Rs Aggarwal 2021 2022 for Class 10 Maths Chapter 11 - T Ratios Of Some Particular Angles

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Rs Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 11 T Ratios Of Some Particular Angles are provided here with simple step-by-step explanations. These solutions for T Ratios Of Some Particular Angles are extremely popular among Class 10 students for Maths T Ratios Of Some Particular Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2021 2022 Book of Class 10 Maths Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 554:

Answer:

$3 (x^{2} - \frac{1}{x^{2}}) = \frac{9}{3} (x^{2} - \frac{1}{x^{2}}) = \frac{1}{3} (9 x^{2} - \frac{9}{x^{2}}) = \frac{1}{3} [{(3 x)}^{2} - {(\frac{3}{x})}^{2}]$
$= \frac{1}{3} [{(cosec θ)}^{2} - {(\cot θ)}^{2}] = \frac{1}{3} ({cosec}^{2} θ - \cot^{2} θ) = \frac{1}{3} (1) = \frac{1}{3}$

Page No 554:

Answer:

From the given right-angled triangle, we have:
$\frac{B C}{A C} = \sin 30^{o} \Rightarrow \frac{B C}{20} = \frac{1}{2} \Rightarrow B C = \frac{20}{2} = 10 cm Also, \frac{A B}{A C} = \cos 30^{o} \Rightarrow \frac{A B}{20} = \frac{\sqrt{3}}{2} \Rightarrow A B = (20 \times \frac{\sqrt{3}}{2}) = 10 \sqrt{3} cm ∴ BC = 10 cm and AB = 10 \sqrt{3} cm$

Page No 554:

Answer:

From the given right-angled triangle, we have:

$\frac{B C}{A B} = \tan 30^{o} \Rightarrow \frac{6}{A B} = \frac{1}{\sqrt{3}} \Rightarrow A B = 6 \sqrt{3} cm Also, \frac{B C}{A C} = \sin 30^{o} \Rightarrow \frac{6}{A C} = \frac{1}{2} \Rightarrow A C = (2 \times 6) = 12 cm ∴ A B = 6 \sqrt{3} cm and A C = 12 cm$

Page No 554:

Answer:

From right-angled ∆ABC, we have:

$\frac{B C}{A C} = \sin 45^{o} \Rightarrow \frac{B C}{3 \sqrt{2}} = \frac{1}{\sqrt{2}} \Rightarrow B C = 3 cm Also, \frac{A B}{A C} = \cos 45^{o} \Rightarrow \frac{A B}{3 \sqrt{2}} = \frac{1}{\sqrt{2}} \Rightarrow A B = 3 cm ∴ B C = 3 cm and A B = 3 cm$

Page No 572:

Answer:

On substituting the values of various T-ratios, we get:
sin 60^o cos 30^o + cos 60^o sin 30^o
$= (\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2}) = (\frac{3}{4} + \frac{1}{4}) = \frac{4}{4} = 1$

Page No 572:

Answer:

Ans

Page No 572:

Answer:

On substituting the values of various T-ratios, we get:
cos 60^o cos 30^o − sin 60^o sin 30^o
$= (\frac{1}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \times \frac{1}{2}) = (\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4}) = 0$

Page No 572:

Answer:

On substituting the values of various T-ratios, we get:
cos 45^o cos 30^o + sin 45^o sin 30^o
$= (\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2}) = (\frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}}) = (\frac{\sqrt{3} + 1}{2 \sqrt{2}})$

Page No 572:

Answer:

As we know that,
$\tan 30 ° = \frac{1}{\sqrt{3}} \sec 30 ° = \frac{2}{\sqrt{3}} cosec 60 ° = \frac{2}{\sqrt{3}} \tan 60 ° = \sqrt{3} By substituting these values, we get \tan 30 ° cosec 60 ° + \tan 60 ° \sec 30 ° = \frac{1}{\sqrt{3}} \times \frac{2}{\sqrt{3}} + \sqrt{3} \times \frac{2}{\sqrt{3}} = \frac{2}{3} + 2 = \frac{2 + 6}{3} = \frac{8}{3} Hence, \tan 30 ° cosec 60 ° + \tan 60 ° \sec 30 ° = \frac{8}{3}$

Page No 572:

Answer:

On substituting the values of various T-ratios, we get:
2 cos² 60^o + 3 sin² 45^o − 3 sin² 30^o + 2 cos² 90^o
$= 2 \times {(\frac{1}{2})}^{2} + 3 \times {(\frac{1}{\sqrt{2}})}^{2} - 3 \times {(\frac{1}{2})}^{2} + 2 \times {(0)}^{2} = 2 \times \frac{1}{4} + 3 \times \frac{1}{2} - 3 \times \frac{1}{4} + 0 = (\frac{1}{2} + \frac{3}{2} - \frac{3}{4}) = (\frac{2 + 6 - 3}{4}) = \frac{5}{4}$

Page No 572:

Answer:

As we know that,
$\sin 30 ° = \frac{1}{2} \cos 45 ° = \frac{1}{\sqrt{2}} \tan 30 ° = \frac{1}{\sqrt{3}} \sin 90 ° = 1 \cot 60 ° = \frac{1}{\sqrt{3}} By substituting these values, we get \sin^{2} 30 ° \cos^{2} 45 ° + 4 \tan^{2} 30 ° + \frac{1}{2} \sin^{2} 90 ° + \frac{1}{8} \cot^{2} 60 ° = {(\frac{1}{2})}^{2} \times {(\frac{1}{\sqrt{2}})}^{2} + 4 {(\frac{1}{\sqrt{3}})}^{2} + \frac{1}{2} {(1)}^{2} + \frac{1}{8} {(\frac{1}{\sqrt{3}})}^{2} = (\frac{1}{4}) \times (\frac{1}{2}) + 4 (\frac{1}{3}) + \frac{1}{2} (1) + \frac{1}{8} (\frac{1}{3}) = \frac{1}{8} + \frac{4}{3} + \frac{1}{2} + \frac{1}{24} = \frac{3 + 32 + 12 + 1}{24} = \frac{48}{24} = 2 Hence, \sin^{2} 30 ° \cos^{2} 45 ° + 4 \tan^{2} 30 ° + \frac{1}{2} \sin^{2} 90 ° + \frac{1}{8} \cot^{2} 60 ° = 2 .$

Page No 572:

Answer:

On substituting the values of various T-ratios, we get:
cot² 30^o − 2 cos² 30^o − $\frac{3}{4}$ sec² 45^o + $\frac{1}{4}$ cosec² 30^o
$= {(\sqrt{3})}^{2} - 2 \times {(\frac{\sqrt{3}}{2})}^{2} - \frac{3}{4} \times {(\sqrt{2})}^{2} + \frac{1}{4} \times {(2)}^{2} = 3 - 2 \times \frac{3}{4} - \frac{3}{4} \times 2 + \frac{1}{4} \times 4 = 3 - \frac{3}{2} - \frac{3}{2} + 1 = 4 - (\frac{3}{2} + \frac{3}{2}) = 4 - 3 = 1$

Page No 572:

Answer:

As we know that,
$\cos 0 ° = 1 \cos 45 ° = \frac{1}{\sqrt{2}} \sin 30 ° = \frac{1}{2} \sin 90 ° = 1 \cos 60 ° = \frac{1}{2} By substituting these values, we get (\cos 0 ° + \cos 45 ° + \sin 30 °) (\sin 90 ° - \cos 45 ° + \cos 60 °) = (1 + \frac{1}{\sqrt{2}} + \frac{1}{2}) (1 - \frac{1}{\sqrt{2}} + \frac{1}{2}) = ((1 + \frac{1}{2}) + (\frac{1}{\sqrt{2}})) ((1 + \frac{1}{2}) - (\frac{1}{\sqrt{2}})) = {(1 + \frac{1}{2})}^{2} - {(\frac{1}{\sqrt{2}})}^{2} = 1 + \frac{1}{4} + 2 (1) (\frac{1}{2}) - \frac{1}{2} = 1 + \frac{1}{4} + 1 - \frac{1}{2} = \frac{4 + 1 + 4 - 2}{4} = \frac{7}{4} Hence, (\cos 0 ° + \cos 45 ° + \sin 30 °) (\sin 90 ° - \cos 45 ° + \cos 60 °) = \frac{7}{4} .$

Page No 572:

Answer:

