SELINA Solution Class 9 Chapter 22 Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals] Exercise 22 A

Question 1

From the following figure, find the values of :
(i) sin A
(ii) cos A
(iii) cot A
(iv) sec C
(v) cosec C
(vi) tan C.

Sol:

Given angle  ABC = 90°

⇒ AC²  = AB² + BC²  ...( AC is hypotenuse )
⇒ AC² = 3² + 4² 
∴ AC² = 9 + 16 = 25 and AC = 5

(i) sin A = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{4}{5}}$

(ii) cos A = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{3}{5}}$

(iii) cot A = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{BC}}=\frac{3}{4}}$

(iv) sec C = ${\displaystyle \frac{\text{hypotenuse}}{\text{base}}=\frac{\text{AC}}{\text{BC}}=\frac{5}{4}}$

(v) cosec C = ${\displaystyle \frac{\text{hypotenuse}}{\text{perpendicular}}=\frac{\text{AC}}{\text{AB}}=\frac{5}{3}}$

(vi) tan C = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{AB}}{\text{BC}}=\frac{3}{4}}$

Question 2

Form the following figure, find the values of :
(i) cos B
(ii) tan C
(iii) sin²B + cos²B
(iv) sin B. cos C + cos B. sin C

Sol:

Given angle  BAC = 90°

⇒ BC² = AB² + AC²          ...( BC is hypotenuse )

⇒ 17² = 8² + AC²

∴ AC² = 289 - 64 = 225 and AC = 15

(i) cos B = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{BC}}=\frac{8}{17}}$

(ii) tan C = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{AB}}{\text{AC}}=\frac{8}{15}}$

(iiii) sin B = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{AC}}{\text{BC}}=\frac{15}{17}}$

cos B = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{BC}}=\frac{8}{17}}$

sin² B+ cos² B = ${\displaystyle {\left(\frac{15}{17}\right)}^2+{\left(\frac{18}{17}\right)}^2}$

= ${\displaystyle \frac{225+64}{289}}$

= ${\displaystyle \frac{289}{289}}$

=  1

(iv) sin B = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{AC}}{\text{BC}}=\frac{15}{17}}$

cos B = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{BC}}=\frac{8}{17}}$

sin C = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{BC}}=\frac{8}{17}}$

cos C = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\text{AC}}{\text{BC}}=\frac{15}{17}}$

sin B · cos C + cos B · sin C

= ${\displaystyle \frac{15}{17}.\frac{15}{17}+\frac{8}{17}.\frac{8}{17}}$

= ${\displaystyle \frac{225+64}{289}}$

= ${\displaystyle \frac{289}{289}}$

= 1

Question 3

From the following figure, find the values of
(i) cos A 
(ii) cosec A
(iii) tan²A - sec²A 
(iv) sin C
(v) sec C 
(vi) cot² C - ${\displaystyle \frac{1}{{\sin }^2\text{c}}}$

Sol:

Consider the diagram as

Given angle  ADB = 90° and BDC = 90°
⇒ AB² = AD² + BD²         ...( AB is hypotenuse in ΔABD )

⇒ AB² = 3² + 4²

∴ AB² = 9 + 16 = 25 and AB = 5

⇒ BC² = BD² + DC²       ...( BC is hypotenuse in ΔBDC )

⇒ DC² = 12² - 4²

∴ DC² = 144 - 16 = 128 aand DC = 8${\displaystyle \sqrt{2}}$ 

(i) cos A = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\text{AD}}{\text{AB}}=\frac{3}{5}}$

(ii) cosec A = ${\displaystyle \frac{\text{hypotenuse}}{\text{perpendicular}}=\frac{\text{AB}}{\text{BD}}=\frac{5}{4}}$

(iii) tan A = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{BD}}{\text{AD}}=\frac{4}{3}}$

sec A = ${\displaystyle \frac{\text{hypotenuse}}{\text{base}}=\frac{\text{AB}}{\text{AD}}=\frac{5}{3}}$

tan² A - sec² A = ${\displaystyle {\left(\frac{4}{3}\right)}^2-{\left(\frac{5}{3}\right)}^2}$

