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

Question 1.1

Given A = 60° and B = 30°,
prove that : sin (A + B) = sin A cos B + cos A sin B

Sol:

Given A = 60° and B = 30°

LHS = sin(A + B)
= sin (60° + 30°)
= sin 90°
= 1

RHS = sin A cos B + cos A sin B
= sin 60° cos 30° + cos 60° sin 30°

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

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

= 1
LHS = RHS

Question 1.2

Given A = 60° and B = 30°,
prove that : cos (A + B) = cos A cos B - sin A sin B

Sol:

Given A = 60° and B = 30°

LHS = cos(A+B)
= cos(60° + 30°)
= cos90°
=0

RHS = cos A cos B – sin A sin B
= cos 60° cos 30° – sin 60° sin 30°

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

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

= 0
LHS  = RHS

Question 1.3

Given A = 60° and B = 30°,
prove that : cos (A - B) = cos A cos B + sin A sin B

Sol:

Given A = 60° and B = 30°

LHS = cos(A – B)

= cos (60° – 30°)

= cos 30°

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

RHS = cos A cos B + sin A sin B

= cos 60° cos 30° + sin 60° sin 30°

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

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

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

LHS = RHS

Question 1.4

Given A = 60° and B = 30°,
prove that : tan (A - B) = ${\displaystyle \frac{\tan \text{A}–\tan \text{B}}{1+\tan \text{A}.\tan \text{B}}}$

Sol:

LHS = tan(A – B) 

= tan (60° – 30°)

= tan30°

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

RHS = ${\displaystyle \frac{\tan \text{A}– \tan \text{B}}{1+\tan 60°.\tan 30°}}$

= ${\displaystyle \frac{\tan 60°–\tan 30°}{1+\tan 60°.\tan 30°}}$

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

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

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

LHS = RHS

Question 2.1

If A =30^o, then prove that :
sin 2A = 2sin A cos A =  ${\displaystyle \frac{2\tan \text{A}}{1+{\tan }^2\text{A}}}$

Sol:

Given A = 30°

sin 2A = sin 2(30°) = sin60° = ${\displaystyle \frac{\sqrt{3}}{2}}$

2sin A cos A = 2sin 30° cos 30°

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

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

${\displaystyle \frac{2\tan \text{A}}{1+{\tan }^{2}30°}=\frac{2\tan 30°}{1+{\tan }^{2}30°}}$

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

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

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

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

∴ sin 2A = 2sin A cos A = ${\displaystyle \frac{2\tan \text{A}}{1+{\tan }^2\text{A}}}$

Question 2.2

If A =30^o, then prove that :
cos 2A = cos²A - sin²A = ${\displaystyle \frac{1–{\tan }^2\text{A}}{1+{\tan }^2\text{A}}}$

Sol:

Given A = 30°

cos2A = cos 2 (30°) = cos 60° = ${\displaystyle \frac{1}{2}}$

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

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

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

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

= ${\displaystyle \frac{2}{4}}$

= '(1)/(2)`

∴ cos 2A = ${\displaystyle {\cos }^{\text{A}}–{\sin }^2\text{A}=\frac{1–{\tan }^2\text{A}}{1+{\tan }^2\text{A}}}$

Question 2.3

If A =30^o, then prove that :
2 cos² A - 1 = 1 - 2 sin²A

Sol:

Given A = 30°

2 cos² A – 1 = 2 cos² 30° – 1

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

= ${\displaystyle \frac{3}{2}–1}$

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

∴ 2 cos²A – 1 = 1 – 2 sin²A

Question 2.4

If A =30^o, then prove that :
sin 3A = 3 sin A - 4 sin³A.

Sol:

Given A = 30°

sin 3A = sin 3(30°)
= sin 90°
=1

3 sin A – 4 sin³A = 3 sin 30° – 4 sin³30°

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

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

= 1

∴ sin 3A = 3 sin A – 4 sin³A

Question 3.1

If A = B = 45° ,
show that:
sin (A - B) = sin A cos B - cos A sin B

Sol:

Given that A = B = 45°

LHS = sin (A – B)

= sin ( 45° – 45°)

= sin 0°

= 0

RHS = sin A cos B – cos A sin B

= sin 45° cos 45° – cos 45° sin 45°

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

= 0

LHS = RHS

Question 3.2

If A = B = 45° ,
show that:
cos (A + B) = cos A cos B - sin A sin B

Sol:

Given that A = B = 45°

LHS = cos (A +  B)

= cos ( 45° + 45°)

