SELINA Solution Class 9 Chapter 23 Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios] Exercise 23 A

Question 1.1

find the value of: sin 30° cos 30°

Sol:

sin 30° cos 30° = ${\displaystyle \frac{1}{2}.\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}}$ 

Question 1.2

find the value of: tan 30° tan 60°

Sol:

tan 30° tan 60° = ${\displaystyle \frac{1}{\sqrt{3}}\left(\sqrt{3}\right)=1}$

Question 1.3

find the value of: cos² 60° + sin² 30°

Sol:

cos² 60° + sin² 30° = ${\displaystyle {\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}}$

Question 1.4

find the value of: cosec² 60° - tan² 30°

Sol:

cosec² 60° – tan² 30° = ${\displaystyle {\left(\frac{2}{\sqrt{3}}\right)}^2–{\left(\frac{1}{\sqrt{3}}\right)}^2=\frac{4}{3}–\frac{1}{3}=1}$

Question 1.5

find the value of: sin² 30° + cos² 30°+ cot² 45°

Sol:

sin² 30° + cos²30° + cot² 45° = ${\displaystyle {\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2+1^2=\frac{1}{4}+\frac{3}{4}+1=2}$

Question 1.6

find the value of: cos² 60° + sec² 30° + tan² 45°

Sol:

cos²  60° + sec² 30° + tan² 45° = ${\displaystyle {\left(\frac{1}{2}\right)}^2+{\left(\frac{2}{\sqrt{3}}\right)}^2+1^2}$

= ${\displaystyle \frac{1}{4}+\frac{4}{3}+1}$

= ${\displaystyle \frac{3+16+12}{12}}$

= ${\displaystyle \frac{31}{12}}$

= ${\displaystyle 2\frac{7}{12}}$

Question 2.1

find the value of :
tan² 30° + tan² 45° + tan² 60°

Sol:

tan² 30° + tan² 45° + tan² 60° = ${\displaystyle {\left(\frac{1}{\sqrt{3}}\right)}^2+1^2+{\left(\sqrt{3}\right)}^2=\frac{1}{3}+1+3=\frac{13}{3}=4\frac{1}{3}}$

Question 2.2

find the value of :

${\displaystyle \frac{\tan 45°}{\cos ec30°}+\frac{\sec 60°}{co45°}–\frac{5\sin 90°}{2\cos 0°}}$

Sol:

${\displaystyle \frac{\tan 45°}{\cos ec30°}+\frac{\sec 60°}{co45°}–\frac{5\sin 90°}{2\cos 0°}=\frac{1}{2}+\frac{2}{1}+\frac{5}{2}}$

${\displaystyle =\frac{1+4–5}{2}=0}$

Question 2.3

find the value of :

3sin² 30° + 2tan² 60° - 5cos² 45°

Sol:

3 sin² 30° + 2 tan² 60° – 5 cos² 45° 

= ${\displaystyle 3{\left(\frac{1}{2}\right)}^2+2{\left(\sqrt{3}\right)}^2–5{\left(\frac{1}{2}\right)}^2=\frac{3}{4}+6–\frac{5}{2}=\frac{3+24–10}{4}=4\frac{1}{4}}$

Question 3.1

Prove that:

sin 60° cos 30° + cos 60° . sin 30°  = 1

Sol:

LHS =sin 60° cos 30° + cos 60°. sin 30°

= ${\displaystyle \frac{\sqrt{3}}{2}\frac{\sqrt{3}}{2}+\frac{1}{2}\frac{1}{2}=\frac{3}{4}+\frac{1}{4}=1=RHS}$

Question 3.2

Prove that:

cos 30° . cos 60° - sin 30° . sin 60°  = 0

Sol:

LHS=cos 30°. cos 60° - sin 30°. sin 60°

= ${\displaystyle \frac{\sqrt{3}}{2}\frac{1}{2}–\frac{1}{2}\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}–\frac{\sqrt{3}}{4}=0=RHS}$

Question 3.3

Prove that:

cosec² 45°  - cot² 45°  = 1

Sol:

LHS= cosec² 45° - cot² 45°

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

Question 3.4

Prove that:

cos² 30°  - sin² 30° = cos 60°

Sol:

${\displaystyle {\left(\frac{\sqrt{3}}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2=\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}=\cos 60°}$

Question 3.5

Prove that:

${\displaystyle {\left(\frac{\tan 60° +1}{\tan 60° –1}\right)}^2=\frac{1+\cos 30°}{1–\cos 30°}}$

Sol:

LHS = ${\displaystyle {\left(\frac{\tan 60°+1}{\tan 60°–1}\right)}^2}$

= ${\displaystyle {\left(\frac{\sqrt{3}+1}{\sqrt{3}–1}\right)}^2=\frac{4+2\sqrt{3}}{4–2\sqrt{3}}=\frac{1+\frac{\sqrt{3}}{2}}{1–\frac{\sqrt{3}}{2}}=\frac{1+\cos 30°}{1–\cos 30°}=RHS}$