(i)
$LHS = \frac{1 - \sin 60^{o}}{\cos 60^{o}} = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{(\frac{2 - \sqrt{3}}{2})}{\frac{1}{2}} = (\frac{2 - \sqrt{3}}{2}) \times 2 = 2 - \sqrt{3} RHS = \frac{\tan 60^{o} - 1}{\tan 60^{o} + 1} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{{(\sqrt{3} - 1)}^{2}}{{(\sqrt{3})}^{2} - 1^{2}} = \frac{3 + 1 - 2 \sqrt{3}}{3 - 1} = \frac{4 - 2 \sqrt{3}}{2} = 2 - \sqrt{3}$

Hence, LHS = RHS

∴ $\frac{1 - \sin 60^{o}}{\cos 60^{o}} = \frac{\tan 60^{o} - 1}{\tan 60^{o} + 1}$

(ii)
$LHS = \frac{\cos 30^{o} + \sin 60^{o}}{1 + \sin 30^{o} + \cos 60^{o}} = \frac{(\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2})}{(1 + \frac{1}{2} + \frac{1}{2})} = \frac{\frac{\sqrt{3} + \sqrt{3}}{2}}{\frac{2 + 1 + 1}{2}} = \frac{\sqrt{3}}{2} Also, RHS = \cos 30^{o} = \frac{\sqrt{3}}{2}$

Hence, LHS = RHS

∴ $\frac{\cos 30^{o} + \sin 60^{o}}{1 + \sin 30^{o} + \cos 60^{o}} = \cos 30^{o}$ 1−sin60°cos60°

Page No 572:

Answer:

(i) sin 60^o cos 30^o − cos 60^o sin 30^o
$= (\frac{\sqrt{3}}{2}) \times (\frac{\sqrt{3}}{2}) - (\frac{1}{2}) \times (\frac{1}{2}) = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} Also, \sin 30^{o} = \frac{1}{2}$
∴ sin 60^o cos 30^o − cos 60^o sin 30^o = sin 30^o

(ii) cos 60^o cos 30^o + sin 60^o sin 30^o

$= (\frac{1}{2}) \times (\frac{\sqrt{3}}{2}) + (\frac{\sqrt{3}}{2}) \times (\frac{1}{2}) = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2} Also, \cos 30^{o} = \frac{\sqrt{3}}{2}$
∴ cos 60^o cos 30^o + sin 60^o sin 30^o = cos 30^o

(iii) 2 sin 30^o cos 30^o
$= 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} Also, \sin 60^{o} = \frac{\sqrt{3}}{2}$
∴ 2 sin 30^o cos 30^o = sin 60^o

(iv) 2 sin 45^o cos 45^o
$= 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 1$
Also, sin 90^o = 1
∴ 2 sin 45^o cos 45^o = sin 90^o

Page No 572:

Answer:

A = 45^o
⇒ 2A = 2 $\times$ 45^o = 90^o

(i) sin 2A = sin 90^o = 1
2 sin A cos A = 2 sin 45^o cos 45^o = $2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 1$
∴ sin 2A = 2 sin A cos A

(ii) cos 2A = cos 90^o = 0
2 cos² A − 1 = 2 cos² 45^o − 1 = $2 \times {(\frac{1}{\sqrt{2}})}^{2} - 1 = 2 \times \frac{1}{2} - 1 = 1 - 1 = 0$
Now, 1 − 2 sin² A = $1 - 2 \times {(\frac{1}{\sqrt{2}})}^{2} = 1 - 2 \times \frac{1}{2} = 1 - 1 = 0$
∴ cos 2A = 2 cos² A − 1 = 1 − 2 sin² A

Page No 572:

Answer:

A = 30^o
⇒ 2A = 2 $\times$ 30^o = 60^o

(i) sin 2A = sin 60^o = $\frac{\sqrt{3}}{2}$
$\frac{2 \tan A}{1 + \tan^{2} A} = \frac{2 \tan 30^{o}}{1 + \tan^{2} 30^{o}} = \frac{2 \times (\frac{1}{\sqrt{3}})}{1 + {(\frac{1}{\sqrt{3}})}^{2}} = \frac{(\frac{2}{\sqrt{3}})}{1 + \frac{1}{3}} = \frac{(\frac{2}{\sqrt{3}})}{\frac{4}{3}} = (\frac{2}{\sqrt{3}}) \times \frac{3}{4} = \frac{\sqrt{3}}{2}$
∴ $\sin 2 A = \frac{2 \tan A}{1 + \tan^{2} A}$

