\text{Perpendicular}}{\text{Base}}.
\displaystyle \text{Let perpendicular }=5x\text{ and base }=12x.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Hypotenuse}=\sqrt{(5x)^2+(12x)^2}=\sqrt{169x^2}=13x.

\displaystyle \text{(i) }\cos A=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{12x}{13x}=\frac{12}{13}.

\displaystyle \text{(ii) }\sin A=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{5x}{13x}=\frac{5}{13}.

\displaystyle \text{(iii) }\frac{\cos A+\sin A}{\cos A-\sin A}=\frac{\frac{12}{13}+\frac{5}{13}}{\frac{12}{13}-\frac{5}{13}}
\displaystyle =\frac{\frac{17}{13}}{\frac{7}{13}}=\frac{17}{7}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Given: }\sin\theta=\frac{p}{q},\text{ find }\cos\theta+\sin\theta\text{ in terms of }p\text{ and }q.
\displaystyle \text{Answer:}
\displaystyle \sin\theta=\frac{p}{q}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}.
\displaystyle \text{Let perpendicular }=p\text{ and hypotenuse }=q.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Base}=\sqrt{q^2-p^2}.
\displaystyle \therefore \cos\theta=\frac{\sqrt{q^2-p^2}}{q}.
\displaystyle \therefore \cos\theta+\sin\theta=\frac{\sqrt{q^2-p^2}}{q}+\frac{p}{q}
\displaystyle =\frac{\sqrt{q^2-p^2}+p}{q}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }\cos A=\frac12\text{ and }\sin B=\frac{1}{\sqrt2},\text{ find the value of:}
\displaystyle \frac{\tan A-\tan B}{1+\tan A\tan B}
\displaystyle \text{Answer:}
\displaystyle \cos A=\frac12=\frac{\text{Base}}{\text{Hypotenuse}}.
\displaystyle \text{Let base }=x\text{ and hypotenuse }=2x.
\displaystyle \text{By Pythagoras theorem, perpendicular}=\sqrt{(2x)^2-x^2}=\sqrt3x.
\displaystyle \therefore \tan A=\frac{\sqrt3x}{x}=\sqrt3.
\displaystyle \sin B=\frac{1}{\sqrt2}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}.
\displaystyle \text{Let perpendicular }=x\text{ and hypotenuse }=\sqrt2x.
\displaystyle \text{By Pythagoras theorem, base}=\sqrt{(\sqrt2x)^2-x^2}=x.
\displaystyle \therefore \tan B=\frac{x}{x}=1.
\displaystyle \therefore \frac{\tan A-\tan B}{1+\tan A\tan B}=\frac{\sqrt3-1}{1+\sqrt3}
\displaystyle =\frac{(\sqrt3-1)(\sqrt3-1)}{(1+\sqrt3)(\sqrt3-1)}
\displaystyle =\frac{4-2\sqrt3}{2}=2-\sqrt3.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }5\cot\theta=12,\text{ find the value of:}
\displaystyle \mathrm{cosec}\,\theta+\sec\theta
\displaystyle \text{Answer:}
\displaystyle 5\cot\theta=12\Rightarrow\cot\theta=\frac{12}{5}=\frac{\text{Base}}{\text{Perpendicular}}.
\displaystyle \text{Let base }=12x\text{ and perpendicular }=5x.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Hypotenuse}=\sqrt{(12x)^2+(5x)^2}=\sqrt{169x^2}=13x.
\displaystyle \therefore \mathrm{cosec}\,\theta=\frac{13}{5},\quad\sec\theta=\frac{13}{12}.
\displaystyle \therefore \mathrm{cosec}\,\theta+\sec\theta=\frac{13}{5}+\frac{13}{12}
\displaystyle =\frac{156+65}{60}=\frac{221}{60}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }\tan x=1\frac13,\text{ find the value of:}
\displaystyle 4\sin^2x-3\cos^2x+2
\displaystyle \text{Answer:}
\displaystyle \tan x=1\frac13=\frac43=\frac{\text{Perpendicular}}{\text{Base}}.
\displaystyle \text{Let perpendicular }=4a\text{ and base }=3a.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Hypotenuse}=\sqrt{(4a)^2+(3a)^2}=\sqrt{25a^2}=5a.
\displaystyle \therefore \sin x=\frac45,\quad\cos x=\frac35.
\displaystyle \therefore 4\sin^2x-3\cos^2x+2
\displaystyle =4\left(\frac45\right)^2-3\left(\frac35\right)^2+2
\displaystyle =\frac{64}{25}-\frac{27}{25}+\frac{50}{25}
\displaystyle =\frac{87}{25}.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }\mathrm{cosec}\,\theta=\sqrt5,\text{ find the value of:}
\displaystyle \text{(i) }2-\sin^2\theta-\cos^2\theta
\displaystyle \text{(ii) }2+\frac{1}{\sin^2\theta}-\frac{\cos^2\theta}{\sin^2\theta}
\displaystyle \text{Answer:}
\displaystyle \mathrm{cosec}\,\theta=\sqrt5\Rightarrow\sin\theta=\frac{1}{\sqrt5}.
\displaystyle \therefore \sin^2\theta=\frac15.
\displaystyle \cos^2\theta=1-\sin^2\theta=1-\frac15=\frac45.