Question 3.6

Prove that:

3 cosec² 60°  - 2 cot² 30°  + sec² 45°  = 0

Sol:

LHS =3 cosec²60° – 2 cot²30° + sec²45°

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

= ${\displaystyle 3\times \frac{4}{3}-2\times 3+2}$

= 4 – 6 + 2

= 0

= RHS

Question 4.1

prove that:

sin (2 x 30°) = ${\displaystyle \frac{2\tan 30°}{1+{\tan }^{2}30°}}$

Sol:

RHS = ${\displaystyle \frac{2\tan 30°}{1+{\tan }^{2}30°}=\frac{2\times \frac{1}{\sqrt{3}}}{1+{\left(\frac{1}{\sqrt{3}}\right)}^2}=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}=\frac{\sqrt{3}}{2}}$

LHS = sin (2 x 30°) = sin 60° = ${\displaystyle \frac{\sqrt{3}}{2}}$

∴ LHS = RHS

Question 4.2

prove that:

cos (2 x 30°) = ${\displaystyle \frac{1–{\tan }^{2}30°}{1+{\tan }^{2}30°}}$

Sol:

RHS,

${\displaystyle \frac{1–{\tan }^{2}30°}{1+{\tan }^{2}30°}=\frac{1–\frac{1}{3}}{1+\frac{1}{3}}=\frac{1}{2}}$

LHS,

cos (2 x 30°) = ${\displaystyle \cos 60°=\frac{1}{2}}$

LHS = RHS

Question 4.3

prove that:

tan (2 x 30°) = ${\displaystyle \frac{2\tan 30°}{1–{\tan }^{2}30°}}$

Sol:

RHS,

${\displaystyle \frac{2{\tan }^{2}30°}{1–{\tan }^{2}30°}=\frac{2\frac{1}{\sqrt{3}}}{1–\frac{1}{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}=\sqrt{3}}$

LHS,
tan (2 x 30°) = tan 60° = ${\displaystyle \sqrt{3}}$
LHS = RHS

Question 5.1

ABC is an isosceles right-angled triangle. Assuming of AB = BC = x, find the value of each of the following trigonometric ratios: sin 45°

Sol:

Given that AB = BC = x

∴ AC = ${\displaystyle \sqrt{A{B}^2+B{C}^2}=\sqrt{x^2+x^2}=x\sqrt{2}}$

sin 45° = ${\displaystyle \frac{\text{AB}}{\text{AC}}=\frac{x}{x\sqrt{2}}=\frac{1}{\sqrt{2}}}$

Question 5.2

ABC is an isosceles right-angled triangle. Assuming of AB = BC = x, find the value of each of the following trigonometric ratio: cos 45°

Sol:

Given that AB = BC = x

∴ AC = ${\displaystyle \sqrt{A{B}^2+B{C}^2}=\sqrt{x^2+x^2}=x\sqrt{2}}$

cos 45° =${\displaystyle \frac{\text{BC}}{\text{AC}}=\frac{x}{x\sqrt{2}}=\frac{1}{\sqrt{2}}}$

Question 5.3

ABC is an isosceles right-angled triangle. Assuming of AB = BC = x, find the value of each of the following trigonometric ratios: tan 45°

Sol:

Given that AB = BC = x

∴ AC = ${\displaystyle \sqrt{A{B}^2+B{C}^2}=\sqrt{x^2+x^2}=x\sqrt{2}}$

tan 45°  = ${\displaystyle \frac{\text{AB}}{\text{BC}}=\frac{x}{x}=1}$

Question 6.1

Prove that:
sin 60° = 2 sin 30° cos 30°

Sol:

LHS = sin 60° = ${\displaystyle \frac{\sqrt{3}}{2}}$

RHS = 2 sin 60° cos 60° = ${\displaystyle 2\times \frac{\sqrt{3}}{2}\times \frac{1}{2}=\frac{\sqrt{3}}{2}}$

LHS = RHS

Question 6.2

Prove that:
4 (sin⁴ 30° + cos⁴ 60°) -3 (cos² 45° - sin² 90°) = 2

Sol:

LHS = ${\displaystyle 4({\sin }^{4}30°+{\cos }^{4}60°)-3({\cos }^{2}45°–{\sin }^{2}90°)}$

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

= ${\displaystyle 4[\frac{1}{16}+\frac{1}{16}]–3[\frac{1}{2}–1]=\frac{4\times 2}{16}+3\times \frac{1}{2}=2}$ 

RHS = 2
LHS = RHS

Question 7.1

If sin x = cos x and x is acute, state the value of x

Sol:

The angle, x is acute and hence we have, 0 < x
We know that
cos²x + sin² x = 1
⇒ 2sin² x = 1

⇒ sin x = ${\displaystyle \frac{1}{\sqrt{2}}}$

⇒ x = 45°

Question 7.2

If sec A = cosec A and 0° ∠A ∠90°, state the value of A

Sol:

sec A = cosec A
cos A = sin A
cos²A = sin²A
cos² A = 1 – cos²A
2cos²A = 1

cos A = ${\displaystyle \frac{1}{\sqrt{2}}}$
A = 45°

Question 7.3

If tan θ = cot θ and 0°∠θ ∠90°, state the value of θ

Sol:

tan θ = cotθ 

tan θ  = ${\displaystyle \frac{1}{\tan \theta }}$

tan² θ  = 1
tan θ  = 1
tan θ  = tan 45°
θ  = 45°

Question 7.4

If sin x = cos y; write the relation between x and y, if both the angles x and y are acute.

