ML Aggarwal Solution Class 9 Chapter 17 Trigonometric Ratios Exercise 17.1

 Exercise 17.1

Question 1

Sol :

(a)

By Pythagoras theorem

$O P^{2}=O M^{2}+P M^{2}$

$15^{2}=12^{2}+P M^{2}$

$225=144+\mathrm{PM}^{2}$

$P M^{2}=225-144$

$P M^{2}=81$

$P M=\sqrt{81}$

PM=9 

(i) $\sin \theta=\frac{P M}{O P}$

$=\frac{9}{15}$

$\sin \theta=\frac{3}{5}$

(ii) $\cos \theta=\frac{O M}{O P}$

$\cos \theta=\frac{12}{15}$

$\cos \theta=\frac{4}{5}$

(iii) $\operatorname{tan} \theta=\frac{P M}{O M}$

$=\frac{9}{12}$

$\tan \theta=\frac{3}{4}$

(iv) $\cot \theta=\frac{O M}{P M}$

$=\frac{12}{9}$

$\cot \theta=\frac{4}{3}$

(v) $\sec \theta=\frac{O P}{O M}$

$=\frac{15}{12}$

$\sec \theta=\frac{5}{4}$

(vi) $\operatorname{cosec} \theta=\frac{OP}{P M}$

$=\frac{15}{9}$

$\operatorname{cosec} \theta=\frac{5}{3}$

(b) By Pythagoras theorem 

$A B^{2}=A C^{2}+B C^{2}$

$=12^{2}+5^{2}$

=144+25

$A B^{2}=169$

$A B=\sqrt{169}$

AB=13

(i) Sin A$=\frac{B C}{A B}$

Sin A$=\frac{5}{13}$

(ii) Cos A$=\frac{A C}{A B}$

Cos A$=\frac{12}{13}$

(iii) $\sin ^{2} A+\cos ^{2} A=\left(\frac{5}{13}\right)^{2}+\left(\frac{12}{13}\right)^{2}$

$=\frac{25}{169}+\frac{144}{169}$

$=\frac{25+144}{169}$

$=\frac{169}{169}$

=1

$\therefore \sin ^{2} A+\cos ^{2} A=1$

(iv) $\sec ^{2} A-\tan ^{2} A$

$\because \sec A=\frac{1}{\cos A}$

$=\frac{1}{\frac{12}{13}}$

$\because \tan A=\frac{\sin A}{\cos A}$

$=\frac{\left(\frac{5}{13}\right)}{\left(\frac{12}{13}\right)}$

$\operatorname{TanA}=\frac{5}{12}$

$\therefore \quad \sec ^{2} A-\tan ^{2} A=\left(\frac{13}{12}\right)^{2}-\left(\frac{5}{12}\right)^{2}$

$=\frac{169}{144}-\frac{25}{144}$

$=\frac{169-25}{144}$

$=\frac{144}{144}$

=1

$=\sec ^{2} A-\operatorname{tan}^{2} A=1$

Question 2

Sol :

(a) 

By Pythagoras theorem

$\because$ hypotenuse $=B C$

$\therefore B C^{2}=A B^{2}+A C^{2}$

$10^{2}=6^{2}+A C^{2}$

$100=36+A C^{2}$

$A C^{2}=100-36$

$A C^{2}=64$

$A C=\sqrt{64}$

AC=8

(i) Sin B$=\frac{A C}{B C}$

$=\frac{8}{10}$

Sin B$=\frac{4}{5}$

(ii) Cos C$=\frac{A C}{B C}$

$=\frac{8}{10}$

$\cos C=\frac{4}{5}$

(iii) Sin B+ Sin C

$\because \sin C=\frac{A B}{B C}$

$=\frac{6}{10}$

$=\frac{3}{5}$

$\therefore \sin B=\frac{4}{5}$

$\therefore \quad \sin B+\sin C=\frac{4}{5}+\frac{3}{5}$

$=\frac{4+3}{5}$

Sin B + Sin C$=\frac{7}{5}$

(iv) Sin B. Cos C+ Sin C. cos B

Cos B$=\frac{A B}{B C}=\frac{6}{10}$

$=\frac{3}{5}$

$\therefore \sin B \cos C+\sin C \cdot \cos B$

$\Rightarrow \frac{4}{5} \cdot \frac{4}{5}+\frac{3}{5} \cdot \frac{3}{5}$

$\Rightarrow \frac{16}{25}+\frac{9}{25}$

$\Rightarrow \frac{16+9}{25}$

$\Rightarrow \frac{25}{25}$

⇒1

(b) From Figure ΔADC

$A D^{2}+C D^{2}=A C^{2}$

$A D^{2}+5^{2}=13^{2}$

$A D^{2}+25=169$

$A D^{2}=169-25$

$A D^{2}=144$

AD=√144=12

From figure ΔABD

⇒BD=BC-CD

=21-5

BD=16

∴$A B^{2}=B D^{2}+A D^{2}$

$A B^{2}=16^{2}+12^{2}$

$A B^{2}=256+144$

$A B^{2}=400$

$A B=\sqrt{400}$

AB=20

∴AB=20, AD=12

BD=16

AC=13

CD=5

(i) tan x$=\frac{C D}{A D}$

$=\frac{5}{12}$

(ii) cos y $=\frac{B D}{A B}$

$=\frac{16}{20}$

Cos y$=\frac{4}{5}$

(iii) $\operatorname{cosec}^{2} y-\cot ^{2} y$

$\Rightarrow \operatorname{cosec} y=\frac{A B}{A D}$

$=\frac{20}{12}$

$\Rightarrow  \cot y=\frac{B D}{A D}$

$=\frac{16}{12}$

$\cot y=\frac{4}{3}$

$\therefore \operatorname{cosec}^{2} y-\cot ^{2} y=\left(\frac{5}{3}\right)^{2}-\left(\frac{4}{3}\right)^{2}$