\displaystyle \text{(i) }2-\sin^2\theta-\cos^2\theta=2-\frac15-\frac45
\displaystyle =2-1=1.

\displaystyle \text{(ii) }2+\frac{1}{\sin^2\theta}-\frac{\cos^2\theta}{\sin^2\theta}
\displaystyle =2+\frac{1}{1/5}-\frac{4/5}{1/5}
\displaystyle =2+5-4=3.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }\sec A=\sqrt2,\text{ find the value of:}
\displaystyle \frac{3\cos^2A+5\tan^2A}{4\tan^2A-\sin^2A}
\displaystyle \text{Answer:}
\displaystyle \sec A=\sqrt2\Rightarrow\cos A=\frac{1}{\sqrt2}.
\displaystyle \therefore \cos^2A=\frac12.
\displaystyle \sin^2A=1-\cos^2A=1-\frac12=\frac12.
\displaystyle \tan^2A=\frac{\sin^2A}{\cos^2A}=\frac{1/2}{1/2}=1.
\displaystyle \therefore \frac{3\cos^2A+5\tan^2A}{4\tan^2A-\sin^2A}
\displaystyle =\frac{3\left(\frac12\right)+5(1)}{4(1)-\frac12}
\displaystyle =\frac{\frac32+5}{4-\frac12}=\frac{13/2}{7/2}=\frac{13}{7}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }\cot\theta=1,\text{ find the value of:}
\displaystyle 5\tan^2\theta+2\sin^2\theta-3
\displaystyle \text{Answer:}
\displaystyle \cot\theta=1=\frac{\text{Base}}{\text{Perpendicular}}.
\displaystyle \text{Let base }=x\text{ and perpendicular }=x.
\displaystyle \text{By Pythagoras theorem, hypotenuse}=\sqrt{x^2+x^2}=x\sqrt2.
\displaystyle \therefore \tan\theta=\frac{x}{x}=1,\quad\sin\theta=\frac{x}{x\sqrt2}=\frac{1}{\sqrt2}.
\displaystyle \therefore 5\tan^2\theta+2\sin^2\theta-3
\displaystyle =5(1)^2+2\left(\frac{1}{\sqrt2}\right)^2-3
\displaystyle =5+1-3=3.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{In the following figure, }AD\perp BC,\ AC=26,\ CD=10,\ BC=42,
\displaystyle \angle DAC=x\text{ and }\angle B=y.\text{ Find the value of:}
\displaystyle \text{(i) }\cot x\qquad\text{(ii) }\frac{1}{\sin^2y}-\frac{1}{\tan^2y}
\displaystyle \text{(iii) }\frac{6}{\cos x}-\frac{5}{\cos y}+8\tan y \displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ADC,\ AC=26,\ CD=10.
\displaystyle AD=\sqrt{AC^2-CD^2}=\sqrt{26^2-10^2}
\displaystyle =\sqrt{676-100}=\sqrt{576}=24.
\displaystyle DB=BC-CD=42-10=32.
\displaystyle \text{In right-angled }\triangle ADB,
\displaystyle AB=\sqrt{AD^2+DB^2}=\sqrt{24^2+32^2}
\displaystyle =\sqrt{1600}=40.

\displaystyle \text{(i) }\cot x=\frac{AD}{CD}=\frac{24}{10}=\frac{12}{5}.

\displaystyle \text{(ii) }\sin y=\frac{AD}{AB}=\frac{24}{40}=\frac35,
\displaystyle \tan y=\frac{AD}{DB}=\frac{24}{32}=\frac34.
\displaystyle \therefore \frac{1}{\sin^2y}-\frac{1}{\tan^2y}
\displaystyle =\frac{1}{(3/5)^2}-\frac{1}{(3/4)^2}
\displaystyle =\frac{25}{9}-\frac{16}{9}=1.

\displaystyle \text{(iii) }\cos x=\frac{AD}{AC}=\frac{24}{26}=\frac{12}{13},
\displaystyle \cos y=\frac{DB}{AB}=\frac{32}{40}=\frac45,\quad\tan y=\frac34.
\displaystyle \therefore \frac{6}{\cos x}-\frac{5}{\cos y}+8\tan y
\displaystyle =\frac{6}{12/13}-\frac{5}{4/5}+8\left(\frac34\right)
\displaystyle =\frac{13}{2}-\frac{25}{4}+6
\displaystyle =\frac{26-25+24}{4}=\frac{25}{4}.
\displaystyle \\

\displaystyle \textbf{Exercise - 22(B)}

\displaystyle \textbf{Question 1: }\text{From the following figure, find:}
\displaystyle \text{(i) }y\qquad\text{(ii) }\sin x^\circ
\displaystyle \text{(iii) }(\sec x^\circ-\tan x^\circ)(\sec x^\circ+\tan x^\circ) \displaystyle \text{Answer:}
\displaystyle \text{In the right-angled triangle, hypotenuse}=2\text{ and one side}=1.

\displaystyle \text{(i) By Pythagoras theorem,}
\displaystyle y^2+1^2=2^2
\displaystyle y^2=4-1=3
\displaystyle \therefore y=\sqrt3.

\displaystyle \text{(ii) }\sin x^\circ=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{y}{2}
\displaystyle =\frac{\sqrt3}{2}.