Sol:

sin x = cos y = sin (90° – y )
if x and y are acute angles,
x = 90° – y
⇒ x + y = 90
Hence x and y are complement angles 

Question 8.1 

 MCQ 

True or False

If sin x = cos y, then x + y = 45° ; write true of false

False

sin x = cosy = sin${\displaystyle (\frac{x}{2}–y)}$

if x and y are acute angles,

x = ${\displaystyle \frac{x}{2}–y}$

x + y = ${\displaystyle \frac{x}{2}}$

∴ x + y = 45° is false.

Question 8.2

secθ . Cot θ= cosecθ ; write true or false

Sol:

True

sec θ . cot θ = ${\displaystyle \frac{1}{\cos \theta }\frac{\cos \theta }{\sin \theta }=\frac{1}{\sin \theta }=\cos ec\theta }$

Secθ . cot θ = cosec θ is true

Question 8.3

For any angle θ, state the value of : sin² θ + cos² θ

Sol:

sin² θ  =cos² θ = sin² θ + 1 – sin²θ = 1

Question 9.1

State for any acute angle θ whether sin θ increases or decreases as θ increases

Increase Sol:

For acute angles, remember what sine means: opposite over hypotenuse. If we increase the angle, then the opposite side gets larger. That means "opposite/hypotenuse" gets larger or increases.

Question 9.2

State for any acute angle θ whether cos θ increases or decreases as θ increases.

Increase Sol:

For acute angles, remember what cosine means: base over hypotenuse. If we increase the angle, then the hypotenuse side gets larger. That means "base/hypotenuse" gets smaller or decreases.

Question 9.3

State for any acute angle θ whether tan θ increases or decreases as θ decreases.

Decrease Sol: For Acute angles, remember what tangent means: Opposite over base. If we decrease the angle, then the opposite side gets smaller. That Means "Opposite/Base" Decreases.

Question 10.1

If ${\displaystyle \sqrt{3}}$ = 1.732, find (correct to two decimal place)  the value of sin 60^o

Sol:

sin 60° = ${\displaystyle \frac{\sqrt{3}}{2}=\frac{1.732}{2}=0.87}$

Question 10.2

If ${\displaystyle \sqrt{3}}$ = 1.732, find (correct to two decimal place)  the value of  ${\displaystyle \frac{2}{\tan 30°}}$

Sol:

${\displaystyle \frac{2}{\tan 30°}=\frac{2}{\frac{1}{\sqrt{3}}}=2\sqrt{3}=2\times 1.732=3.46}$

Question 11.1

Evaluate : 
${\displaystyle \frac{\cos 3\text{A}–2\cos 4\text{A}}{\sin 3\text{A}+2\sin 4\text{A}}}$ , when A = 15°

Sol:

(i) Given that A= 15°

${\displaystyle \frac{\cos 3\text{A}–2\cos 4\text{A}}{\sin 3\text{A}+2\sin 4\text{A}}=\frac{\cos (3x15°)–2\cos (4x15°)}{\sin (3x15°)+2\sin (4x15°)}}$

= ${\displaystyle \frac{\cos 45°–2\cos 60°}{\sin 45°+2\sin 60°}}$

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

= ${\displaystyle \frac{\frac{1}{\sqrt{2}}–1}{\frac{1}{\sqrt{2}}+3}}$

= ${\displaystyle \frac{1–\sqrt{2}}{1+\sqrt{6}}}$

= ${\displaystyle \frac{1}{5}(\sqrt{6}–1–2\sqrt{3}+\sqrt{2})}$

Question 11.2

Evaluate : 

${\displaystyle \frac{3\sin 3\text{B}+2\cos (2\text{B}+5°)}{2\cos 3\text{B}–\sin (2\text{B}–10°)}}$ ; when "B" = 20°.

Sol:

Given that B = 20°

${\displaystyle \frac{3\sin 3\text{B}+2\cos (2\text{B}+5°)}{2\cos 3\text{B}–\sin (2\text{B}–10°)}}$ = ${\displaystyle \frac{3\sin 3\times 20°+2\cos (2\times 20°+5°)}{2\cos 3\times 20°–\sin (2\times 20°–10°)}}$

= ${\displaystyle \frac{3\sin 60°+2\cos 45°}{2\cos 60°–\sin 30°}}$

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

= ${\displaystyle \frac{3\frac{\sqrt{3}}{2}+\sqrt{2}}{2}}$

= ${\displaystyle 3\sqrt{3} +2\sqrt{2}}$

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