$=\frac{25}{9}-\frac{16}{9}$

$=\frac{25-16}{9}$

$=\frac{9}{9}$

$\therefore \operatorname{cosec}^{2} y-\cot ^{2} y=1$

Question 3

Sol :

From figure

BD=BC-CD

BD=21-5=16

From ΔADC

$A D^{2}+D C^{2}=A C^{2}$

$A D^{2}+5^{2}=13^{2}$

$A D^{2}+25=169$

$A D^{2}=169-25$

$A D^{2}=144$

$A D=\sqrt{144}$

AD=12

$\therefore  \operatorname{Sec} \theta=\frac{A B}{B D}$

From figure ΔABD

$A B^{2}=B D^{2}+A D^{2}$

$=16^{2}+12^{2}$

=256+144=400

$A B^{2}=400$

$A B=\sqrt{400}$

AB=20

$\therefore \sec \theta=\frac{20}{16}$

$=\frac{5}{4}$

(b) From figure ΔABC

$A C^{2}=A B^{2}+B C^{2}$

$=3^{2}+4^{2}$

=9+16

$A C^{2}=2 b$

$A C=\sqrt{25}$

AC=5

From figure ΔBCD

$C D^{2}=B C^{2}+B D^{2}$

$12^{2}=4^{2}+B D^{2}$

$144=16+B D^{2}$

$B D^{2}=144-16$

=128

$B D^{2}=128$

$B D=\sqrt{128}$

$=\sqrt{16 \times 4 \times 2}$

$=4 \times 2 \sqrt{2}$

$B D=8 \sqrt{2}$

(i) Sin x$=\frac{B C}{A C}$

$\sin x=\frac{4}{5}$

(ii) $\cot x=\frac{A B}{B C}$

$=\frac{3}{4}$

(iii) $\cot ^{2} x-\operatorname{cosec}^{2} x$

$\because \cot x=\frac{3}{4}$

$\operatorname{cosec} x=\frac{A C}{B C}$

$=\frac{5}{4}$

$\therefore \cot ^{2} x-\operatorname{cosec}^{2} x=\left(\frac{3}{4}\right)^{2}-\left(\frac{5}{4}\right)^{2}$

$=\frac{9}{16}-\frac{25}{4}$

$\therefore \cot ^{2} x-\operatorname{cosec}^{2} x=\frac{9-25}{16}$

$=\frac{-16}{16}$

$=-1$

(iv) $\sec y=\frac{C D}{B D}$

$=\frac{12}{8 \sqrt{2}}$

secy $=\frac{3}{2 \sqrt{2}}$

(v) $\tan ^{2} y-\frac{1}{\cos ^{2} y}$

$\because \sec ^{2} y=\frac{1}{\cos ^{2} y}$

$=\left(\frac{3}{2 \sqrt{2}}\right)^{2}$

$=\frac{9}{4 \times 2}$

$\frac{1}{\cos ^{2} y}=\frac{9}{8}$

$\therefore \operatorname{Tan}^{2} y=\left(\frac{B C}{B D}\right)^{2}$

$=\left(\frac{4}{8 \sqrt{2}}\right)^{2}$

$=\frac{16}{64 \times 2}$

$\operatorname{Tan}^{2} y=\frac{1}{8}$

$\therefore \tan ^{2} y-\frac{1}{\cos ^{2} y}=\frac{1}{8}-\frac{9}{8}$

$=\frac{1-9}{8}$

$=\frac{-8}{8}$

=-1

Question 4

Sol :

(a) 