\displaystyle \text{(iii) }\sec x^\circ=\frac{2}{1}=2,\quad\tan x^\circ=\frac{\sqrt3}{1}=\sqrt3.
\displaystyle \therefore (\sec x^\circ-\tan x^\circ)(\sec x^\circ+\tan x^\circ)
\displaystyle =(2-\sqrt3)(2+\sqrt3)=4-3=1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Use the given figure to find:}
\displaystyle \text{(i) }\sin x^\circ\qquad\text{(ii) }\cos y^\circ
\displaystyle \text{(iii) }3\tan x^\circ-2\sin y^\circ+4\cos y^\circ \displaystyle \text{Answer:}
\displaystyle \text{For the larger right-angled triangle, hypotenuse}=17\text{ and perpendicular}=8.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Base}=\sqrt{17^2-8^2}=\sqrt{289-64}=\sqrt{225}=15.
\displaystyle \text{For the smaller right-angled triangle, base}=6\text{ and perpendicular}=8.
\displaystyle \text{Hypotenuse}=\sqrt{6^2+8^2}=\sqrt{100}=10.

\displaystyle \text{(i) }\sin x^\circ=\frac{8}{17}.

\displaystyle \text{(ii) }\cos y^\circ=\frac{6}{10}=\frac35.

\displaystyle \text{(iii) }\tan x^\circ=\frac8{15},\quad\sin y^\circ=\frac8{10}=\frac45,
\displaystyle \cos y^\circ=\frac6{10}=\frac35.
\displaystyle \therefore 3\tan x^\circ-2\sin y^\circ+4\cos y^\circ
\displaystyle =3\left(\frac8{15}\right)-2\left(\frac45\right)+4\left(\frac35\right)
\displaystyle =\frac85-\frac85+\frac{12}{5}=\frac{12}{5}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In the diagram given below, triangle }ABC\text{ is right-angled at }B
\displaystyle \text{and }BD\text{ is perpendicular to }AC.\text{ Find:}
\displaystyle \text{(i) }\cos\angle DBC\qquad\text{(ii) }\cot\angle DBA

\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ABC,\ BC=5\text{ cm and }BA=12\text{ cm}.
\displaystyle AC=\sqrt{BC^2+BA^2}=\sqrt{5^2+12^2}=\sqrt{169}=13\text{ cm}.
\displaystyle \text{Since }BD\perp AC,\ \triangle BDC\sim\triangle ABC\text{ and }\triangle BDA\sim\triangle ABC.

\displaystyle \text{(i) }\angle DBC=\angle A.
\displaystyle \therefore \cos\angle DBC=\cos A=\frac{BA}{AC}=\frac{12}{13}.

\displaystyle \text{(ii) }\angle DBA=\angle C.
\displaystyle \therefore \cot\angle DBA=\cot C=\frac{BC}{BA}=\frac{5}{12}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In the given figure, triangle }ABC\text{ is right-angled at }B.
\displaystyle D\text{ is the foot of the perpendicular from }B\text{ to }AC.\text{ Given that }BC=3\text{ cm}
\displaystyle \text{and }AB=4\text{ cm. Find:}
\displaystyle \text{(i) }\tan\angle DBC\qquad\text{(ii) }\sin\angle DBA

\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ABC,\ AB=4\text{ cm and }BC=3\text{ cm}.
\displaystyle AC=\sqrt{AB^2+BC^2}=\sqrt{4^2+3^2}=\sqrt{25}=5\text{ cm}.
\displaystyle \text{Since }BD\perp AC,\ \triangle BDC\sim\triangle ABC\text{ and }\triangle BDA\sim\triangle ABC.

\displaystyle \text{(i) }\angle DBC=\angle A.
\displaystyle \therefore \tan\angle DBC=\tan A=\frac{BC}{AB}=\frac34.

\displaystyle \text{(ii) }\angle DBA=\angle C.
\displaystyle \therefore \sin\angle DBA=\sin C=\frac{AB}{AC}=\frac45.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In triangle }ABC,\ AB=AC=15\text{ cm and }BC=18\text{ cm,}
\displaystyle \text{find }\cos\angle ABC.
\displaystyle \text{Answer:}
\displaystyle \text{Draw }AD\perp BC.
\displaystyle \text{Since }AB=AC,\ AD\text{ bisects }BC.
\displaystyle \therefore BD=DC=\frac{18}{2}=9\text{ cm}.
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle \cos\angle ABC=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{BD}{AB}
\displaystyle =\frac{9}{15}=\frac35.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In the figure given below, }ABC\text{ is an isosceles triangle with }BC=8\text{ cm}
\displaystyle \text{and }AB=AC=5\text{ cm. Find:}
\displaystyle \text{(i) }\sin B\qquad\text{(ii) }\tan C\qquad\text{(iii) }\sin^2B+\cos^2B
\displaystyle \text{(iv) }\tan C-\cot B \displaystyle \text{Answer:}
\displaystyle \text{Draw }AD\perp BC.
\displaystyle \text{Since }AB=AC,\ AD\text{ bisects }BC.
\displaystyle \therefore BD=DC=\frac{8}{2}=4\text{ cm}.
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle AD=\sqrt{AB^2-BD^2}=\sqrt{5^2-4^2}=\sqrt9=3\text{ cm}.

\displaystyle \text{(i) }\sin B=\frac{AD}{AB}=\frac35.