(ii) cos 2A = cos 60^o = $\frac{1}{2}$

\frac{1 - \tan^{2} A}{1 + \tan^{2} A} = \frac{1 - \tan^{2} 30^{o}}{1 + \tan^{2} 30^{o}} = \frac{1 - {(\frac{1}{\sqrt{3}})}^{2}}{1 + {(\frac{1}{\sqrt{3}})}^{2}} = \frac{(1 - \frac{1}{3})}{(1 + \frac{1}{3})} = \frac{(\frac{2}{3})}{\frac{4}{3}} = (\frac{2}{3}) \times \frac{3}{4} = \frac{1}{2}

∴

\cos 2 A = \frac{1 - \tan^{2} A}{1 + \tan^{2} A}

(iii) tan 2A = tan 60^o =

\sqrt{3}

\frac{2 \tan A}{1 - \tan^{2} A} = \frac{2 \tan 30^{o}}{1 - \tan^{2} 30^{o}} = \frac{2 \times (\frac{1}{\sqrt{3}})}{1 - {(\frac{1}{\sqrt{3}})}^{2}} = \frac{(\frac{2}{\sqrt{3}})}{1 - \frac{1}{3}} = \frac{(\frac{2}{\sqrt{3}})}{\frac{2}{3}} = (\frac{2}{\sqrt{3}}) \times \frac{3}{2} = \sqrt{3}

∴

\tan 2 A = \frac{2 \tan A}{1 - \tan^{2} A}

=2tanA1+tan2A

Page No 572:

Answer:

A = 60^o and B = 30^o
Now, A + B = 60^o + 30^o = 90^o
Also, A − B = 60^o − 30^o = 30^o

(i) sin (A + B) = sin 90^o = 1
sin A cos B + cos A sin B = sin 60^o cos 30^o + cos 60^o sin 30^o
= $(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2}) = (\frac{3}{4} + \frac{1}{4}) = 1$
∴ sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos 90^o = 0
cos A cos B − sin A sin B =cos 60^o cos 30^o − sin 60^o sin 30^o $= (\frac{1}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \times \frac{1}{2}) = (\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4}) = 0$
∴ cos (A + B) = cos A cos B − sin A sin B

Page No 573:

Answer:

(i) sin (A − B) = sin 30^o = $\frac{1}{2}$
sin A cos B − cos A sin B = sin 60^o cos 30^o − cos 60^o sin 30^o
= $(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{2}) = (\frac{3}{4} - \frac{1}{4}) = \frac{2}{4} = \frac{1}{2}$
∴ sin (A − B) = sin A cos B − cos A sin B

(ii) cos (A − B) = cos 30^o = $\frac{\sqrt{3}}{2}$
cos A cos B + sin A sin B = cos 60^o cos 30^o + sin 60^o sin 30^o
= $(\frac{1}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \times \frac{1}{2}) = (\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4}) = 2 \times \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$
∴ cos (A − B) = cos A cos B + sin A sin B

(iii) tan (A − B) = tan 30^o = $\frac{1}{\sqrt{3}}$
$\frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{\tan 60^{o} - \tan 30^{o}}{1 + \tan 60^{o} \tan 30^{o}} = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + (\sqrt{3} \times \frac{1}{\sqrt{3}})} = \frac{1}{2} \times (\frac{3 - 1}{\sqrt{3}}) = \frac{1}{\sqrt{3}}$
∴ tan (A − B) = $\frac{\tan A - \tan B}{1 + \tan A \tan B}$

Page No 573:

Answer:

Given:
$\tan A = \frac{1}{3} and \tan B = \frac{1}{2} \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} On substituting these values in RHS of the expression, we get : \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{(\frac{1}{3} + \frac{1}{2})}{(1 - \frac{1}{3} \times \frac{1}{2})} = \frac{(\frac{5}{6})}{(1 - \frac{1}{6})} = \frac{(\frac{5}{6})}{(\frac{5}{6})} = 1 \Rightarrow \tan (A + B) = 1 = \tan 45^{o} [∵ \tan 45^{o} = 1] ∴ A + B = 45^{o}$