From figure ΔBCD

$B C^{2}=B D^{2}+C D^{2}$

$B^{2}=9^{2}+12^{2}$

=81+144

$B C^{2}=225$

$B C=\sqrt{225}$

BC=15

ΔABC⇒$A B^{2}+B C^{2}=A C^{2}$

$A B^{2}+15^{2}=25^{2}$

$A B^{2}=625-225$

$A B^{2}=400$

$A B=\sqrt{400}=20$

(i) 2sin y-cos y

$\because \sin y=\frac{B D}{B C}$

$=\frac{9}{15}$

$\therefore$ Siny $=\frac{3}{5}$

$\cos y=\frac{CD}{BC}$

$=\frac{12}{15}$

$\therefore \cos y=\frac{4}{5}$

$\therefore 2 \sin y-\cos y=2 \cdot \frac{3}{5}-\frac{4}{5}$

$=\frac{6}{5}-\frac{4}{5}$

$=\frac{6-4}{5}=\frac{2}{5}$

(ii) 2sin x-cos x

$\sin x=\frac{B C}{A C}=\frac{15}{25}=\frac{3}{5}$

$\cos x=\frac{A B}{A C}=\frac{20}{25}=\frac{4}{5}$

$\therefore  2 \sin x-\cos x=2 \cdot \frac{3}{5}-\frac{4}{5}$

$=\frac{6}{5}-\frac{4}{5}$

$=\frac{6-4}{5}=\frac{2}{5}$

(iii) 1-sin x+cos y

$\because \sin x=\frac{3}{5}$

$\cos y=\frac{4}{5}$

$\therefore 1-\sin x+\cos y$

$=1-\frac{3}{5}+\frac{4}{5}$

$=\frac{5-3+4}{5}$

$\frac{6}{5}$

(iv) 2cosx-3siny+4tanx 

$\therefore \sin x=\frac{3}{5}$

$\cos x=\frac{4}{5}$

$\sin y=\frac{3}{5}$

$\therefore \operatorname{tan} x=\frac{\sin x}{\cos x}$

$=\frac{\frac{3}{5}}{\frac{4}{5}}$

$\tan x=\frac{3}{4}$

∴2cosx-3siny+4tanx

$\Rightarrow 2 \cdot \frac{4}{5}-3 \cdot \frac{3}{5}+4 \cdot \frac{3}{4}$

$\Rightarrow  \frac{8}{5}-\frac{9}{5}+3$

$\Rightarrow  \frac{8-9+15}{5}$

$\Rightarrow \frac{14}{5}$

(b) By Pythagoras theorem

(ii) $A C^{2}=B C^{2}+A B^{2}$

$5^{2}=3^{2}+A B^{2}$

$25=9+A B^{2}$

$A B^{2}=25-9$

$A B^{2}=16$

$A B=\sqrt{16}$

AB=4

∴AB=y=4

(i) Sinx $=\frac{B C}{A C}$

$\sin x=\frac{3}{5}$

Question 5

Sol :

Given : $\operatorname{Tan} A=\frac{y z}{x y}=\frac{5}{12}$

By Pythagoras theorem

$x y^{2}+y z^{2}=x z^{2}$

$12^{2}+5^{2}=x z^{2}$

∴144+25=$xz^{2}$

$xz^{2}=169$

xz=13

(i) cos A$=\frac{x y}{xz}$

$=\frac{12}{13}$

(ii) cosec A-cot A

$\operatorname{cosec} A=\frac{xz}{yz}$

$=\frac{13}{5}$

cot A$=\frac{x y}{y z}$

$=\frac{12}{5}$

∴cosec A-cot A$=\frac{13}{5}-\frac{12}{5}$

$=\frac{13-12}{5}=\frac{1}{5}$

Question 6

Sol :

(a) Given : AB=7

BC-AC=1

By Pythagoras theorem

$B C^{2}=A B^{2}+A C^{2}$

∵BC=1+AC

$\therefore(1+A C)^{2}=7^{2}+A C^{2}$

$1+A C^{2}+2 A C=49+AC^{2}$

2AC=49-1

2AC=48

AC=24

(i) Sin C$=\frac{A B}{B C}$

∵From figure 

BC-AC=1

BC-24=1

BC=1+24

BC=25

Sin C$=\frac{7}{25}$

(i) tan B$=\frac{A C}{A B}$

$=\frac{24}{7}$

(b)

Given PQ=40

PR+QR=50

By Pythagoras theorem

$P R^{2}=P Q^{2}+Q R^{2}$

$(50-Q R)^{2}=P Q^{2}+Q R^{2}$

$2500+Q R^{2}-100 Q R=40^{2}+Q R^{2}$

2500-1600=100QR

100QR=900

$Q R=\frac{900}{100}$

QR=9

∴Given 

PR+QR=50

PR+9=50

PR=50-9=41

(i) Sin P$=\frac{Q R}{P R}$

$=\frac{9}{41}$

(ii) Cos P $=\frac{P Q}{P R}$

$=\frac{40}{41}$

(iii) tan R$=\frac{\sin R}{\cos R} .$

∵$\operatorname{Sin} R=\frac{P Q}{P R}$

$=\frac{40}{41}$

Cos R$=\frac{Q R}{P R}$

$=\frac{9}{41}$

∴tan R $=\frac{\frac{40}{41}}{\frac{9}{41}}$

$=\frac{40}{9}$

∴tan R$=\frac{40}{9}$

Question 7

Sol :

(i)

cos < ABC 

$=\frac{B D}{B A}$

$=\frac{9}{15}=\frac{3}{5}$

(ii)

sin < ACB$=\frac{A D}{A C}$

$=\frac{12}{15}=\frac{4}{5}$

Question 8

Sol :

(a) 

Given : AB=AC=5 cm

BC=6

(i) sin C$=\frac{A B}{A C}$

$=\frac{4}{5}$

(ii) tan B$=\frac{A D}{B D}$

$=\frac{4}{3}$

(iii) tan C-cot B

∵tan C $=\frac{A D}{D C}=\frac{4}{3}$

cot B $=\frac{B D}{A D}=\frac{3}{4}$

⇒tan C-cot B$=\frac{4}{3}-\frac{3}{4}$

$=\frac{16-9}{12}=\frac{7}{12}$

(b) 

Given AB=2

BC=1

∴sin θ$=\frac{A B}{A C}$

tan θ$=\frac{A B}{B C}$

∴By Pythagoras theorem

$A C^{2}=A B^{2}+B C^{2}$

$A C^{2}=2^{2}+1^{2}$

$A C^{2}=4+1$

$A C^{2}=5$

$A C=\sqrt{5}$

∴$\sin ^{2} \theta=\left(\frac{A B}{A C}\right)^{2}=\left(\frac{2}{\sqrt{5}}\right)^{2}=\frac{4}{5}$

$\operatorname{tan}^{2} \theta=\left(\frac{A B}{B C}\right)^{2}=\left(\frac{2}{1}\right)^{2}=4$