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

= ${\displaystyle \frac{–9}{9}}$

= – 1

(iv) sin C = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{BD}}{\text{BC}}=\frac{4}{12}=\frac{1}{3}}$

(v) sec C = ${\displaystyle \frac{\text{hypotenuse}}{\text{base}}=\frac{\text{BC}}{\text{DC}}=\frac{12}{8\sqrt{2}}=\frac{3}{2\sqrt{2}}=\frac{3\sqrt{2}}{4}}$

(vi) cot C = ${\displaystyle \frac{\text{base}}{\text{perpendicular}}=\frac{\text{DC}}{\text{BD}}=\frac{8\sqrt{2}}{4}=2\sqrt{2}}$

sin C = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{BD}}{\text{BC}}=\frac{4}{12}=\frac{1}{3}}$

cot² C –  ${\displaystyle \frac{1}{{\sin }^{2}C}={\left(2\sqrt{2}\right)}^2-\frac{1}{{\left(\frac{1}{3}\right)}^2}}$

= 8 - 9
=  – 1                        

Question 4

From the following figure, find the values of
(i) sin B 
(ii) tan C
(iii) sec² B - tan²B 
(iv) sin²C + cos²C

Sol:

Given angle  ADB = 90° and ADC = 90°

⇒ AB² = AD² + BD²            ...( AB is hypotenuse in ΔABD)

⇒  13² = AD² + 5²

∴ AD² = 169 – 25 = 144 and AD = 12

⇒  AC² = AD² + DC²        ...( AC is hypotenuse in ΔADC)

⇒ AC² = 12² + 16²

∴ AC² = 144 + 256 = 400 and AC = 20

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

(ii) tan C = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{AD}}{\text{DC}}=\frac{12}{16}=\frac{3}{4}}$

(iii) sec B = ${\displaystyle \frac{\text{hypotenuse}}{\text{base}}=\frac{\text{AB}}{\text{BD}}=\frac{13}{5}}$

tan B = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{AD}}{\text{BD}}=\frac{12}{5}}$

sec² B –  tan² B = ${\displaystyle {\left(\frac{13}{5}\right)}^2–{\left(\frac{12}{5}\right)}^2}$

= ${\displaystyle \frac{169–144}{25}}$

= ${\displaystyle \frac{25}{25}}$

= 1

(iv) sin C = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{AD}}{\text{AC}}=\frac{12}{20}=\frac{3}{5}}$

cos C = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\text{DC}}{\text{AC}}=\frac{16}{20}=\frac{4}{5}}$

sin² C + cos² C = ${\displaystyle {\left(\frac{3}{5}\right)}^2+{\left(\frac{4}{5}\right)}^2}$

= ${\displaystyle \frac{9+16}{25}}$

= ${\displaystyle \frac{25}{25}}$

= 1

Question 5

Given : sin A = ${\displaystyle \frac{3}{5}}$ , find : (i) tan A (ii) cos A 

Sol:

Consider the diagram below : 

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

i .e. ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{3}{5}\Rightarrow \frac{\text{BC}}{\text{AC}}=\frac{3}{5}}$

Therefore if length of BC = 3x, length of AC = 5x 
Since

AB² + BC² = AC²          ...[ Using Pythagoras Theorem ]
AB² + (3x)² = (5x)²
AB² = 25x² – 9x² = 16x²
∴ AB = 4x (base)

Now
(i) tan A = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{3x}{4x}=\frac{3}{4}}$

(ii) cos A = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{4x}{5x}=\frac{4}{5}}$

Question 6

From the following figure, find the values of :
(i) sin A
(ii) sec A
(iii) cos² A + sin²A

Sol:

Given angle, ABC = 90° in the figure

⇒ AC² = AB² + BC²       ...(AC is hypotenuse in Δ ABC )
⇒ AC² = a² + a²
∴ AC² = 2a² and AC =${\displaystyle \sqrt{2}a}$

Now

(i) sin A = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{a}{\sqrt{2}a}=\frac{1}{\sqrt{2}}}$