\tan A = \frac{1}{3} a n d \tan B = \frac{1}{2} By substituting the values, we get; \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{(\frac{1}{3} + \frac{1}{2})}{(1 - \frac{1}{3} \times \frac{1}{2})} = \frac{(\frac{5}{6})}{(1 - \frac{1}{6})} = \frac{(\frac{5}{6})}{(\frac{5}{6})} = 1 \tan (A + B) = 1 = \tan 45^{o} Hence, A + B = 45

Page No 573:

Answer:

A = 30^o
⇒ 2A = 2 $\times$ 30^o = 60^o

By substituting the value of the given T-ratio, we get:
$\tan 2 A = \frac{2 \tan A}{1 - \tan^{2} A} \Rightarrow \tan 60^{o} = \frac{2 \tan 30^{o}}{1 - \tan^{2} 30^{o}} = \frac{2 \times \frac{1}{\sqrt{3}}}{1 - {(\frac{1}{\sqrt{3}})}^{2}} = \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \sqrt{3}$
∴ tan 60^o = $\sqrt{3}$

Page No 573:

Answer:

A = 30^o
⇒ 2A = 2 $\times$ 30^o = 60^o

By substituting the value of the given T-ratio, we get:
$\cos A = \sqrt{\frac{1 + \cos 2 A}{2}} \Rightarrow \cos 30^{o} = \sqrt{\frac{1 + \cos 60^{o}}{2}} = \sqrt{\frac{1 + \frac{1}{2}}{2}} = \sqrt{\frac{\frac{3}{2}}{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$
∴ cos 30^o = $\frac{\sqrt{3}}{2}$

Page No 573:

Answer:

A = 30^o
⇒ 2A = 2 $\times$ 30^o = 60^o

By substituting the value of the given T-ratio, we get:

$\sin A = \sqrt{\frac{1 - \cos 2 A}{2}} \Rightarrow \sin 30^{o} = \sqrt{\frac{1 - \cos 60^{o}}{2}} = \sqrt{\frac{1 - \frac{1}{2}}{2}} = \sqrt{\frac{\frac{1}{2}}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
∴ sin 30^o = $\frac{1}{2}$

Page No 573:

Answer:

$Let A = 45 ° and B = 30 °$

$(i) As, \sin (A + B) = sinA cosB + cosA sinB \Rightarrow \sin (45 ° + 30 °) = \sin 45 ° \cos 30 ° + \cos 45 ° \sin 30 ° \Rightarrow \sin (75 °) = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} \Rightarrow \sin 75 ° = \frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}} ∴ \sin 75 ° = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

$(ii) As, \cos (A - B) = cosA cosB + sinA sinB \Rightarrow \cos (45 ° - 30 °) = \cos 45 ° \cos 30 ° + \sin 45 ° \sin 30 ° \Rightarrow \cos (15 °) = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} \Rightarrow \cos 15 ° = \frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}} ∴ \cos 15 ° = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

Disclaimer: $\cos 15 °$ can also be calculated by taking $A = 60 ° and B = 45 °$ .

Page No 573:

Answer:

As we know that,
$\tan 45 ° = 1 Thus, if \tan (x + 30 °) = 1 \Rightarrow x + 30 ° = 45 ° \Rightarrow x = 45 ° - 30 ° \Rightarrow x = 15 ° Hence, the value of x is 15 ° .$

Page No 573:

Answer:

$Given : \frac{\cot θ - 1}{\cot θ + 1} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \frac{\cot θ - 1}{\cot θ + 1} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \Rightarrow \frac{\frac{1}{\tan θ} - 1}{\frac{1}{\tan θ} + 1} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} (∵ \cot θ = \frac{1}{\tan θ}) \Rightarrow \frac{\frac{1 - \tan θ}{\tan θ}}{\frac{1 + \tan θ}{\tan θ}} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \Rightarrow \frac{1 - \tan θ}{1 + \tan θ} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} On comparing LHS and RHS, we get \tan θ = \sqrt{3} \Rightarrow θ = 60 ° (∵ \tan 60 ° = \sqrt{3}) Hence, the acute angle θ is equal to 60 ° .$

Page No 573:

Answer:

As we know that,
$\sin 90 ° = 1 Thus, if \sin (A + B) = 1 \Rightarrow A + B = 90 ° . . . (1) and \tan 30 ° = \frac{1}{\sqrt{3}} Thus, if \tan (A - B) = \frac{1}{\sqrt{3}} \Rightarrow A - B = 30 ° . . . (2) Solving (1) and (2), we get A = 60 ° and B = 30 ° Hence, the values of A and B are 60 ° and 30 °, respectively .$