$\therefore \sin ^{2} \theta+\tan ^{2} \theta=\frac{4}{5}+4$

$=\frac{4+20}{5}$

$=\frac{24}{5}$

$\sin ^{2} \theta+\tan ^{2} \theta=4 \frac{4}{5}$

(c) 

Given BD=15

Sin B$=\frac{4}{5}$

tan C=1

$\because \quad \sin B=\frac{4}{5} \times \frac{5}{5}$

$\operatorname{Sin} B=\frac{A D}{A B}=\frac{20}{25}$

(i) AD=20

$\because \tan C \Rightarrow \frac{\sin C}{\cos C}=1$

$\Rightarrow \frac{A D}{D C}=1$

∴AD=DC

∴DC=20

In ΔACD

By Pythagoras theroem

$A C^{2}=A D^{2}+D C^{2}$

$=20^{2}+20^{2}$

=400+400

$A C^{2}=800$

$A C=\sqrt{800}$

$=\sqrt{400 \times 2}$

$A C=20 \sqrt{2}$

(ii) $\tan ^{2} B-\frac{1}{\cos ^{2} B}=-1$

$\therefore  L H S \Rightarrow \tan ^{2} B-\frac{1}{\cos ^{2} B}$

$\tan ^{2} B=\left(\frac{A D}{B D}\right)^{2}=\left(\frac{.20}{15}\right)^{2}=\left(\frac{4}{3}\right)^{2}$

$=\frac{16}{9}$

$\cos ^{2} B=\left(\frac{B D}{A B}\right)^{2}=\left(\frac{15}{25}\right)^{2}=\left(\frac{3}{5}\right)^{2}=\frac{9}{25}$

$\therefore \tan ^{2} B-\frac{1}{\cos ^{2} B} \Rightarrow \frac{16}{9}-\frac{1}{\left(\frac{9}{25}\right)}$

$\Rightarrow \frac{16}{9}-\frac{25}{9}$

$\Rightarrow \frac{16-25}{9}$

$\Rightarrow -1$

∴LHS=RHS

Hence proved

Question 9

Sol :

Given : $\sin \theta=\frac{3}{5}$

(i) ∴From Pythagoras theorem

$b{c}^{2}=a b^{2}+a c^{2}$

$5^{2}=3^{2}+a c^{2}$

$25=9+a c^{2}$

$a c^{2}=25-9$

$a c^{2}=16$

$a c=\sqrt{16}$

ac=4

∴$\cos \theta=\frac{a c}{b c}$

$=\frac{4}{5}$

(ii) $\tan \theta=\frac{\sin \theta}{\cos \theta}$

$=\frac{\frac{3}{8}}{\frac{4}{8}}$

$=\frac{3}{4}$

Question 10

Sol :

Given that : $\tan \theta=\frac{5}{12}$

By Pythagoras theorem

$b c^{2}=a b^{2}+a c^{2}$

$=5^{2}+12^{2}$

=25+144

$b c^{2}=169$

$b c=\sqrt{169}$

∴bc=13

$\sin \theta=\frac{a b}{b c}$

$=\frac{5}{13}$

$\cos \theta=\frac{a c}{b c}$

$=\frac{12}{13}$

Question 11

Sol :

Given : $\sin \theta=\frac{6}{10}$

By Pythagoras theorem

$b c^{2}=a b^{2}+a c^{2}$

$10^{2}=6^{2}+a c^{2}$

$100=36+a c^{2}$

$a c^{2}=100-36$

$a c^{2}=64$

$a c=\sqrt{64}$

ac=8

$\therefore  \cos \theta=\frac{a c}{b c}=\frac{8}{10}$

$\tan \theta=\frac{a b}{a c}=\frac{6}{8}$

$\therefore \cos \theta+\tan \theta=\frac{8}{10}+\frac{6}{8}$

$=\frac{64+60}{80}$

$=\frac{124}{80} \Rightarrow \frac{31}{20}$

Question 12

Sol :

By Pythagoras theorem

$a b^{2}+a c^{2}=b c^{2}$

$4^{2}+3^{2}=b c^{2}$

$16+9=b c^{2}$

$b c^{2}=25$

$b c=\sqrt{25}$

bc=5

$\therefore \sin \theta=\frac{a b}{b c}=\frac{4}{5}$

$\cos \theta=\frac{a c}{b c}=\frac{3}{5}$

$\therefore \sin \theta+\cos \theta=\frac{4}{5}+\frac{3}{5}$

$=\frac{4+3}{5}$

$=\frac{7}{5}$

Question 13

Sol :

Given : cosecθ=√5

$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{b c}{a b}=\frac{\sqrt{5}}{1}$

$\therefore  bc=\sqrt{5}$ ; ab=1

By Pythagoras theorem

$b c^{2}=a b^{2}+a c^{2}$

$(\sqrt{5})^{2}=1^{2}+a c^{2}$

$5=1+a c^{2}$

$a c^{2}=5-1$

$a c^{2}=4$

$ac=\sqrt{4}$

ac=2

$\therefore \cot \theta-\cos \theta=\frac{a c}{a b}-\frac{a c}{b c}$

$=\frac{2}{1}-\frac{2}{\sqrt{5}}$

$=\frac{2 \sqrt{5}-2}{\sqrt{5}}$

$=\frac{2(\sqrt{5}-1)}{\sqrt{5}}$

Question 14

Sol :