(ii) sec A = ${\displaystyle \frac{\text{hypotenuse}}{\text{base}}=\frac{\text{AC}}{\text{AB}}=\frac{\sqrt{2}a}{a}=\sqrt{2}}$

(iii) sin A = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{a}{\sqrt{2}a}=\frac{1}{\sqrt{2}}}$

cos A = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{a}{\sqrt{2}a}=\frac{1}{\sqrt{2}}}$

cos²A + sin²A  = ${\displaystyle {\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2}$

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

= 1

Question 8

Given: sec A = ${\displaystyle \frac{29}{21},\text{evaluate : sin A}-\frac{1}{\tan \text{A}}}$

Sol:

Consider the diagram below :

sec A = ${\displaystyle \frac{29}{21}}$

i.e. ${\displaystyle \frac{\text{hypotenuse}}{\text{base}}=\frac{29}{21}\Rightarrow \frac{\text{AC}}{\text{AB}}=\frac{29}{21}}$

Therefore if length of AB = 21x , length of AC = 29x
Since

AB² + BC² = AC²         ...[ Using Pythagoras Theorem ]

(21x)² + BC² = ( 29x)²

BC² = 841x² – 441x² =400x²

BC = 20x                    ...(perpendicular ) 

Now

sin A = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{20x}{29x}=\frac{20}{29}}$

tan A = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{20x}{21x}=\frac{20}{21}}$

Therefore

sin A – ${\displaystyle \frac{1}{\tan \text{A}}}$

= ${\displaystyle \frac{20}{29}–\frac{1}{\frac{20}{21}}}$

= ${\displaystyle \frac{20}{29}– \frac{21}{20}}$

= – ${\displaystyle \frac{209}{580}}$

Question 9

Given: tan A = ${\displaystyle \frac{4}{3},\text{find}:\frac{\cos ec\text{A}}{\cot \text{A}–\sec \text{A}}}$

Sol:

Consider the diagram below :

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

i.e. ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{4}{3}\Rightarrow \frac{\text{BC}}{\text{AB}}=\frac{4}{3}}$

Therefore if length of AB = 3x, length of BC = 4x

Since
AB² + BC² = AC²    ... [ Using Pythagoras Theorem ]

( 3x )² + (4x)² = AC²

AC² = 9x² + 16x² = 25x²

AC = 5x                   ...( hypotenuse )

Now

sec A = ${\displaystyle \frac{\text{hypotenuse}}{\text{base}}=\frac{\text{AC}}{\text{AB}}=\frac{5x}{3x}=\frac{5}{3}}$

cot A = ${\displaystyle \frac{\text{base}}{\text{perpendicular}}=\frac{\text{AB}}{\text{BC}}=\frac{3x}{4x}=\frac{3}{4}}$

cosec A = ${\displaystyle \frac{\text{hypotenuse}}{\text{perpendicular}}=\frac{\text{AC}}{\text{BC}}=\frac{5x}{4x} =\frac{5}{4}}$

Therefore

${\displaystyle \frac{\cos ec\text{A}}{\cot \text{A}–\sec \text{A}}}$

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

= ${\displaystyle \frac{\frac{5}{4}}{–\frac{11}{12}}}$

= ${\displaystyle –\frac{60}{44}}$

= ${\displaystyle –\frac{15}{11}}$

Question 10

Given: 4 cot A = 3
find :
(i) sin A
(ii) sec A
(iii) cosec²A - cot²A.

