Rs Aggarwal 2021 2022 for Class 10 Maths Chapter 10 - Trignometric Ratios

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Rs Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 10 Trignometric Ratios are provided here with simple step-by-step explanations. These solutions for Trignometric Ratios are extremely popular among Class 10 students for Maths Trignometric Ratios 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 10 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 546:

Answer:

Let us first draw a right $∆$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that sin $θ$ = $\frac{perpendicular}{hypotenuse}$ = $\frac{A B}{A C}$ = $\frac{\sqrt{3}}{2}$ .

So, if AB = $\sqrt{3} k$ , then AC = 2k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ BC² = AC² $-$ AB² = (2k)² $-$ ( $\sqrt{3} k$ )²
⇒ BC² = 4k² $-$ 3k² = k²
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:
cos $θ$ = $\frac{B C}{A C}$ = $\frac{k}{2 k} = \frac{1}{2}$
tan $θ$ = $\frac{A B}{B C} = \frac{\sqrt{3} k}{k} = \sqrt{3}$

∴ cot $θ$ = $\frac{1}{\tan θ} = \frac{1}{\sqrt{3}}$ , cosec $θ$ = $\frac{1}{\sin θ} = \frac{2}{\sqrt{3}}$ and sec $θ$ = $\frac{1}{\cos θ} = 2$

Page No 546:

Answer:

Let us first draw a right $∆$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that cos $θ$ = $\frac{Base}{hypotenuse}$ = $\frac{B C}{A C}$ = $\frac{7}{25}$ .

So, if BC = 7k, then AC = 25k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC² = (25k)² $-$ (7k)².
⇒ AB² = 625k² $-$ 49k² = 576k²
⇒ AB = 24k
Now, finding the other trigonometric ratios using their definitions, we get:
sin $θ$ = $\frac{A B}{A C}$ = $\frac{24 k}{25 k} = \frac{24}{25}$
tan $θ$ = $\frac{A B}{B C} = \frac{24 k}{7 k} = \frac{24}{7}$
∴ cot $θ$ = $\frac{1}{\tan θ} = \frac{7}{24}$ , cosec $θ$ = $\frac{1}{\sin θ} = \frac{25}{24}$ and sec $θ$ = $\frac{1}{\cos θ} = \frac{25}{7}$

Page No 546:

Answer:

Let us first draw a right $∆$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that tan $θ$ = $\frac{Perpendicular}{Base}$ = $\frac{A B}{B C}$ = $\frac{15}{8}$ .

So, if BC = 8k, then AB = 15k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC² = (15k)² + (8k)²
⇒ AC² = 225k² + 64k² = 289k²
⇒ AC = 17k

Now, finding the other T-ratios using their definitions, we get:
sin $θ$ = $\frac{A B}{A C}$ = $\frac{15 k}{17 k} = \frac{15}{17}$
cos $θ$ = $\frac{B C}{A C} = \frac{8 k}{17 k} = \frac{8}{17}$

∴ cot $θ$ = $\frac{1}{\tan θ} = \frac{8}{15}$ , cosec $θ$ = $\frac{1}{\sin θ} = \frac{17}{15}$ and sec $θ$ = $\frac{1}{\cos θ} = \frac{17}{8}$

Page No 546:

Answer:

Let us first draw a right $∆$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that cot $θ$ = $\frac{base}{Perpendicular}$ = $\frac{B C}{A B}$ = 2.

So, if BC = 2k, then AB = k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC² = (2k)² + (k)²
⇒ AC² = 4k² + k² = 5k²
⇒ AC = $\sqrt{5}$ k
Now, finding the other T-ratios using their definitions, we get:
sin $θ$ = $\frac{A B}{A C}$ = $\frac{k}{\sqrt{5} k} = \frac{1}{\sqrt{5}}$
cos $θ$ = $\frac{B C}{A C} = \frac{2 k}{\sqrt{5} k} = \frac{2}{\sqrt{5}}$

∴ tan $θ$ = $\frac{1}{c o t θ} = \frac{1}{2}$ , cosec $θ$ = $\frac{1}{\sin θ} = \sqrt{5}$ and sec $θ$ = $\frac{1}{\cos θ} = \frac{\sqrt{5}}{2}$

Page No 546:

Answer:

Let us first draw a right $∆$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that cosec $θ$ = $\frac{Hypotenuse}{Perpendicular}$ = $\frac{A C}{A B}$ = $\frac{\sqrt{10}}{1}$ .