Given : $\sin \theta=\frac{p}{q}$

By Pythagoras theorem

$q^{2}=p^{2}+a c^{2}$

$a c^{2}=a^{2}-p^{2}$

$a c=\sqrt{q^{2}-p^{2}}$

cos θ$=\frac{a c}{b c}=\frac{\sqrt{-p^{2}+q^{2}}}{q}$

∴cos θ +sin θ$=\frac{\sqrt{q^{2}-p^{2}}}{q}+\frac{p}{q}$

$=\frac{p+\sqrt{q^{2}-p^{2}}}{q}$

Question 15

Sol :

Given $\tan \theta=\frac{8}{15}$

By Pythagoras theorem

$b c^{2}=a b^{2}+a c^{2}$

$b c^{2}=8^{2}+15^{2}$

$b c^{2}=64+225$

$b c^{2}=289$

$b c=\sqrt{289}$

bc=17

∴ab=8 ; ac=15 ; bc=17

sec θ$=\frac{b c}{a c}=\frac{17}{15}$

cosec θ$=\frac{b c}{a b}=\frac{17}{8}$

∴secθ+cosecθ$=\frac{17}{15}+\frac{17}{8}$

$=\frac{17 \times 8+17 \times 15}{120}$

$=\frac{391}{120}$

$=3 \frac{31}{120}$

Question 16

Sol :

Given : 

13 sin A=5

$\sin A=\frac{5}{13}$

By Pythagoras theorem

$B A^{2}=C B^{2}+A C^{2}$

$13^{2}=5^{2}+A C^{2}$

$169=25+AC^{2}$

$AC^{2}=169-25$

$AC^{2}=144$

$AC=\sqrt{144}$

AC=12

∴CB=5 ; BA=13 ; AC=12

$\sin A=\frac{5}{13}$

$\cos A=\frac{12}{13}$

$\operatorname{tan} A=\frac{\sin A}{\cos A}$

$=\frac{\left(\frac{5}{13}\right)}{\left(\frac{12}{13}\right)}$

$\Rightarrow  \frac{5}{12}$

∴$\frac{5 \sin A-2 \cos A}{\tan A}=\frac{5\left(\frac{5}{13}\right)-2\left(\frac{12}{13}\right)}{\left(\frac{5}{12}\right)}$

$=\frac{\frac{25}{13}-\frac{24}{13}}{\frac{5}{12}}$

$=\frac{\frac{25-24}{13}}{\frac{5}{12}}$

$=\frac{1}{13} \times \frac{12}{5}=\frac{12}{65}$

Question 17

Sol :

Given : cosec A=√2

$\operatorname{cosec} A=\frac{1}{\sin A}=\sqrt{2}$

∴$\sin A=\frac{1}{\sqrt{2}}$

By Pythagoras theorem

$bc^{2}=a b^{2}+a c^{2}$

$(\sqrt{2})^{2}=1^{2}+a c^{2}$

$2=1+a c^{2}$

$a c^{2}=2-1$

$a c^{2}=1$

$a c=\sqrt{1}$

ac=1

∴ac=1 ; bc=√2 ; ab=1

∴sin A$=\frac{1}{\sqrt{2}}$

cos A$=\frac{1}{\sqrt{2}}$

tan A$=\frac{\sin A}{\cos A}$

$=\frac{\left(\frac{1}{\sqrt{2}}\right)}{\left(\frac{1}{\sqrt{2}}\right)}=1$

cot A$\frac{1}{\tan A}$

$=\frac{1}{1}$

cot A=1

$\therefore \frac{2 \sin ^{2} A+3 \cot ^{2} A}{\tan ^{2} A-\cos ^{2} A}$

$\Rightarrow \frac{2 \cdot\left(\frac{1}{\sqrt{2}}\right)^{2}+3(1)^{2}}{(1)^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2}}$

$\Rightarrow \frac{2 \cdot \frac{1}{2}+3 \cdot(1)}{1-\frac{1}{2}}$

$\Rightarrow \frac{1+3}{\left(\frac{2-1}{2}\right)}$

$\Rightarrow \frac{4 \times 2}{1}$

⇒8

Question 18

Sol :

Given : ABCD is a rhombus

AC=8 ; BD=6

From figure ΔOBC

By Pythagoras theorem

$B C^{2}=O C^{2}+O B^{2}$

$=3^{2}+4^{2}$

=9+16=25

$B C=\sqrt{25}=5$

sin < OCB$=\frac{O B}{B C}$

$=\frac{3}{5}$

Question 19

Sol :

Given : $\tan \theta=\frac{5}{12}$

By Pythagoras theorem

$bc^{2}=a b^{2}+a c^{2}$

$b c^{2}=5^{2}+12^{2}$

$b c^{2}=25+144$

$b c^{2}=169$

$b c=\sqrt{109}$

bc=13

∴bc=13 ; ab=5 ; ac=12

∴$\sin \theta=\frac{a b}{b c}=\frac{5}{13}$

$\cos \theta=\frac{a c}{b c}=\frac{12}{13}$

∴$\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}=\frac{\frac{12}{13}+\frac{5}{13}}{\frac{12}{13}-\frac{5}{13}}$

$=\frac{\frac{12+5}{13}}{\frac{12-5}{13}}$

$=\frac{17}{78}=2 \frac{3}{7}$

Question 20

Sol :