Sol:

Consider the diagram below :

4 cot A = 3

cot A = ${\displaystyle \frac{3}{4}}$

i.e.${\displaystyle \frac{\text{base}}{\text{perpendicular}}=\frac{3}{4}\Rightarrow \frac{\text{AB}}{\text{BC}}=\frac{3}{4}}$

Therefore if length of AB = 3x, length of BC = 4x

Since
AB² + BC² = AC²             ...[ Using Pythagoras Theorem ]

(3x)² + (4x)² = AC²

AC² = 9x² + 16x² = 25x²

∴ AC = 5x  ...( hypotenuse )

(i) sin  A = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{4x}{5x}=\frac{4}{5}}$

(ii) sec A = ${\displaystyle \frac{\text{hypotenuse}}{\text{base}}=\frac{\text{AC}}{\text{AB}}=\frac{5x}{3x}=\frac{5}{3}}$

(iii) cosec A = ${\displaystyle \frac{\text{hypotenuse}}{\text{perpendicular}}=\frac{\text{AC}}{\text{BC}}=\frac{5x}{4x}=\frac{5}{4}}$

cot A = ${\displaystyle \frac{3}{4}}$

cosec² A – cot² A 

=${\displaystyle {\left(\frac{5}{4}\right)}^2–{\left(\frac{3}{4}\right)}^2}$

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

= ${\displaystyle \frac{16}{16}}$

= 1

Question 11

Given: cos A = 0.6; find all other trigonometrical ratios for angle A.

Sol :

Consider the diagram below:

cos A = 0 . 6
cos A = ${\displaystyle \frac{6}{10}=\frac{3}{5}}$

i .e. ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{3}{5}\Rightarrow \frac{\text{AB}}{\text{AC}}=\frac{3}{5}}$

Therefore if length of AB = 3x, length of AC = 5x
Since

AB² + BC² = AC²               ...[ Using Pythagoras Theorem ]
(3x)² + BC² = (5x)²
BC² = 25x² - 9x² = 16x²
∴ BC = 4x                         ...( perpendicular )

Now all other trigonometric ratios are

sin A = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{4x}{5x}=\frac{4}{5}}$

cosec A = ${\displaystyle \frac{\text{hypotenuse}}{\text{perpendicular}}=\frac{\text{AC}}{\text{BC}}=\frac{5x}{4x}=\frac{5}{4}}$

sec A = ${\displaystyle \frac{\text{hypotenuse}}{\text{Base}}=\frac{\text{AC}}{\text{AB}}=\frac{5x}{3x}=\frac{5}{3}}$

tan A = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{4x}{3x}=\frac{4}{3}}$

cot A = ${\displaystyle \frac{\text{base}}{\text{perpendicular}}=\frac{3x}{4x}=\frac{3}{4}}$

Question 12

In a right-angled triangle, it is given that A is an acute angle and tan A = ${\displaystyle \frac{5}{12}}$.

find the value of :
(i) cos A
(ii) sin A
(iii)  ${\displaystyle \frac{\cos \text{A}+\sin \text{A}}{\cos \text{A}–\sin \text{A}}}$

Sol:

Consider the diagram below : 

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

i.e.${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{5}{12}\Rightarrow \frac{\text{BC}}{\text{AB}}=\frac{5}{12}}$

Therefore if length of AB = 12x, length of BC = 5x

Since
AB² + BC² = AC²           ...[ Using Pythagoras Theorem ]

(12x)² + (5x)² = AC²

AC² = 144x² + 25x² = 169x²

∴ AC = 13x                    ...( hypotenuse)

(i) cos A = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{12x}{13x}=\frac{12}{13}}$

(ii) sin A = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{5x}{13x}=\frac{5}{3}}$

(iii) ${\displaystyle \frac{\cos \text{A}+\sin \text{A}}{\cos \text{A}–\sin \text{A}}}$

= ${\displaystyle \frac{\frac{12}{13}+\frac{5}{13}}{\frac{12}{13}–\frac{5}{13}}}$ 

= ${\displaystyle \frac{\frac{17}{13}}{\frac{17}{7}}}$

= ${\displaystyle \frac{17}{7}}$

= ${\displaystyle 2\frac{3}{7}}$

Question 13

Given: sin θ = ${\displaystyle \frac{p}{q}}$.
Find cos θ + θ sin in terms of p and q.