Page No 573:

Answer:

As we know that,
$\sin 60 ° = \frac{\sqrt{3}}{2} Thus, if \sin (A + B) = \frac{\sqrt{3}}{2} \Rightarrow A + B = 60 ° . . . (1) and \cos 30 ° = \frac{\sqrt{3}}{2} Thus, if \cos (A - B) = \frac{\sqrt{3}}{2} \Rightarrow A - B = 30 ° . . . (2) Solving (1) and (2), we get A = 45 ° and B = 15 ° Hence, the values of A and B are 45 ° and 15 °, respectively .$

Page No 573:

Answer:

Here, tan (A − B) = $\frac{1}{\sqrt{3}}$
⇒ tan (A − B) = tan 30^o   [∵ tan 30^o = $\frac{1}{\sqrt{3}}$ ]
⇒ A − B = 30^o                     ...(i)

Also, tan (A + B) = $\sqrt{3}$
⇒ tan (A + B) = tan 60^o      [∵ tan 60^o = $\sqrt{3}$ ]
⇒ A + B = 60^o...(ii)

Solving (i) and (ii), we get:
A = 45^o and B = 15^o

Page No 574:

Answer:

As we know that,
$cosec 90 ° = 1 Thus, if cosec (A + B) = 1 \Rightarrow A + B = 90 ° . . . (1) and cosec 30 ° = 2 Thus, if cosec (A - B) = 2 \Rightarrow A - B = 30 ° . . . (2) Solving (1) and (2), we get A = 60 ° and B = 30 ° . . . (3) Now, (i) \sin 60 ° = \frac{\sqrt{3}}{2} \cos 30 ° = \frac{\sqrt{3}}{2} \cos 60 ° = \frac{1}{2} \sin 30 ° = \frac{1}{2} On substituting these values, we get \sin A \cos B + \cos A \sin B = \sin 60 ° \cos 30 ° + \cos 60 ° \sin 30 ° = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1 Hence, \sin A \cos B + \cos A \sin B = 1 . (ii) \tan 60 ° = \sqrt{3} \tan 30 ° = \frac{1}{\sqrt{3}} On substituting these values, we get \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{\tan 60 ° - \tan 30 °}{1 + \tan 60 ° \tan 30 °} = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} (\frac{1}{\sqrt{3}})} = \frac{\frac{3 - 1}{\sqrt{3}}}{1 + 1} = \frac{2}{2 \sqrt{3}} = \frac{1}{\sqrt{3}} Hence, \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{1}{\sqrt{3}} .$

Page No 574:

Answer:

(i) As we know that,
$\tan 45 ° = 1 \cot 60 ° = \frac{1}{\sqrt{3}} \sin 30 ° = \frac{1}{2} cosec 60 ° = \frac{2}{\sqrt{3}} On substituting these values, we get x \tan 45 ° \cot 60 ° = \sin 30 ° cosec 60 ° \Rightarrow x (1) (\frac{1}{\sqrt{3}}) = (\frac{1}{2}) (\frac{2}{\sqrt{3}}) \Rightarrow \frac{x}{\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow x = 1 Hence, the value of x is 1 .$

(ii) As we know that,
$\tan 30 ° = \frac{1}{\sqrt{3}} \sin 60 ° = \frac{\sqrt{3}}{2} cosec 30 ° = 2 On substituting these values, we get 2 {cosec}^{2} 30 ° + x \sin^{2} 60 ° - \frac{3}{4} \tan^{2} 30 ° = 10 \Rightarrow 2 {(2)}^{2} + x {(\frac{\sqrt{3}}{2})}^{2} - \frac{3}{4} {(\frac{1}{\sqrt{3}})}^{2} = 10 \Rightarrow 2 (4) + x (\frac{3}{4}) - \frac{3}{4} (\frac{1}{3}) = 10 \Rightarrow 8 + \frac{3 x}{4} - \frac{1}{4} = 10 \Rightarrow \frac{3 x}{4} - \frac{1}{4} = 10 - 8 \Rightarrow \frac{3 x - 1}{4} = 2 \Rightarrow 3 x - 1 = 8 \Rightarrow 3 x = 9 \Rightarrow x = 3 Hence, the value of x is 3 .$