Given : 5cos A-12 sin A=0

5cos A=12sin A

$\frac{5}{12}=\frac{\sin A}{\cos A}$

$\frac{5}{12}=\tan A$

By Pythagoras theorem

$b c^{2}=a b^{2}+a c^{2}$

$b c^{2}=5^{2}+12^{2}$

$b c^{2}=25+144$

$b c^{2}=169$

$b c=\sqrt{169}$

bc=13

∴ab=5 ; bc=13 ; ac=12

Sin A$=\frac{a b}{b c}=\frac{5}{13}$

cos A$=\frac{a c}{b c}=\frac{12}{13}$

∴$\frac{\sin A+\cos A}{2 \cos A-\sin A}$

$=\frac{\frac{5}{13}+\frac{12}{13}}{2 \cdot \frac{12}{13}-\frac{5}{13}}$

$=\frac{(5+12) / 13}{(24-5) / 13}$

$=\frac{17}{19}$

Question 21

Sol :

Given : $\tan \theta=\frac{p}{q}$

By Pythagoras theorem

$b c^{2}=a b^{2}+a c^{2}$

$b c^{2}=p^{2}+q^{2}$

$b c=\sqrt{p^{2}+q^{2}}$

∴sin θ$=\frac{a b}{b c}=\frac{p}{\sqrt{p^{2}+q^{2}}}$

cos θ$=\frac{a c}{b c}=\frac{q}{\sqrt{p^{2}+q^{2}}}$

∴$\frac{p \sin \theta-q \cos \theta}{p \sin \theta+q \cos \theta}$

$=\frac{p \frac{p}{\sqrt{p^{2}+q^{2}}}-q \cdot \frac{q}{\sqrt{p^{2}+q^{2}}}}{p \cdot \frac{p}{\sqrt{p^{2}+q^{2}}}+q \frac{q}{\sqrt{p^{2}+q^{2}}}}$

$=\frac{\frac{p^{2}-q^{2}}{\sqrt{p^{2}+q^{2}}}}{\frac{p^{2}+q^{2}}{\sqrt{p^{2}+q^{2}}}}$

$\frac{p^{2}-a^{2}}{p^{2}+q^{2}}$

Question 22

Sol :

Given : 3cotθ=4

$\cot \theta=\frac{4}{3}$

By Pythagoras theorem

$b c^{2}=a b^{2}+a c^{2}$

$b c^{2}=3^{2}+4^{2}$

$b c^{2}=9+16$

$b c^{2}=25$

$b c=\sqrt{25}$

bc=5

∴$\sin \theta=\frac{a b}{b c}=\frac{3}{5}$

$\cos \theta=\frac{a c}{b c}=\frac{4}{5}$

∴$\frac{5 \sin \theta-3 \cos \theta}{5 \sin \theta+3 \cos \theta}$

$=\frac{5.\frac{3}{5}-3 \cdot \frac{4}{5}}{5 \cdot \frac{3}{5}+3 \cdot \frac{4}{5}}$

$=\frac{(15-12) / 5}{(15+12) / 5}$

$=\frac{3}{27}=\frac{1}{9}$

Question 23

Sol :

(i)

Given : 5cosθ-12sinθ=0

⇒5cosθ=12sinθ

⇒$\frac{5}{12}=\frac{\sin \theta}{\cos \theta}$

⇒$\frac{5}{12}=\tan \theta$

By Pythagoras theorem

$bc^{2}=a b^{2}+a c^{2}$

$b c^{2}=5^{2}+12^{2}$

$b c^{2}=25+144$

$b c^{2} =169$

$b c=\sqrt{169}$

bc=13

∴sinθ$=\frac{a b}{b c}=\frac{5}{13}$

cosθ$=\frac{a c}{b c}=\frac{12}{13}$

∴$\frac{\sin \theta+\cos \theta}{2 \cos \theta-\sin \theta}$

$=\frac{\frac{5}{13}+\frac{12}{13}}{2 \cdot \frac{12}{13}-\frac{5}{13}}$

$=\frac{\frac{5+12}{13}}{\frac{24-5}{13}}$

$=\frac{17}{16}$

(ii)

Given : cosecθ$=\frac{13}{12}$

$\Rightarrow \operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{13}{12}$

$\therefore \sin \theta=\frac{12}{13}$

By Pythagoras theorem

$b c^{2}=a b^{2}+a c^{2}$

$13^{2}=12^{2}+a c^{2}$

$169=144+a c^{2}$

$169-144=a c^{2}$

$25=a c^{2}$

$a c=\sqrt{25}$

ac=5

∴sinθ$=\frac{12}{13}$

cosθ$=\frac{a c}{b c}=\frac{5}{13}$

∴$\frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9\cos \theta}$

$=\frac{2 \cdot \frac{12}{13}-3 \cdot \frac{5}{13}}{4 \cdot \frac{12}{13}-9 \cdot \frac{5}{13}}$

$=\frac{\frac{24-15}{13}}{\frac{48-45}{13}}$

$=\frac{9}{3}$

=3

Question 24

Sol :

Given : 

5sinθ=3

sinθ$=\frac{3}{5}$

By Pythagoras theorem

$b c^{2}=a b^{2}+a c^{2}$

$5^{2}=3^{2}+a c^{2}$

$25=a+a c^{2}$

$a c^{2}=25-9$

$a c^{2}=16$

$a c=\sqrt{16}$

ac=4

cosθ$=\frac{a c}{b c}=\frac{4}{5}$

∴secθ$=\frac{1}{\cos \theta}=\frac{5}{4}$

$\operatorname{tan} \theta=\frac{\sin \theta}{\cos \theta}=\frac{3 / 5}{4 / 5}=\frac{3}{4}$