So, if AC = ( $\sqrt{10}$ )k, then AB = k, where k is a positive number.
Now, by using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ BC² = AC² $-$ AB² = 10k² $-$ k²
⇒ BC² = 9k²
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
tan $θ$ = $\frac{A B}{B C}$ = $\frac{k}{3 k} = \frac{1}{3}$

cos $θ$ = $\frac{B C}{A C} = \frac{3 k}{\sqrt{10} k} = \frac{3}{\sqrt{10}}$

∴ $\sin θ = \frac{1}{\cos e c θ} = \frac{1}{\sqrt{10}}$ , cot $θ$ = $\frac{1}{\tan θ} = 3$ and sec $θ$ = $\frac{1}{\cos θ} = \frac{\sqrt{10}}{3}$

Page No 546:

Answer:

We have $\sin θ = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$ ,

As,

$\cos^{2} θ = 1 - \sin^{2} θ = 1 - {(\frac{a^{2} - b^{2}}{a^{2} + b^{2}})}^{2} = \frac{1}{1} - \frac{{(a^{2} - b^{2})}^{2}}{{(a^{2} + b^{2})}^{2}} = \frac{{(a^{2} + b^{2})}^{2} - {(a^{2} - b^{2})}^{2}}{{(a^{2} + b^{2})}^{2}} = \frac{[(a^{2} + b^{2}) - (a^{2} - b^{2})] [(a^{2} + b^{2}) + (a^{2} - b^{2})]}{{(a^{2} + b^{2})}^{2}}$
$= \frac{[a^{2} + b^{2} - a^{2} + b^{2}] [a^{2} + b^{2} + a^{2} - b^{2}]}{{(a^{2} + b^{2})}^{2}} = \frac{[2 b^{2}] [2 a^{2}]}{{(a^{2} + b^{2})}^{2}} \Rightarrow \cos^{2} θ = \frac{4 a^{2} b^{2}}{{(a^{2} + b^{2})}^{2}} \Rightarrow \cos θ = \sqrt{\frac{4 a^{2} b^{2}}{{(a^{2} + b^{2})}^{2}}} \Rightarrow \cos θ = \frac{2 a b}{(a^{2} + b^{2})}$

Also,

$\tan θ = \frac{\sin θ}{\cos θ} = \frac{(\frac{a^{2} - b^{2}}{a^{2} + b^{2}})}{(\frac{2 a b}{a^{2} + b^{2}})} = \frac{a^{2} - b^{2}}{2 a b}$

Now,

$cosec θ = \frac{1}{\sin θ} = \frac{1}{(\frac{a^{2} - b^{2}}{a^{2} + b^{2}})} = \frac{a^{2} + b^{2}}{a^{2} - b^{2}}$

Also,

$\sec θ = \frac{1}{\cos θ} = \frac{1}{(\frac{2 a b}{a^{2} + b^{2}})} = \frac{a^{2} + b^{2}}{2 a b}$

And,

$\cot θ = \frac{1}{\tan θ} = \frac{1}{(\frac{a^{2} - b^{2}}{2 a b})} = \frac{2 a b}{a^{2} - b^{2}}$

Page No 546:

Answer:

$Given : \sin θ = \frac{c}{\sqrt{c^{2} + d^{2}}} Since, \sin θ = \frac{P}{H} \Rightarrow P = c and H = \sqrt{c^{2} + d^{2}} Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow c^{2} + B^{2} = c^{2} + d^{2} \Rightarrow B^{2} = d^{2} \Rightarrow B = d Therefore, \cos θ = \frac{B}{H} = \frac{d}{\sqrt{c^{2} + d^{2}}} \tan θ = \frac{P}{B} = \frac{c}{d} Hence, \cos θ = \frac{d}{\sqrt{c^{2} + d^{2}}} and \tan θ = \frac{c}{d} .$

Page No 546:

Answer:

$Given : \sqrt{3} \tan θ = 1 \Rightarrow \tan θ = \frac{1}{\sqrt{3}} Since, \tan θ = \frac{P}{B} \Rightarrow P = 1 and B = \sqrt{3} Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow 1^{2} + {(\sqrt{3})}^{2} = H^{2} \Rightarrow H^{2} = 1 + 3 = 4 \Rightarrow H = 2 Therefore, \sin θ = \frac{P}{H} = \frac{1}{2} \cos θ = \frac{B}{H} = \frac{\sqrt{3}}{2} \cos^{2} θ - \sin^{2} θ = {(\frac{\sqrt{3}}{2})}^{2} - {(\frac{1}{2})}^{2} = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} Hence, \cos^{2} θ - \sin^{2} θ = \frac{1}{2} .$

Page No 546:

Answer:

$Given : 4 \tan θ = 3 \Rightarrow \tan θ = \frac{3}{4} Since, \tan θ = \frac{P}{B} \Rightarrow P = 3 and B = 4 Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow 3^{2} + 4^{2} = H^{2} \Rightarrow H^{2} = 9 + 16 = 25 \Rightarrow H = 5 Therefore, \sin θ = \frac{P}{H} = \frac{3}{5} \cos θ = \frac{B}{H} = \frac{4}{5} \sin θ \times \cos θ = \frac{3}{5} \times \frac{4}{5} = \frac{12}{25} Hence, \sin θ \times \cos θ = \frac{12}{25} .$

Page No 546:

Answer:

$LHS = (\sec θ + \tan θ) = \frac{1}{\cos θ} + \frac{\sin θ}{\cos θ} = \frac{1 + \sin θ}{\cos θ} = \frac{1 + \sin θ}{\sqrt{1 - \sin^{2} θ}} = \frac{(1 + \frac{a}{b})}{\sqrt{1 - {(\frac{a}{b})}^{2}}}$
$= \frac{(\frac{1}{1} + \frac{a}{b})}{\sqrt{\frac{1}{1} - \frac{a^{2}}{b^{2}}}} = \frac{(\frac{b + a}{b})}{\sqrt{\frac{b^{2} - a^{2}}{b^{2}}}} = \frac{(\frac{b + a}{b})}{(\frac{\sqrt{b^{2} - a^{2}}}{b})} = \frac{(b + a)}{\sqrt{(b + a) (b - a)}}$
$= \frac{(b + a)}{\sqrt{(b + a)} \sqrt{(b - a)}} = \frac{\sqrt{(b + a)}}{\sqrt{(b - a)}} = \sqrt{\frac{b + a}{b - a}} = RHS$

Page No 547:

Answer:

It is given that tan $θ = \frac{a}{b}$ .

LHS = $\frac{a \sin θ - b \cos θ}{a \sin θ + b \cos θ}$
Dividing the numerator and denominator by cos $θ$ , we get:

$\frac{a \tan θ - b}{a \tan θ + b}$ (∵ tan $θ = \frac{\sin θ}{\cos θ}$ )
Now, substituting the value of tan $θ$ in the above expression, we get:
$\frac{a (\frac{a}{b}) - b}{a (\frac{a}{b}) + b} = \frac{\frac{a^{2}}{b} - b}{\frac{a^{2}}{b} + b} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}} = RHS$
i.e., LHS = RHS

Hence proved.

Page No 547:

Answer:

$Given : \sin θ = \frac{12}{13} Since, \sin θ = \frac{P}{H} \Rightarrow P = 12 and H = 13 Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow 12^{2} + B^{2} = 13^{2} \Rightarrow B^{2} = 169 - 144 \Rightarrow B^{2} = 25 \Rightarrow B = 5 Therefore, \cos θ = \frac{B}{H} = \frac{5}{13} Now, (\frac{2 \sin θ - 3 \cos θ}{4 \sin θ - 9 \cos θ}) = (\frac{2 (\frac{12}{13}) - 3 (\frac{5}{13})}{4 (\frac{12}{13}) - 9 (\frac{5}{13})}) = (\frac{\frac{24}{13} - \frac{15}{13}}{\frac{48}{13} - \frac{45}{13}}) = (\frac{24 - 15}{48 - 45}) = \frac{9}{3} = 3 Hence, (\frac{2 \sin θ - 3 \cos θ}{4 \sin θ - 9 \cos θ}) = 3 .$

Page No 547:

Answer:

$Given : \tan θ = \frac{1}{2} Since, \tan θ = \frac{P}{B} \Rightarrow P = 1 and B = 2 Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow 1^{2} + 2^{2} = H^{2} \Rightarrow H^{2} = 1 + 4 \Rightarrow H^{2} = 5 \Rightarrow H = \sqrt{5} Therefore, \sin θ = \frac{P}{H} = \frac{1}{\sqrt{5}} \cos θ = \frac{B}{H} = \frac{2}{\sqrt{5}} Now, (\frac{\cos θ}{\sin θ} + \frac{\sin θ}{1 + \cos θ}) = (\frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}} + \frac{\frac{1}{\sqrt{5}}}{1 + \frac{2}{\sqrt{5}}}) = (\frac{2}{1} + \frac{\frac{1}{\sqrt{5}}}{\frac{\sqrt{5} + 2}{\sqrt{5}}}) = (\frac{2}{1} + \frac{1}{\sqrt{5} + 2}) = (2 + (\frac{1}{\sqrt{5} + 2} \times \frac{\sqrt{5} - 2}{\sqrt{5} - 2})) = (2 + (\frac{\sqrt{5} - 2}{5 - 4})) = (2 + \sqrt{5} - 2) = \sqrt{5} Hence, (\frac{\cos θ}{\sin θ} + \frac{\sin θ}{1 + \cos θ}) = \sqrt{5} .$

Page No 547:

Answer:

$LHS = (3 \cos α - 4 \cos^{3} α) = \cos α (3 - 4 \cos^{2} α) = \sqrt{1 - \sin^{2} α} [3 - 4 (1 - \sin^{2} α)] = \sqrt{1 - {(\frac{1}{2})}^{2}} [3 - 4 (1 - {(\frac{1}{2})}^{2})] = \sqrt{\frac{1}{1} - \frac{1}{4}} [3 - 4 (\frac{1}{1} - \frac{1}{4})] = \sqrt{\frac{3}{4}} [3 - 4 (\frac{3}{4})] = \sqrt{\frac{3}{4}} [3 - 3] = \sqrt{\frac{3}{4}} [0] = 0 = RHS$

Page No 547:

Answer:

It is given that cot $θ = \frac{2}{3}$ .

LHS = $\frac{4 \sin θ - 3 \cos θ}{2 \sin θ + 6 \cos θ}$
Dividing the above expression by sin $θ$ , we get:
$\frac{4 - 3 c o t θ}{2 + 6 c o t θ}$ [∵ cot $θ = \frac{\cos θ}{\sin θ}$ ]
Now, substituting the values of cot $θ$ in the above expression, we get:
$\frac{4 - 3 (\frac{2}{3})}{2 + 6 (\frac{2}{3})} = \frac{4 - 2}{2 + 4} = \frac{2}{6} = \frac{1}{3}$
i.e., LHS = RHS

Hence proved.

Page No 547:

Answer:

It is given that sec $θ$ = $\frac{17}{8}$ .

Let us consider a right $△$ ABC right angled at B and $∠ C = θ$ .
We know that cos $θ$ = $\frac{1}{s e c θ} = \frac{8}{17} = \frac{B C}{A C}$

So, if BC = 8k, then AC = 17k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC² = (17k)² $-$ (8k)²
⇒ AB² = 289k² $-$ 64k² = 225k²
⇒ AB = 15k.

Now, tan $θ$ = $\frac{A B}{B C} = \frac{15}{8}$ and sin $θ$ = $\frac{A B}{A C} = \frac{15 k}{17 k} = \frac{15}{17}$

The given expression is $\frac{3 - 4 \sin^{2} θ}{4 \cos^{2} θ - 3} = \frac{3 - \tan^{2} θ}{1 - 3 \tan^{2} θ}$ .

Substituting the values in the above expression, we get:
$LHS = \frac{3 - 4 {(\frac{15}{17})}^{2}}{4 {(\frac{8}{17})}^{2} - 3} = \frac{3 - \frac{900}{289}}{\frac{256}{289} - 3} = \frac{867 - 900}{256 - 867} = \frac{- 33}{- 611} = \frac{33}{611}$

$RHS = \frac{3 - {(\frac{15}{8})}^{2}}{1 - 3 {(\frac{15}{8})}^{2}} = \frac{3 - \frac{225}{64}}{1 - \frac{675}{64}} = \frac{192 - 225}{64 - 675} = \frac{- 33}{- 611} = \frac{33}{611}$

∴ LHS = RHS
Hence proved.