Page No 576:

Answer:

As we know that,
$\sec 60 ° = 2 By substituting the value, we get (\sec^{2} 60 ° - 1) = {(2)}^{2} - 1 = 4 - 1 = 3 Hence, the correct option is (c) .$

Page No 576:

Answer:

As we know that,
$\sin 30 ° = \frac{1}{2} \sec 60 ° = 2 \cot 45 ° = 1 By substituting these values, we get (\sin^{2} 30 ° - \sec^{2} 60 ° + 4 \cot^{2} 45 °) = {(\frac{1}{2})}^{2} - {(2)}^{2} + 4 {(1)}^{2} = \frac{1}{4} - 4 + 4 = \frac{1}{4} Hence, the correct option is (d) .$

Page No 576:

Answer:

As we know that,
$\sin 45 ° = \frac{1}{\sqrt{2}} \cos 60 ° = \frac{1}{2} \cot 30 ° = \sqrt{3} By substituting these values, we get (3 \cos^{2} 60 ° + 2 \cot^{2} 30 ° - 5 \sin^{2} 45 °) = 3 {(\frac{1}{2})}^{2} + 2 {(\sqrt{3})}^{2} - 5 {(\frac{1}{\sqrt{2}})}^{2} = 3 (\frac{1}{4}) + 2 (3) - 5 (\frac{1}{2}) = \frac{3}{4} + 6 - \frac{5}{2} = \frac{3 + 24 - 10}{4} = \frac{17}{4} Hence, the correct option is (c) .$

Page No 576:

Answer:

As we know that,
$\cos 30 ° = \frac{\sqrt{3}}{2} \cos 45 ° = \frac{1}{\sqrt{2}} \sec 60 ° = 2 \cos 90 ° = 0 \tan 60 ° = \sqrt{3} By substituting these values, we get (\cos^{2} 30 ° \cos^{2} 45 ° + 4 \sec^{2} 60 ° + \frac{1}{2} \cos^{2} 90 ° - 2 \tan^{2} 60 °) = {(\frac{\sqrt{3}}{2})}^{2} {(\frac{1}{\sqrt{2}})}^{2} + 4 {(2)}^{2} + \frac{1}{2} {(0)}^{2} - 2 {(\sqrt{3})}^{2} = (\frac{3}{4}) (\frac{1}{2}) + 4 (4) - 2 (3) = \frac{3}{8} + 16 - 6 = \frac{3}{8} + 10 = \frac{3 + 80}{8} = \frac{83}{8} Hence, the correct option is (b) .$

Page No 576:

Answer:

As we know that,
$\cos 0 ° = 1 \cos 45 ° = \frac{1}{\sqrt{2}} \sin 30 ° = \frac{1}{2} \sin 90 ° = 1 \cos 60 ° = \frac{1}{2} \sin 45 ° = \frac{1}{\sqrt{2}} By substituting these values, we get (\cos 0 ° + \sin 30 ° + \sin 45 °) (\sin 90 ° + \cos 60 ° - \cos 45 °) = (1 + \frac{1}{2} + \frac{1}{\sqrt{2}}) (1 + \frac{1}{2} - \frac{1}{\sqrt{2}}) = ((1 + \frac{1}{2}) + (\frac{1}{\sqrt{2}})) ((1 + \frac{1}{2}) - (\frac{1}{\sqrt{2}})) = {(1 + \frac{1}{2})}^{2} - {(\frac{1}{\sqrt{2}})}^{2} = 1 + \frac{1}{4} + 2 (1) (\frac{1}{2}) - \frac{1}{2} = 1 + \frac{1}{4} + 1 - \frac{1}{2} = \frac{4 + 1 + 4 - 2}{4} = \frac{7}{4} Hence, the correct option is (a) .$

Page No 576:

Answer:

As we know that,
$\cos 45 ° = \frac{1}{\sqrt{2}} \cos 30 ° = \frac{\sqrt{3}}{2} \tan 45 ° = 1 \sin 45 ° = \frac{1}{\sqrt{2}} By substituting these values, we get \tan^{2} 45 ° - \cos^{2} 30 ° = x \sin 45 ° \cos 45 ° \Rightarrow {(1)}^{2} - {(\frac{\sqrt{3}}{2})}^{2} = x (\frac{1}{\sqrt{2}}) (\frac{1}{\sqrt{2}}) \Rightarrow 1 - \frac{3}{4} = \frac{x}{2} \Rightarrow \frac{4 - 3}{4} = \frac{x}{2} \Rightarrow \frac{x}{2} = \frac{1}{4} \Rightarrow x = \frac{1}{2} Hence, the correct option is (c) .$