∴$\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}$

$=\frac{\frac{5}{4}-\frac{3}{4}}{\frac{5}{4}+\frac{3}{4}}$

$=\frac{\frac{5-3}{4}}{\frac{5+3}{4}}=\frac{2}{8}=\frac{1}{4}$

Question 25

Sol :

Given :

sinθ=cosθ

Then $2 \tan ^{2} \theta+\sin ^{2} \theta-1$

$\Rightarrow 2\left(\frac{\sin \theta}{\cos \theta}\right)^{2}+\sin ^{2} \theta-1$

$\Rightarrow 2\left(\frac{\sin \theta}{\sin \theta}\right)^{2}+\sin ^{2} \theta-1$

$\Rightarrow 2(1)^{2}+\sin ^{2} \theta-1$

$\Rightarrow 2+\sin ^{2} \theta-1$

$\Rightarrow \sin ^{2} \theta+1$

From given sinθ=cosθ

So, θ=45

∴sin45=cos45

$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}$

∴$2 \tan ^{2} \theta+\sin ^{2} \theta-1$

⇒$\sin ^{2} \theta+1$

⇒$\left(\frac{1}{\sqrt{2}}\right)^{2}+1$

⇒$\frac{1}{2}+1=\frac{3}{2}$

∴$\operatorname{2tan}^{2} \theta+\sin ^{2} \theta-1=\frac{3}{2}$

Question 26

Sol :

(i)

L.H.S⇒cosθ.tanθ

⇒cosθ$\frac{\sin \theta}{\cos \theta}$ $\left(\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right)$

⇒sinθ

R.H.S=sinθ

∴L.H.S=R.H.S

(ii)

LHS⇒sinθ.cotθ

⇒$\sin \theta \cdot \frac{\cos \theta}{\sin \theta}$ [$\because \cot \theta=\frac{\cos \theta}{\sin \theta}$]

⇒cosθ

∴L.H.S=R.H.S

(iii)

LHS⇒$\Rightarrow \frac{\sin ^{2} \theta}{\cos \theta}+\cos ^{2} \theta$

⇒$\frac{\sin ^{2} \theta+\cos \theta \cdot \cos \theta}{\cos \theta}$

$\Rightarrow \frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos \theta}$ $\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

⇒$\frac{1}{\cos \theta}$

∴L.H.S=R.H.S

Question 27

Sol :

Given : ∠C=90°

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

By Pythagoras theorem

$A B^{2}=B C^{2}+A C^{2}$

$A B^{2}=3^{2}+4^{2}$

$A B^{2}=9+16$

$A B^{2}=25$

$A B=\sqrt{25}$

AB=5

$\therefore \sin A=\frac{B C}{A B}=\frac{3}{5}$

$\cos \cdot A=\frac{A C}{A B}=\frac{4}{5}$

$\cos B=\frac{B C}{A B}=\frac{3}{5}$

$\operatorname{Sin} B=\frac{A C}{A B}=\frac{4}{5}$

∴sinA cosB+cosAsinB

⇒$\frac{3}{5} \cdot \frac{3}{5}+\frac{4}{5} \cdot \frac{4}{5}$

⇒$\frac{9}{25}+\frac{16}{25}$

⇒$\frac{9+16}{25}=\frac{25}{25}$

⇒1

∴L.H.S=R.H.S

Question 28

Sol :

(a)

Given : 

In ΔABC, ΔBRS

AB=18 cm

BC=7.5 cm

RS=5 cm

∠BSR=x°

∠SAB=y°

From Hint ; AR=12 cm

RB=6 cm

AC=19.5 cm

(i) $\tan x=\frac{R B}{S R}=\frac{6}{5}$

(ii) $\sin y=\frac{B C}{A C}$

$=\frac{7.5}{19.5} \times \frac{10}{10}$

$=\frac{75}{195}=\frac{5}{13}$

∴sin y$=\frac{5}{13}$

(b)

By Pythagoras theorem

$A C^{2}=A B^{2}+B C^{2}$

$A C^{2} =12^{2}+5^{2}$

$A C^{2} =144+2$

$A C^{2} =169$

$A C =\sqrt{169}$

AC=13

∴cos < CBD$=\frac{A B}{A C}$

$=\frac{12}{13}$

(ii) cot<ABD$=\frac{B C}{A C}$

$=\frac{5}{13}$

Question 29

Sol :

Given :

ABCD is rectangle

AC=15

∠ACD=ɑ ; cot ɑ$=\frac{3}{2}$

$\cot \alpha=\frac{C D}{A D}=\frac{3}{2}$

$\therefore C D=\frac{3}{2} A D$

From ΔACD

$A C^{2} =A D^{2}+C D^{2}$

$15^{2} =A D^{2}+\left(\frac{3}{2} A D\right)^{2}$

$225 =A D^{2}+\frac{9 A D^{2}}{4}$

$225 =\frac{4 A D^{2}+9 A D^{2}}{4}$

$225 \times 4 =13 A D^{2}$

$A D^{2} =\frac{225 \times 4}{13}$

$A D =\sqrt{\frac{225 \times{4}}{13}}$

$A D =\frac{15 \times 2}{\sqrt{13}}$

$A D =\frac{30}{\sqrt{13}}$

∵CD$=\frac{3}{2} A D$

$=\frac{3}{2}\times \frac{30}{\sqrt{3}}$

$=\frac{45}{\sqrt{3}}$

∴Area of ΔACD ≠Area of ΔABC=Area of rectangle ABCD

∴$C D=\frac{45}{\sqrt{13}}=A B$

$A D=\frac{30}{\sqrt{13}}=B C$

∴Area of 🗆ABCD=CD×AD

$=\frac{45}{\sqrt{13}} \times \frac{30}{\sqrt{13}}$

Area of $=\frac{45 \times 30}{\sqrt{13} \cdot \sqrt{13}}$

$=\frac{1350}{13}$

$=103 \frac{11}{13}$ cm²

∴Perimeter=(AB+BC)