Sol :

Consider the diagram below :

sin θ = ${\displaystyle \frac{p}{q}}$

i.e.${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{p}{q}}$

Therefore if length of perpendicular = px, length of hypotenuse = qx

Since
base² + perpendicular² = hypotenuse²   ...[ Using Pythagoras Theorem]

base² + (px)² = (qx)²

base² = q²x² – p²x² = (q² – p²)x²

∴ base =  ${\displaystyle \sqrt{q^2–p^{2}x}}$

Now

cosθ = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\sqrt{q^2–p^2}}{q}}$

Therefore
cosθ + sinθ

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

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

Question 14

If cos A = ${\displaystyle \frac{1}{2}}$ and sin B = ${\displaystyle \frac{1}{\sqrt{2}}}$, find the value of :  ${\displaystyle \frac{\tan \text{A}–\tan \text{B}}{1+\tan \text{A}\tan \text{B}}}$.
Are angles A and B from the same triangle? Explain.

Sol:

Consider the diagram below: 

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

i.e.${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{1}{2}\Rightarrow \frac{\text{AB}}{\text{AC}}=\frac{1}{2}}$

Therefore if length of AB = x, length of AC = 2x

Since
AB² + BC² = AC²   ...[ Using Pythagoras Theorem ]

(x)² + BC² = (2x)²

BC² = 4x² – x² = 3x²

∴ BC = ${\displaystyle \sqrt{3}x}$     ...(perpendicular)

Consider the diagram below: 

sin B = ${\displaystyle \frac{1}{\sqrt{2}}}$

i.e.${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{1}{\sqrt{2}}\Rightarrow \frac{\text{AC}}{\text{BC}}=\frac{1}{\sqrt{2}}}$

Therefore if length of AC = x, length of BC = ${\displaystyle \sqrt{2}}$
Since

AB² + AC² = BC²   ...[ Using Pythagoras Theorem]
AB² + x² = (${\displaystyle \sqrt{2}x{)}^2}$
AB² = 2x² – x² = x²
∴ AB = x (base)

Now

tan A = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\sqrt{3}x}{x}=\sqrt{3}}$ 

tan B =${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{x}{x}=1}$

Thererfore

${\displaystyle \frac{\tan \text{A}–\tan \text{B}}{1+\tan \text{A}\tan \text{B}}}$

= ${\displaystyle \frac{\sqrt{3}–1}{1+\sqrt{3}}}$

= ${\displaystyle \frac{\sqrt{3}–1}{1+\sqrt{3}}.\frac{\sqrt{3}–1}{\sqrt{3}–1}}$

= ${\displaystyle \frac{4–2\sqrt{3}}{2}}$

= ${\displaystyle 2 –\sqrt{3}}$  

Question 15

If 5 cot θ = 12, find the value of : Cosec θ+ sec θ 

Sol:

Consider the diagram below : 

5cot θ = 12
cot θ = ${\displaystyle \frac{12}{5}}$

i.e.${\displaystyle \frac{\text{base}}{\text{perpendicular}}=\frac{12}{5}}$

Therefore if length of base = 12x, length of perpendicular = 5x

Since
base² + perpendicular² = hypotenuse²  ...[ Using Pythagoras Theorem]

(12x)² + (5x)² = hypotenuse²

hypotenuse² = 144x² + 25x² = 169x²

∴ hypotenuse = 13x

Now

cosec θ = ${\displaystyle \frac{\text{hypotenuse}}{\text{perpendicular}}=\frac{13x}{5x}=\frac{13}{5}}$

sec θ = ${\displaystyle \frac{\text{hypotenuse}}{\text{base}}=\frac{13x}{12x}=\frac{13}{12}}$

Therefore
cosec θ+sec θ

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

= ${\displaystyle \frac{221}{60}}$

= ${\displaystyle 3\frac{41}{60}}$

Question 16

If tan x = ${\displaystyle 1\frac{1}{3}}$, find the value of : 4 sin²x - 3 cos²x + 2

Sol:

Consider the diagram below :

tan x = ${\displaystyle 1\frac{1}{3}}$

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

i.e.${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{4}{3}}$

Therefore if length of base = 3x, length of perpendicular = 4x

Since

base²+ perpendicular² = hypotenuse²  ..[ Using Pythagoras Theorem]