\displaystyle \text{(ii) }\tan C=\frac{AD}{DC}=\frac34.

\displaystyle \text{(iii) }\cos B=\frac{BD}{AB}=\frac45.
\displaystyle \therefore \sin^2B+\cos^2B=\left(\frac35\right)^2+\left(\frac45\right)^2
\displaystyle =\frac9{25}+\frac{16}{25}=1.

\displaystyle \text{(iv) }\cot B=\frac{BD}{AD}=\frac43.
\displaystyle \therefore \tan C-\cot B=\frac34-\frac43=-\frac7{12}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In triangle }ABC,\ \angle ABC=90^\circ,\ \angle CAB=x^\circ,
\displaystyle \tan x^\circ=\frac34\text{ and }BC=15\text{ cm. Find the measures of }AB\text{ and }AC.
\displaystyle \text{Answer:}
\displaystyle \tan x^\circ=\frac{BC}{AB}=\frac34.
\displaystyle \frac{15}{AB}=\frac34
\displaystyle 3AB=60
\displaystyle \therefore AB=20\text{ cm}.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle AC=\sqrt{AB^2+BC^2}=\sqrt{20^2+15^2}
\displaystyle =\sqrt{400+225}=\sqrt{625}=25\text{ cm}.
\displaystyle \therefore AB=20\text{ cm and }AC=25\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Using the measurements given in the following figure:}
\displaystyle \text{(i) Find the values of }\sin\phi\text{ and }\tan\theta.
\displaystyle \text{(ii) Write an expression for }AD\text{ in terms of }\theta. \displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle BCD,\ BD=13,\ BC=12.
\displaystyle CD=\sqrt{BD^2-BC^2}=\sqrt{13^2-12^2}
\displaystyle =\sqrt{169-144}=\sqrt{25}=5.
\displaystyle \text{Since }AB=14,\text{ the vertical distance between }A\text{ and }D=14-5=9.

\displaystyle \text{(i) }\sin\phi=\frac{CD}{BD}=\frac5{13}.
\displaystyle \tan\theta=\frac{12}{9}=\frac43.

\displaystyle \text{(ii) }\sin\theta=\frac{12}{AD}.
\displaystyle \therefore AD=\frac{12}{\sin\theta}=12\,\mathrm{cosec}\,\theta.
\displaystyle \text{Equivalently, }AD=9\sec\theta.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In the given figure, }BC=15\text{ cm and }\sin B=\frac45.
\displaystyle \text{(i) Calculate the measures of }AB\text{ and }AC.
\displaystyle \text{(ii) Now, if }\tan\angle ADC=1,\text{ calculate the measures of }CD\text{ and }AD.
\displaystyle \text{Also, show that: }\tan^2B-\frac{1}{\cos^2B}=-1. \displaystyle \text{Answer:}
\displaystyle \text{Since }AC\perp BD,\ \triangle ABC\text{ is right-angled at }C.

\displaystyle \text{(i) }\sin B=\frac{AC}{AB}=\frac45.
\displaystyle \text{Let }AC=4x\text{ and }AB=5x.
\displaystyle BC=\sqrt{AB^2-AC^2}=\sqrt{(5x)^2-(4x)^2}=3x.
\displaystyle 3x=15\Rightarrow x=5.
\displaystyle \therefore AC=4\times5=20\text{ cm}.
\displaystyle \therefore AB=5\times5=25\text{ cm}.

\displaystyle \text{(ii) In right-angled }\triangle ACD,
\displaystyle \tan\angle ADC=\frac{AC}{CD}=1.
\displaystyle \frac{20}{CD}=1\Rightarrow CD=20\text{ cm}.
\displaystyle AD=\sqrt{AC^2+CD^2}=\sqrt{20^2+20^2}
\displaystyle =\sqrt{800}=20\sqrt2\text{ cm}.

\displaystyle \text{Also, }\tan B=\frac{AC}{BC}=\frac{20}{15}=\frac43,
\displaystyle \cos B=\frac{BC}{AB}=\frac{15}{25}=\frac35.
\displaystyle \therefore \tan^2B-\frac{1}{\cos^2B}
\displaystyle =\left(\frac43\right)^2-\frac{1}{(3/5)^2}
\displaystyle =\frac{16}{9}-\frac{25}{9}=-1.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }\sin A+\mathrm{cosec}\,A=2,\text{ find the value of}
\displaystyle \sin^2A+\mathrm{cosec}^2A.
\displaystyle \text{Answer:}
\displaystyle \sin A+\mathrm{cosec}\,A=2.
\displaystyle \text{Squaring both sides,}
\displaystyle \sin^2A+\mathrm{cosec}^2A+2\sin A\,\mathrm{cosec}\,A=4.
\displaystyle \text{Since }\sin A\,\mathrm{cosec}\,A=1,
\displaystyle \sin^2A+\mathrm{cosec}^2A+2=4.
\displaystyle \therefore \sin^2A+\mathrm{cosec}^2A=2.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }\tan A+\cot A=5,\text{ find the value of}
\displaystyle \tan^2A+\cot^2A.
\displaystyle \text{Answer:}
\displaystyle \tan A+\cot A=5.
\displaystyle \text{Squaring both sides,}
\displaystyle \tan^2A+\cot^2A+2\tan A\cot A=25.
\displaystyle \text{Since }\tan A\cot A=1,
\displaystyle \tan^2A+\cot^2A+2=25.
\displaystyle \therefore \tan^2A+\cot^2A=23.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Given: }4\sin\theta=3\cos\theta,\text{ find the value of:}
\displaystyle \text{(i) }\sin\theta\qquad\text{(ii) }\cos\theta\qquad\text{(iii) }\cot^2\theta-\mathrm{cosec}^2\theta
\displaystyle \text{(iv) }4\cos^2\theta-3\sin^2\theta+2
\displaystyle \text{Answer:}
\displaystyle 4\sin\theta=3\cos\theta.
\displaystyle \frac{\sin\theta}{\cos\theta}=\frac34.
\displaystyle \therefore \tan\theta=\frac34=\frac{\text{Perpendicular}}{\text{Base}}.
\displaystyle \text{Let perpendicular }=3x\text{ and base }=4x.
\displaystyle \text{By Pythagoras theorem, hypotenuse}=\sqrt{(3x)^2+(4x)^2}=5x.