$=2\left(\frac{45}{\sqrt{13}}+\frac{30}{\sqrt{13}}\right.$

$=2\left(\frac{45+30}{\sqrt{13}}\right)$

$=2 \times \frac{75}{\sqrt{13}}$

$=\frac{150}{\sqrt{13}}$

Question 30

Sol :

(a)

From ΔBCD

By Pythagoras theorem

$B D^{2}=B C^{2}+C D^{2}$

$13^{2}=12^{2}+C D^{2}$

$169=144+C D^{2}$

$C D^{2}=169-144$

$C D^{2}=25$

$C D=\sqrt{25}=5$

(i) $\sin \phi=\frac{C D}{B D}$

$=\frac{5}{13}$

(ii) $\operatorname{tan} \theta=\frac{D E}{A E}$

$=\frac{12}{9}=\frac{4}{3}$

(b)

$\operatorname{Tan} \theta=\frac{4}{3}$

$\therefore \sin \theta=\frac{D E}{A D}$

$A D=\frac{12}{\sin \theta}$

$\cos \theta =\frac{A E}{A D}$

$A D=\frac{9}{\cos \theta}$

Question 31

Sol :

(i) 

LHS$\Rightarrow(\sin A+\cos A)^{2}+(\sin A-\cos A)^{2}$

$\Rightarrow \sin ^{2} A+\cos ^{2} A+2 \sin A \cos A+\sin ^{2} A+\cos ^{2} A-2 \sin A \cos A$

$\left(\begin{array}{l}\because(a+b)^{2}=a^{2}+b^{2}+2 a b \\ (a-b)^{2}=a^{2}+b^{2}-2 a b\end{array}\right)$

$\Rightarrow 2 \sin ^{2} A+2 \cos ^{2} A$

$\Rightarrow 2\left(\sin ^{2} A+\cos ^{2} A\right)$

⇒2(1)=2

∴L.HS.=R.H.S

(ii) 

LHS

$\Rightarrow \cot ^{2} A-\frac{1}{\sin ^{2} A}+1$

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

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

$\Rightarrow \frac{\left(\cos ^{2} A+\sin ^{2} A\right)-1}{\sin ^{2} A}$

$\Rightarrow \frac{1-1}{\sin ^{2} A}$

⇒0=RHS

(iii)

LHS

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

$\Rightarrow \frac{1}{1+\frac{\sin ^{2} A}{\cos ^{2} A}}+\frac{1}{1+\frac{\cos ^{2} A}{\sin ^{2} A}}$

$\Rightarrow \frac{1}{\frac{\cos ^{2}+\sin ^{2} A}{\cos ^{2} A}}+\frac{1}{\frac{\sin ^{2} A+\cos ^{2} A}{\sin ^{2} A}}$

$\Rightarrow \frac{\cos ^{2} A}{(1)}+\frac{\sin ^{2} A}{(1)}$

⇒$\cos ^{2} A+\sin ^{2} A$

⇒1=RHS

$x^{2}=a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+2 a b \sin \theta \cos \theta$...(1)

∵y=asinθ-bcosθ

Squaring on both sides

$y^{2}=(a \sin \theta-b \cos \theta)^{2}$

$y^{2}=a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta-2 a b \sin \theta \cos \theta$...(2)

(1)+(2)

$x^{2}+y^{2}= a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+2 a b \sin \theta \cos \theta+ a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta-2 a b \sin \theta \cos \theta$

$=a^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+b^{2}\left(\sin ^{2} \theta+\cos^2 \theta\right)$

$=a^{2}(1)+b^{2}(1)$

$=a^{2}+b^{2}$

∴LHS=RHS

Question 32

Sol :

Given : $\sqrt{\frac{1-\sin ^{2} \theta}{1-\cos ^{2} \theta}}$

$\Rightarrow \sqrt{\frac{\cos ^{2} \theta}{\sin ^{2} \theta}}$ $\left(\because \sin ^{2} \theta+\cos \theta=1\right)$

$\Rightarrow \sqrt{\cot ^{2} \theta}$

$\Rightarrow \quad \cot \theta$

Question 33

Sol :

Given : sinθ+cosecθ=2

Squaring on both sides

(sinθ+cosecθ)²=2²

$\sin ^{2} \theta+\operatorname{cosec}^{2} \theta+2 \cdot \sin \theta \cdot \cos \theta=4$

$\sin ^{2} \theta+\operatorname{cosec}^{2} \theta+2 \sin \theta \frac{1}{\sin \theta}=4$

$\sin ^{2} \theta+\operatorname{cosec}^{2} \theta=4-2$

$\sin ^{2} \theta+\operatorname{cosec}^{2} \theta=2$

Question 34

Sol :

Given : x=acosθ+bsinθ

y=asinθ-bcosθ

Squaring on both sides

$x^{2}=(a \cos \theta+b \sin \theta)^{2}$