(3x)² + (4x)² = hypotenuse²

hypotenuse = 9x²+16x²+ 25x²

∴ hypotenuse = 5x

Now

sin x = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{4x}{5x}=\frac{4}{5}}$

cos x = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{3x}{5x}=\frac{3}{5}}$

Therefore
4 sin² x – 3cos² x +2

= ${\displaystyle 4{\left(\frac{4}{5}\right)}^2–3{\left(\frac{3}{5}\right)}^2+2}$

= ${\displaystyle \frac{64}{25}–\frac{27}{25}+2}$

= ${\displaystyle \frac{87}{25}}$

=${\displaystyle 3\frac{12}{25}}$

Question 17

If cosec θ = ${\displaystyle \sqrt{5}}$ , find the value of :
(i) 2 - sin² θ - cos² θ
(ii) 2+ ${\displaystyle \frac{1}{{\sin }^2\text{θ}}–\frac{{\cos }^2\text{θ}}{{\sin }^2\text{θ}}}$ 

Sol:

Consider the diagram below :

cosec θ =${\displaystyle \sqrt{5}}$

i.e.${\displaystyle \frac{\text{hypotenuse}}{\text{perpendicular}}=\frac{\sqrt{5}}{1}}$

Therefore if length of hypotenuse =${\displaystyle \sqrt{5}}$, length of perpendicular = x

Since
base² + perpendicular² = hypotenuse² ...[ Using Pythagoras Theorem]

base² + (x)² = (${\displaystyle \sqrt{5}x{)}^2}$

base² = 5x² – x² = 4x²

∴ base = 2x

Now

sin θ =${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{x}{\sqrt{5}x}=\frac{1}{\sqrt{5}}}$

cos θ =${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{2}{\sqrt{5}x}=\frac{2}{\sqrt{5}}}$

(i) 2 – sin² θ – cos² θ

= 2 – ${\displaystyle {\left(\frac{1}{\sqrt{5}}\right)}^2–{\left(\frac{2}{\sqrt{5}}\right)}^2}$

= 2 – ${\displaystyle \frac{1}{5}–\frac{4}{5}}$

= ${\displaystyle \frac{5}{5}}$

= 1

(ii) 2+ ${\displaystyle \frac{1}{{\sin }^2\text{θ}}–\frac{{\cos }^2\text{θ}}{{\sin }^2\text{θ}}}$ 

= 2+${\displaystyle \frac{1}{{\left(\sqrt{5}\right)}^2}–\frac{{\left(\frac{2}{\sqrt{5}}\right)}^2}{{\left(1\sqrt{5}\right)}^2}}$

= 2+5 – 4

= 3

Question 18

If sec A = ${\displaystyle \sqrt{2}}$, find the value of :
${\displaystyle \frac{3{\cos }^2\text{A}+5{\tan }^2\text{A}}{4{\tan }^4\text{A}–{\sin }^2\text{A}}}$

Sol :

Consider the diagram below :

sec A =${\displaystyle \sqrt{2}}$

i.e.${\displaystyle \frac{\text{hypotenuse}}{\text{base}}=\frac{\sqrt{2}}{1}\Rightarrow \frac{\text{AC}}{\text{AB}}=\frac{\sqrt{2}}{1}}$

Therefore if length of AB = x, length of AC =${\displaystyle \sqrt{2}}$ 

Since
AB² + BC² = AC²  ...[ Using Pythagoras Theorem]

(x)² + BC² = (${\displaystyle \sqrt{2}\text{x}{)}^2}$

BC² = 2x² – x² = x²

∴ BC = x ...(perpendicular)

Now

tan A = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{x}}{\text{x}}=1}$

sin A = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{x}}{\sqrt{2}\text{x}}=\frac{1}{\sqrt{2}}}$

cos A = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\text{x}}{\sqrt{2}\text{x}}=\frac{1}{\sqrt{2}}}$

Therefore
${\displaystyle \frac{3{\cos }^2\text{A}+5{\tan }^2\text{A}}{4{\tan }^4\text{A}–{\sin }^2\text{A}}}$