\displaystyle \text{(i) }\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{3x}{5x}=\frac35.

\displaystyle \text{(ii) }\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{4x}{5x}=\frac45.

\displaystyle \text{(iii) }\cot\theta=\frac43,\quad\mathrm{cosec}\,\theta=\frac53.
\displaystyle \therefore \cot^2\theta-\mathrm{cosec}^2\theta=\left(\frac43\right)^2-\left(\frac53\right)^2
\displaystyle =\frac{16}{9}-\frac{25}{9}=-1.

\displaystyle \text{(iv) }4\cos^2\theta-3\sin^2\theta+2
\displaystyle =4\left(\frac45\right)^2-3\left(\frac35\right)^2+2
\displaystyle =\frac{64}{25}-\frac{27}{25}+\frac{50}{25}
\displaystyle =\frac{87}{25}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Given: }17\cos\theta=15,\text{ find the value of }\tan\theta+2\sec\theta.
\displaystyle \text{Answer:}
\displaystyle 17\cos\theta=15\Rightarrow\cos\theta=\frac{15}{17}=\frac{\text{Base}}{\text{Hypotenuse}}.
\displaystyle \text{Let base }=15x\text{ and hypotenuse }=17x.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle \text{Perpendicular}=\sqrt{(17x)^2-(15x)^2}=\sqrt{64x^2}=8x.
\displaystyle \therefore \tan\theta=\frac{8}{15},\quad\sec\theta=\frac{17}{15}.
\displaystyle \therefore \tan\theta+2\sec\theta=\frac8{15}+2\left(\frac{17}{15}\right)
\displaystyle =\frac{8+34}{15}=\frac{42}{15}=\frac{14}{5}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Given: }5\cos A-12\sin A=0,\text{ evaluate:}
\displaystyle \frac{\sin A+\cos A}{2\cos A-\sin A}
\displaystyle \text{Answer:}
\displaystyle 5\cos A-12\sin A=0.
\displaystyle 5\cos A=12\sin A.
\displaystyle \frac{\sin A}{\cos A}=\frac5{12}.
\displaystyle \therefore \tan A=\frac5{12}=\frac{\text{Perpendicular}}{\text{Base}}.
\displaystyle \text{Let perpendicular }=5x\text{ and base }=12x.
\displaystyle \text{By Pythagoras theorem, hypotenuse}=\sqrt{(5x)^2+(12x)^2}=13x.
\displaystyle \therefore \sin A=\frac5{13},\quad\cos A=\frac{12}{13}.
\displaystyle \therefore \frac{\sin A+\cos A}{2\cos A-\sin A}
\displaystyle =\frac{\frac5{13}+\frac{12}{13}}{2\left(\frac{12}{13}\right)-\frac5{13}}
\displaystyle =\frac{17/13}{19/13}=\frac{17}{19}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In the given figure, }\angle C=90^\circ\text{ and }D\text{ is the midpoint of }AC.
\displaystyle \text{Find:}
\displaystyle \text{(i) }\frac{\tan\angle CAB}{\tan\angle CDB}\qquad\text{(ii) }\frac{\tan\angle ABC}{\tan\angle DBC} \displaystyle \text{Answer:}
\displaystyle \text{Since }D\text{ is the midpoint of }AC,\ AD=DC.
\displaystyle \therefore AC=AD+DC=2DC.

\displaystyle \text{(i) }\tan\angle CAB=\frac{BC}{AC},\quad\tan\angle CDB=\frac{BC}{DC}.
\displaystyle \therefore \frac{\tan\angle CAB}{\tan\angle CDB}
\displaystyle =\frac{BC}{AC}\times\frac{DC}{BC}=\frac{DC}{AC}=\frac12.

\displaystyle \text{(ii) }\tan\angle ABC=\frac{AC}{BC},\quad\tan\angle DBC=\frac{DC}{BC}.
\displaystyle \therefore \frac{\tan\angle ABC}{\tan\angle DBC}
\displaystyle =\frac{AC}{BC}\times\frac{BC}{DC}=\frac{AC}{DC}=2.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }3\cos A=4\sin A,\text{ find the value of:}
\displaystyle \text{(i) }\cos A\qquad\text{(ii) }3-\cot^2A+\mathrm{cosec}^2A
\displaystyle \text{Answer:}
\displaystyle 3\cos A=4\sin A.
\displaystyle \frac{\sin A}{\cos A}=\frac34.
\displaystyle \therefore \tan A=\frac34=\frac{\text{Perpendicular}}{\text{Base}}.
\displaystyle \text{Let perpendicular }=3x\text{ and base }=4x.
\displaystyle \text{By Pythagoras theorem, hypotenuse}=\sqrt{(3x)^2+(4x)^2}=5x.