Page No 576:

Answer:

As we know that,
$\sin 45 ° = \frac{1}{\sqrt{2}} Thus, if \sqrt{2} \sin (60 ° - α) = 1 \Rightarrow \sin (60 ° - α) = \frac{1}{\sqrt{2}} \Rightarrow 60 ° - α = 45 ° \Rightarrow α = 60 ° - 45 ° \Rightarrow α = 15 ° Hence, the correct option is (a) .$

Page No 576:

Answer:

Given: tanx = 3cotx

$\tan x = 3 \cot x \Rightarrow \tan x = 3 \frac{1}{\tan x} (∵ \cot x = \frac{1}{\tan x}) \Rightarrow \tan^{2} x = 3 \Rightarrow \tan x = \sqrt{3} \Rightarrow x = 60 ° (∵ \tan 60 ° = \sqrt{3}) Hence, the correct option is (a) .$

Page No 576:

Answer:

As we know that,
$\tan 60 ° = \sqrt{3} Thus, if \sqrt{3} \tan 2 θ - 3 = 0 \Rightarrow \sqrt{3} \tan 2 θ = 3 \Rightarrow \tan 2 θ = \frac{3}{\sqrt{3}} \Rightarrow \tan 2 θ = \sqrt{3} \Rightarrow 2 θ = 60 ° \Rightarrow θ = 30 ° Hence, the correct option is (b) .$

Page No 576:

Answer:

As we know that,
$\sin 60 ° = \frac{\sqrt{3}}{2} Thus, if 2 \sin 2 θ = \sqrt{3} \Rightarrow \sin 2 θ = \frac{\sqrt{3}}{2} \Rightarrow 2 θ = 60 ° \Rightarrow θ = 30 ° Hence, the correct option is (a) .$

Page No 576:

Answer:

As we know that,
$\cos 60 ° = \frac{\sqrt{3}}{2} Thus, if 2 \cos 3 θ = 1 \Rightarrow \cos 3 θ = \frac{1}{2} \Rightarrow 3 θ = 60 ° \Rightarrow θ = 20 ° Hence, the correct option is (c) .$

Page No 576:

Answer:

As we know that,
$\tan 45 ° = 1 \cos 60 ° = \frac{1}{2} \sin 60 ° = \frac{\sqrt{3}}{2} \cot 60 ° = \frac{1}{\sqrt{3}} By substituting these values, we get x \tan 45 ° \cos 60 ° = \sin 60 ° \cot 60 ° \Rightarrow x (1) (\frac{1}{2}) = (\frac{\sqrt{3}}{2}) (\frac{1}{\sqrt{3}}) \Rightarrow \frac{x}{2} = \frac{1}{2} \Rightarrow x = 1 Hence, the correct option is (a) .$

Page No 577:

Answer:

As we know that,
$\tan 45 ° = 1 Thus, if \tan (3 x + 30 °) = 1 \Rightarrow 3 x + 30 ° = 45 ° \Rightarrow 3 x = 45 ° - 30 ° \Rightarrow 3 x = 15 ° \Rightarrow x = 5 ° Hence, the correct option is (d) .$

Page No 577:

Answer:

$Given : \sin θ = \cos θ \Rightarrow \sin θ = \sin (90 ° - θ) (∵ \sin (90 ° - θ) = \cos θ) \Rightarrow θ = 90 ° - θ \Rightarrow 2 θ = 90 ° \Rightarrow θ = 45 ° Hence, the correct option is (b) .$

Page No 577:

Answer:

As we know that,
$\sin 30 ° = \frac{1}{2} Thus, if \sin (A - B) = \frac{1}{2} \Rightarrow A - B = 30 ° . . . (1) and \cos 60 ° = \frac{1}{2} Thus, if \cos (A + B) = \frac{1}{2} \Rightarrow A + B = 60 ° . . . (2) Solving (1) and (2), we get A = 45 ° and B = 15 ° Hence, the correct optiom is (c) .$

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