= cos 90°

= 0

RHS = cos A cos B – sin A sin B

= cos 45° cos 45° – sin 45° sin 45°

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

= 0

LHS = RHS

Question 4.1

If A = 30°;
show that:
sin 3 A = 4 sin A sin (60° - A) sin (60° + A)

Sol:

Given that A = 30°

LHS = sin 3 A
= sin 3(30°)
= sin 90°
=1

RHS = 4 sin A sin (60° – A) sin (60° + A)

= 4 sin 30° sin ( 60° – 30°) sin (60° + 30°)

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

= 1

LHS = RHS

Question 4.2

If A = 30°;
show that:
(sinA - cosA)² = 1 - sin2A

Sol:

Given that A = 30°

LHS = ${\displaystyle {(\sin \text{A}–\cos \text{A})}^2}$

=${\displaystyle {(\sin 30°–\cos 30°)}^2}$

=${\displaystyle {(\frac{1}{2}–\frac{\sqrt{3}}{2})}^2}$

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

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

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

RHS = 1 – sin 2A

= 1 – sin 2(30°)

= 1 – sin60°

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

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

LHS = RHS

Question 4.3

If A = 30°;
show that:
cos 2A = cos⁴ A - sin⁴ A

SoL:

Given that A = 30°

LHS = cos 2A

= cos 2(30°)

= cos 60°

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

RHS = ${\displaystyle {\cos }^4\text{A}–{\sin }^4\text{A}}$

= ${\displaystyle {\cos }^{4}30°–{\sin }^{4}30°}$

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

= ${\displaystyle \frac{9}{16}– \frac{1}{16}}$

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

LHS = RHS

Question 4.4

If A = 30°;
show that:
${\displaystyle \frac{1–\cos 2\text{A}}{\sin 2\text{A}}=\tan \text{A}}$

Sol:

Given that A = 30°

LHS = ${\displaystyle \frac{1–\cos 2\text{A}}{\sin 2\text{A}}}$

= ${\displaystyle \frac{1–\cos 2(30°)}{\sin 2(30°)}}$

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

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

RHS = tan A

= tan 30°

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

LHS = RHS

Question 4.5

If A = 30°;
show that:
${\displaystyle \frac{1+\sin 2\text{A}+\cos 2\text{A}}{\sin \text{A}+\cos \text{A}}=2\cos \text{A}}$

Sol:

Given that A = 30°

LHS = ${\displaystyle \frac{1+\sin 2\text{A}+\cos 2\text{A}}{\sin \text{A}+\cos \text{A}}}$

= ${\displaystyle \frac{1+\sin 2(30°)+\cos 2(30°)}{\sin 30°+\cos 30°}}$

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

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

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

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

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

RHS = 2 cos A

= 2 cos (30°)

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

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

Question 4.6

If A = 30°;
show that:
4 cos A cos (60° - A). cos (60° + A) = cos 3A

Sol:

Given that A = 30°

LHS  = 4 cos A cos (60° – A ). cos (60° + A)

= 4 cos 30° cos (60° – 30°). cos (60° + 30°)

= 4 cos 30° cos 30° cos 90°

= ${\displaystyle 4\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\left(0\right)}$

= 0

RHS = cos 3A

= cos3(30°)

= cos 90°

=0

LHS = RHS

Question 4.7

If A = 30°;
show that:
${\displaystyle \frac{{\cos }^3\text{A}–\cos 3\text{A}}{\cos \text{A}}+\frac{{\sin }^3\text{A}+\sin 3\text{A}}{\sin \text{A}}=3}$

Sol:

Given that A = 30°

LHS = ${\displaystyle \frac{{\cos }^3\text{A}–\cos 3\text{A}}{\cos \text{A}}+\frac{{\sin }^3\text{A}+\sin 3\text{A}}{\sin \text{A}}}$

= ${\displaystyle \frac{{\cos }^{3}30°–\cos 3(30°)}{\cos 30°}+\frac{{\sin }^{3}30°+\sin 3(30°)}{\sin 30°}}$

= ${\displaystyle \frac{{\left(\frac{\sqrt{3}}{2}\right)}^3–0}{\frac{\sqrt{3}}{2}}+\frac{{\left(\frac{1}{2}\right)}^3+1}{\frac{1}{2}}}$

= ${\displaystyle {\left(\frac{\sqrt{3}}{2}\right)}^2+\frac{\frac{9}{8}}{\frac{1}{2}}}$

= ${\displaystyle \frac{3}{4}+\frac{9}{4}}$

= ${\displaystyle \frac{12}{4}}$

= 3

= RHS