\displaystyle \text{(i) }\cos A=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{4x}{5x}=\frac45.

\displaystyle \text{(ii) }\cot A=\frac43,\quad\mathrm{cosec}\,A=\frac53.
\displaystyle \therefore 3-\cot^2A+\mathrm{cosec}^2A
\displaystyle =3-\left(\frac43\right)^2+\left(\frac53\right)^2
\displaystyle =3-\frac{16}{9}+\frac{25}{9}=3+1=4.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In triangle }ABC,\ \angle B=90^\circ\text{ and }\tan A=0.75.
\displaystyle \text{If }AC=30\text{ cm, find the lengths of }AB\text{ and }BC.
\displaystyle \text{Answer:}
\displaystyle \tan A=0.75=\frac34=\frac{BC}{AB}.
\displaystyle \text{Let }BC=3x\text{ and }AB=4x.
\displaystyle \text{By Pythagoras theorem,}
\displaystyle AC=\sqrt{(3x)^2+(4x)^2}=5x.
\displaystyle 5x=30\Rightarrow x=6.
\displaystyle \therefore AB=4\times6=24\text{ cm}.
\displaystyle \therefore BC=3\times6=18\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In rhombus }ABCD,\text{ diagonals }AC\text{ and }BD\text{ intersect each other at point }O.
\displaystyle \text{If cosine of angle }CAB\text{ is }0.6\text{ and }OB=8\text{ cm, find the lengths of the side}
\displaystyle \text{and the diagonals of the rhombus.}
\displaystyle \text{Answer:}
\displaystyle \text{The diagonals of a rhombus bisect each other at right angles.}
\displaystyle \therefore \angle AOB=90^\circ.
\displaystyle \cos\angle CAB=0.6=\frac35.
\displaystyle \cos\angle CAB=\frac{AO}{AB}=\frac35.
\displaystyle \text{Let }AO=3x\text{ and }AB=5x.
\displaystyle \text{By Pythagoras theorem, }OB=\sqrt{(5x)^2-(3x)^2}=4x.
\displaystyle 4x=8\Rightarrow x=2.
\displaystyle \therefore AO=3\times2=6\text{ cm and }AB=5\times2=10\text{ cm}.
\displaystyle \therefore AC=2AO=2\times6=12\text{ cm}.
\displaystyle \therefore BD=2OB=2\times8=16\text{ cm}.
\displaystyle \therefore \text{Side of the rhombus}=10\text{ cm, }AC=12\text{ cm and }BD=16\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{In triangle }ABC,\ AB=AC=15\text{ cm and }BC=18\text{ cm. Find:}
\displaystyle \text{(i) }\cos B\qquad\text{(ii) }\sin C\qquad\text{(iii) }\tan^2B-\sec^2B+2
\displaystyle \text{Answer:}
\displaystyle \text{Draw }AD\perp BC.
\displaystyle \text{Since }AB=AC,\ AD\text{ bisects }BC.
\displaystyle \therefore BD=DC=\frac{18}{2}=9\text{ cm}.
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle AD=\sqrt{AB^2-BD^2}=\sqrt{15^2-9^2}
\displaystyle =\sqrt{225-81}=\sqrt{144}=12\text{ cm}.

\displaystyle \text{(i) }\cos B=\frac{BD}{AB}=\frac9{15}=\frac35.

\displaystyle \text{(ii) }\sin C=\frac{AD}{AC}=\frac{12}{15}=\frac45.