= ${\displaystyle \frac{3{\left(\frac{1}{\sqrt{2}}\right)}^2+5{\left(1\right)}^2}{4{\left(1\right)}^2–{\left(\frac{1}{\sqrt{2}}\right)}^2}}$

= ${\displaystyle \frac{\frac{13}{2}}{\frac{7}{2}}}$

= ${\displaystyle \frac{13}{7}}$

=${\displaystyle 1\frac{6}{7}}$

Question 19

If cot θ= 1; find the value of: 5 tan² θ+ 2 sin² θ- 3

Sol :

Consider the diagram below :

cot θ = 1

i.e.${\displaystyle \frac{\text{base}}{\text{perpendicular}}=\frac{1}{1}}$

Therefore if length of base = x, length of perpendicular = x

Since
base² + perpendicular² = hypotenuse² ...[ Using Pythagooras Theorem]

(x)² + (x)² = hypotenuse²

hypotenuse² = x² + x² = 2x

∴ hypotenuse =${\displaystyle \sqrt{2}x}$

Now

sin θ = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{x}{\sqrt{2}x}=\frac{1}{\sqrt{2}}}$

tan θ = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{x}{x}=1}$

Therefore
5tan² θ + 2sin² θ – 3

= ${\displaystyle 5{\left(1\right)}^2+2{\left(\frac{1}{\sqrt{2}}\right)}^2–3}$

= 5 + 1 – 3

=3

Question 20

In the following figure : 
AD⊥BC, AC = 26 CD = 10, BC = 42,
∠DAC = x and ∠B = y.
Find the value of :
(i) cot x
(ii) ${\displaystyle \frac{1}{{\sin }^{2}y}–\frac{1}{{\tan }^{2}y}}$
(iii) ${\displaystyle \frac{6}{\cos x}–\frac{5}{\cos y}+8\tan y}$.

Sol:

Given the angle, DAC = 90°  and ∠ADB = 90° in the figure

⇒ AC² = AD² + DC²  ...(AC is hypotenuse in ΔADC)

⇒ AD² = 26² – 10²

∴ AD² = 576 and AD = 24

Again
⇒ AB² = AD² + BD²  ...(AB is hypotenuse in ΔABD)

⇒ AB² = 24² + 32²

∴ AB² = 1600 and AB = 40

Now
(i) cot x = ${\displaystyle \frac{\text{base}}{\text{perpendicular}}=\frac{\text{AD}}{\text{CD}}=\frac{24}{10}=2.4}$

(ii) sin y = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{AD}}{\text{AB}}=\frac{24}{40}=\frac{3}{5}}$

(iii) tan y = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{AD}}{\text{BD}}=\frac{24}{32}=\frac{3}{4}}$

Therefore

${\displaystyle \frac{1}{{\sin }^{2}y}–\frac{1}{{\tan }^{2}y}}$

= ${\displaystyle \frac{1}{{\left(\frac{3}{2}\right)}^2}–\frac{1}{{\left(\frac{3}{4}\right)}^2}}$

= ${\displaystyle \frac{25}{9}–\frac{16}{9}}$

= ${\displaystyle \frac{9}{9}}$

= 1

(iii) tan y =${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{\text{AD}}{\text{BD}}=\frac{24}{32}=\frac{3}{4}}$

cos x = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\text{AD}}{\text{AC}}=\frac{24}{26}=\frac{12}{13}}$

cos y = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{\text{BD}}{\text{AB}}=\frac{32}{40}=\frac{4}{5}}$  

Therefore
${\displaystyle \frac{6}{\cos x}–\frac{5}{\cos y}+8\tan y}$

= ${\displaystyle \frac{6}{\frac{12}{13}}–\frac{5}{\frac{4}{5}}+8\left(\frac{3}{4}\right)}$

= ${\displaystyle \frac{13}{2}–\frac{25}{4}+6}$

= ${\displaystyle \frac{26–25+24}{4}}$

= ${\displaystyle \frac{25}{4}}$

= ${\displaystyle 6\frac{1}{4}}$