\displaystyle \text{(iii) }\tan B=\frac{AD}{BD}=\frac{12}{9}=\frac43,
\displaystyle \sec B=\frac{AB}{BD}=\frac{15}{9}=\frac53.
\displaystyle \therefore \tan^2B-\sec^2B+2
\displaystyle =\left(\frac43\right)^2-\left(\frac53\right)^2+2
\displaystyle =\frac{16}{9}-\frac{25}{9}+\frac{18}{9}=1.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{In triangle }ABC,\ AD\text{ is perpendicular to }BC.\ \sin B=0.8,\ BD=9\text{ cm}
\displaystyle \text{and }\tan C=1.\text{ Find the lengths of }AB,\ AD,\ AC\text{ and }DC.
\displaystyle \text{Answer:}
\displaystyle \sin B=0.8=\frac45=\frac{AD}{AB}.
\displaystyle \text{Let }AD=4x\text{ and }AB=5x.
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle BD=\sqrt{AB^2-AD^2}=\sqrt{(5x)^2-(4x)^2}=3x.
\displaystyle 3x=9\Rightarrow x=3.
\displaystyle \therefore AD=4\times3=12\text{ cm and }AB=5\times3=15\text{ cm}.
\displaystyle \tan C=\frac{AD}{DC}=1.
\displaystyle \frac{12}{DC}=1\Rightarrow DC=12\text{ cm}.
\displaystyle \text{In right-angled }\triangle ADC,
\displaystyle AC=\sqrt{AD^2+DC^2}=\sqrt{12^2+12^2}
\displaystyle =\sqrt{288}=12\sqrt2\text{ cm}.
\displaystyle \therefore AB=15\text{ cm, }AD=12\text{ cm, }AC=12\sqrt2\text{ cm and }DC=12\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Given: }q\tan A=p,\text{ find the value of:}
\displaystyle \frac{p\sin A-q\cos A}{p\sin A+q\cos A}
\displaystyle \text{Answer:}
\displaystyle q\tan A=p\Rightarrow\tan A=\frac{p}{q}.
\displaystyle \frac{p\sin A-q\cos A}{p\sin A+q\cos A}
\displaystyle =\frac{p\frac{\sin A}{\cos A}-q}{p\frac{\sin A}{\cos A}+q}
\displaystyle =\frac{p\tan A-q}{p\tan A+q}
\displaystyle =\frac{p\left(\frac pq\right)-q}{p\left(\frac pq\right)+q}
\displaystyle =\frac{\frac{p^2-q^2}{q}}{\frac{p^2+q^2}{q}}
\displaystyle =\frac{p^2-q^2}{p^2+q^2}.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If }\sin A=\cos A,\text{ find the value of}
\displaystyle 2\tan^2A-2\sec^2A+5.
\displaystyle \text{Answer:}
\displaystyle \sin A=\cos A.
\displaystyle \frac{\sin A}{\cos A}=1.
\displaystyle \therefore \tan A=1.
\displaystyle \sec^2A=1+\tan^2A=1+1=2.
\displaystyle \therefore 2\tan^2A-2\sec^2A+5
\displaystyle =2(1)^2-2(2)+5
\displaystyle =2-4+5=3.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{In rectangle }ABCD,\text{ diagonal }BD=26\text{ cm and cotangent of}
\displaystyle \text{angle }ABD=1.5.\text{ Find the area and the perimeter of rectangle }ABCD.
\displaystyle \text{Answer:}
\displaystyle \cot\angle ABD=1.5=\frac32=\frac{AB}{AD}.
\displaystyle \text{Let }AB=3x\text{ and }AD=2x.
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle BD^2=AB^2+AD^2.
\displaystyle 26^2=(3x)^2+(2x)^2
\displaystyle 676=9x^2+4x^2=13x^2
\displaystyle x^2=52\Rightarrow x=2\sqrt{13}.
\displaystyle \therefore AB=6\sqrt{13}\text{ cm and }AD=4\sqrt{13}\text{ cm}.
\displaystyle \text{Area of rectangle }ABCD=AB\times AD
\displaystyle =6\sqrt{13}\times4\sqrt{13}=312\text{ cm}^2.
\displaystyle \text{Perimeter of rectangle }ABCD=2(AB+AD)
\displaystyle =2(6\sqrt{13}+4\sqrt{13})=20\sqrt{13}\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If }2\sin x=\sqrt3,\text{ evaluate:}
\displaystyle \text{(i) }4\sin^3x-3\sin x\qquad\text{(ii) }3\cos x-4\cos^3x
\displaystyle \text{Answer:}
\displaystyle 2\sin x=\sqrt3\Rightarrow\sin x=\frac{\sqrt3}{2}.
\displaystyle \text{Let perpendicular }=\sqrt3a\text{ and hypotenuse }=2a.
\displaystyle \text{By Pythagoras theorem, base}=\sqrt{(2a)^2-(\sqrt3a)^2}=a.
\displaystyle \therefore \cos x=\frac{a}{2a}=\frac12.
\displaystyle \text{(i) }4\sin^3x-3\sin x
\displaystyle =4\left(\frac{\sqrt3}{2}\right)^3-3\left(\frac{\sqrt3}{2}\right)
\displaystyle =\frac{3\sqrt3}{2}-\frac{3\sqrt3}{2}=0.
\displaystyle \text{(ii) }3\cos x-4\cos^3x
\displaystyle =3\left(\frac12\right)-4\left(\frac12\right)^3
\displaystyle =\frac32-\frac12=1.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If }\sin A=\frac{\sqrt3}{2}\text{ and }\cos B=\frac{\sqrt3}{2},\text{ find the value of:}
\displaystyle \frac{\tan A-\tan B}{1+\tan A\tan B}
\displaystyle \text{Answer:}
\displaystyle \sin A=\frac{\sqrt3}{2}.
\displaystyle \text{Let perpendicular }=\sqrt3a\text{ and hypotenuse }=2a.
\displaystyle \text{By Pythagoras theorem, base}=a.
\displaystyle \therefore \tan A=\frac{\sqrt3a}{a}=\sqrt3.
\displaystyle \cos B=\frac{\sqrt3}{2}.
\displaystyle \text{Let base }=\sqrt3b\text{ and hypotenuse }=2b.
\displaystyle \text{By Pythagoras theorem, perpendicular}=b.
\displaystyle \therefore \tan B=\frac{b}{\sqrt3b}=\frac1{\sqrt3}.
\displaystyle \therefore \frac{\tan A-\tan B}{1+\tan A\tan B}
\displaystyle =\frac{\sqrt3-\frac1{\sqrt3}}{1+\sqrt3\left(\frac1{\sqrt3}\right)}
\displaystyle =\frac{\frac2{\sqrt3}}{2}=\frac1{\sqrt3}=\frac{\sqrt3}{3}.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Use the information given in the following figure to evaluate:}
\displaystyle \frac{10}{\sin x}+\frac{6}{\sin y}-6\cot y. \displaystyle \text{Answer:}
\displaystyle \text{Let }AD\perp BC.\text{ Given }AD=12,\ AC=20,\ BC=21.
\displaystyle \text{In right-angled }\triangle ADC,
\displaystyle DC=\sqrt{AC^2-AD^2}=\sqrt{20^2-12^2}
\displaystyle =\sqrt{400-144}=\sqrt{256}=16.
\displaystyle BD=BC-DC=21-16=5.
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle AB=\sqrt{AD^2+BD^2}=\sqrt{12^2+5^2}
\displaystyle =\sqrt{169}=13.
\displaystyle \therefore \sin x=\frac{BD}{AB}=\frac5{13}.
\displaystyle \sin y=\frac{AD}{AC}=\frac{12}{20}=\frac35.
\displaystyle \cot y=\frac{DC}{AD}=\frac{16}{12}=\frac43.
\displaystyle \therefore \frac{10}{\sin x}+\frac{6}{\sin y}-6\cot y
\displaystyle =\frac{10}{5/13}+\frac{6}{3/5}-6\left(\frac43\right)
\displaystyle =26+10-8=28.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{If }\sec A=\sqrt2,\text{ find:}
\displaystyle \frac{3\cot^2A+2\sin^2A}{\tan^2A-\cos^2A}.
\displaystyle \text{Answer:}
\displaystyle \sec A=\sqrt2\Rightarrow\cos A=\frac{1}{\sqrt2}.
\displaystyle \therefore \cos^2A=\frac12.
\displaystyle \sin^2A=1-\cos^2A=1-\frac12=\frac12.
\displaystyle \tan^2A=\frac{\sin^2A}{\cos^2A}=\frac{1/2}{1/2}=1.
\displaystyle \cot^2A=\frac{1}{\tan^2A}=1.
\displaystyle \therefore \frac{3\cot^2A+2\sin^2A}{\tan^2A-\cos^2A}
\displaystyle =\frac{3(1)+2\left(\frac12\right)}{1-\frac12}
\displaystyle =\frac{4}{1/2}=8.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{If }5\cos\theta=3,\text{ evaluate:}
\displaystyle \frac{\mathrm{cosec}\,\theta-\cot\theta}{\mathrm{cosec}\,\theta+\cot\theta}.
\displaystyle \text{Answer:}
\displaystyle 5\cos\theta=3\Rightarrow\cos\theta=\frac35=\frac{\text{Base}}{\text{Hypotenuse}}.
\displaystyle \text{Let base }=3x\text{ and hypotenuse }=5x.
\displaystyle \text{By Pythagoras theorem, perpendicular}=\sqrt{(5x)^2-(3x)^2}=4x.
\displaystyle \therefore \mathrm{cosec}\,\theta=\frac54,\quad\cot\theta=\frac34.
\displaystyle \therefore \frac{\mathrm{cosec}\,\theta-\cot\theta}{\mathrm{cosec}\,\theta+\cot\theta}
\displaystyle =\frac{\frac54-\frac34}{\frac54+\frac34}
\displaystyle =\frac{2/4}{8/4}=\frac14.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{If }\mathrm{cosec}\,A+\sin A=5\frac15,\text{ find the value of}
\displaystyle \mathrm{cosec}^2A+\sin^2A.
\displaystyle \text{Answer:}
\displaystyle \mathrm{cosec}\,A+\sin A=5\frac15=\frac{26}{5}.
\displaystyle \text{Squaring both sides,}
\displaystyle \mathrm{cosec}^2A+\sin^2A+2\,\mathrm{cosec}\,A\sin A=\frac{676}{25}.
\displaystyle \text{Since }\mathrm{cosec}\,A\sin A=1,
\displaystyle \mathrm{cosec}^2A+\sin^2A+2=\frac{676}{25}.
\displaystyle \therefore \mathrm{cosec}^2A+\sin^2A=\frac{676}{25}-2
\displaystyle =\frac{676-50}{25}=\frac{626}{25}.
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{If }5\cos\theta=6\sin\theta,\text{ evaluate:}
\displaystyle \text{(i) }\tan\theta\qquad\text{(ii) }\frac{12\sin\theta-3\cos\theta}{12\sin\theta+3\cos\theta}
\displaystyle \text{Answer:}
\displaystyle 5\cos\theta=6\sin\theta.

\displaystyle \text{(i) }\frac{\sin\theta}{\cos\theta}=\frac56.
\displaystyle \therefore \tan\theta=\frac56.

\displaystyle \text{(ii) }\frac{12\sin\theta-3\cos\theta}{12\sin\theta+3\cos\theta}
\displaystyle =\frac{12\frac{\sin\theta}{\cos\theta}-3}{12\frac{\sin\theta}{\cos\theta}+3}
\displaystyle =\frac{12\tan\theta-3}{12\tan\theta+3}
\displaystyle =\frac{12\left(\frac56\right)-3}{12\left(\frac56\right)+3}
\displaystyle =\frac{10-3}{10+3}=\frac7{13}.
\displaystyle \\

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