Page No 547:

Answer:

Let us consider a right $△$ ABC right angled at B and $∠ C = θ$ .
Now, we know that tan $θ$ = $\frac{A B}{B C}$ = $\frac{20}{21}$

So, if AB = 20k, then BC = 21k, where k is a positive number.
Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC²= (20k)²+ (21k)²
⇒ AC² = 841k²
⇒ AC = 29k
Now, sin $θ$ = $\frac{A B}{A C} = \frac{20}{29}$ and cos $θ$ = $\frac{B C}{A C} = \frac{21}{29}$

Substituting these values in the given expression, we get:
$LHS = \frac{1 - \sin θ + \cos θ}{1 + \sin θ + \cos θ} = \frac{1 - \frac{20}{29} + \frac{21}{29}}{1 + \frac{20}{29} + \frac{21}{29}} = \frac{\frac{29 - 20 + 21}{29}}{\frac{29 + 20 + 21}{29}} = \frac{30}{70} = \frac{3}{7} = RHS$
∴ LHS = RHS

Hence proved.

Page No 547:

Answer:

Let us consider a right $△$ ABC, right angled at B and $∠ C = θ$ .
Now it is given that tan $θ$ = $\frac{A B}{B C}$ = $\frac{1}{\sqrt{7}}$ .

So, if AB = k, then BC = $\sqrt{7}$ k, where k is a positive number.
Using Pythagoras theorem, we have:
AC²= AB² + BC²
⇒ AC² = (k)² + ( $\sqrt{7}$ k)²
⇒ AC² = k² + 7k²
⇒ AC = 2 $\sqrt{2}$ k
Now, finding out the values of the other trigonometric ratios, we have:
sin $θ$ = $\frac{A B}{A C} = \frac{k}{2 \sqrt{2} k} = \frac{1}{2 \sqrt{2}}$
cos $θ$ = $\frac{B C}{A C} = \frac{\sqrt{7} k}{2 \sqrt{2} k} = \frac{\sqrt{7}}{2 \sqrt{2}}$
∴ cosec $θ$ = $\frac{1}{\sin θ} = 2 \sqrt{2}$ and sec $θ$ = $\frac{1}{\cos θ} = \frac{2 \sqrt{2}}{\sqrt{7}}$
Substituting the values of cosec $θ$ and sec $θ$ in the given expression, we get:
$\frac{\cos e c^{2} θ - s e c^{2} θ}{\cos e c^{2} θ + s e c^{2} θ} = \frac{(2 \sqrt{2})^{2} - {(\frac{2 \sqrt{2}}{\sqrt{7}})}^{2}}{(2 \sqrt{2})^{2} + {(\frac{2 \sqrt{2}}{\sqrt{7}})}^{2}} = \frac{8 - (\frac{8}{7})}{8 + (\frac{8}{7})} = \frac{\frac{56 - 8}{7}}{\frac{56 + 8}{7}} = \frac{48}{64} = \frac{3}{4} = RHS$
i.e., LHS = RHS

Hence proved.

Page No 547:

Answer:

$LHS = \sqrt{\frac{{cosec}^{2} θ - \cot^{2} θ}{\sec^{2} θ - 1}} = \sqrt{\frac{1}{\tan^{2} θ}} = \sqrt{\cot^{2} θ} = \cot θ = \sqrt{{cosec}^{2} θ - 1} = \sqrt{{(\frac{1}{\sin θ})}^{2} - 1} = \sqrt{{(\frac{1}{(\frac{3}{4})})}^{2} - 1} = \sqrt{{(\frac{4}{3})}^{2} - 1} = \sqrt{\frac{16}{9} - 1} = \sqrt{\frac{16 - 9}{9}} = \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3} = RHS$

Page No 547:

Answer:

(i)
$LHS = \sqrt{\frac{\sec θ - cosec θ}{\sec θ + cosec θ}} = \sqrt{\frac{(\frac{1}{\cos θ} - \frac{1}{\sin θ})}{(\frac{1}{\cos θ} + \frac{1}{\sin θ})}} = \sqrt{\frac{(\frac{\sin θ - \cos θ}{\sin θ \cos θ})}{(\frac{\sin θ + \cos θ}{\sin θ \cos θ})}} = \sqrt{\frac{(\frac{\sin θ - \cos θ}{\sin θ})}{(\frac{\sin θ + \cos θ}{\sin θ})}} = \sqrt{\frac{(\frac{\sin θ}{\sin θ} - \frac{\cos θ}{\sin θ})}{(\frac{\sin θ}{\sin θ} + \frac{\cos θ}{\sin θ})}} = \sqrt{\frac{1 - \cot θ}{1 + \cot θ}} = \sqrt{\frac{(1 - \frac{3}{4})}{(1 + \frac{3}{4})}}$
$= \sqrt{\frac{(\frac{1}{4})}{(\frac{7}{4})}} = \sqrt{\frac{1}{7}} = \frac{1}{\sqrt{7}} = RHS$

(ii)
$Given : 3 \tan A = 4 \Rightarrow \tan A = \frac{4}{3} Since, \tan A = \frac{P}{B} \Rightarrow P = 4 and B = 3 Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow 4^{2} + 3^{2} = H^{2} \Rightarrow H^{2} = 16 + 9 \Rightarrow H^{2} = 25 \Rightarrow H = 5 Therefore, \sin A = \frac{P}{H} = \frac{4}{5} \cos A = \frac{B}{H} = \frac{3}{5} Now, \sqrt{\frac{1 - \sin A}{1 + \cos A}} = \sqrt{\frac{1 - \frac{4}{5}}{1 + \frac{3}{5}}} = \sqrt{\frac{\frac{5 - 4}{5}}{\frac{5 + 3}{5}}} = \sqrt{\frac{\frac{1}{5}}{\frac{8}{5}}} = \sqrt{\frac{1}{8}} = \frac{1}{2 \sqrt{2}} Hence, \sqrt{\frac{1 - \sin A}{1 + \cos A}} = \frac{1}{2 \sqrt{2}} .$

Page No 547:

Answer:

$Given : \cot θ = \frac{15}{8} Since, \cot θ = \frac{B}{P} \Rightarrow P = 8 and B = 15 Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow 8^{2} + 15^{2} = H^{2} \Rightarrow H^{2} = 64 + 225 \Rightarrow H^{2} = 289 \Rightarrow H = 17 Therefore, \sin θ = \frac{P}{H} = \frac{8}{17} \cos θ = \frac{B}{H} = \frac{15}{17} Now, \frac{(1 + \sin θ) (1 - \sin θ)}{(1 + \cos θ) (1 - \cos θ)} = \frac{1 - \sin^{2} θ}{1 - \cos^{2} θ} = \frac{\cos^{2} θ}{\sin^{2} θ} (∵ \sin^{2} θ + \cos^{2} θ = 1) = \cot^{2} θ = {(\frac{15}{8})}^{2} = \frac{225}{64} Hence, \frac{(1 + \sin θ) (1 - \sin θ)}{(1 + \cos θ) (1 - \cos θ)} = \frac{225}{64} .$

Page No 547:

Answer:

$We have, tanA = 1 \Rightarrow \frac{sinA}{cosA} = 1 \Rightarrow sinA = cosA \Rightarrow sinA - cosA = 0 Squaring both sides, we get {(sinA - cosA)}^{2} = 0 \Rightarrow \sin^{2} A + \cos^{2} A - 2 sinA \cdot cosA = 0 \Rightarrow 1 - 2 sinA \cdot cosA = 0 ∴ 2 sinA \cdot cosA = 1$

Page No 547:

Answer:

$Given : In ∆ A B C, A C = 17 cm \sin θ = \frac{8}{17} Since, \sin θ = \frac{P}{H} \Rightarrow P = 8 and H = 17 Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow 8^{2} + B^{2} = 17^{2} \Rightarrow B^{2} = 289 - 64 \Rightarrow B^{2} = 225 \Rightarrow B = 15 Therefore, A B = 8 cm and B C = 15 cm Therefore, (i) Area of rectangle A B C D = A B \times B C = 8 \times 15 = 120 {cm}^{2} (i i) Perimeter of rectangle A B C D = 2 (A B + B C) = 2 (8 + 15) = 2 (23) = 46 cm$

Page No 547:

Answer:

$LHS = {(\frac{2}{x + y})}^{2} + {(\frac{x - y}{2})}^{2} - 1 = {[\frac{2}{(cosecA + cosA) + (cosecA - cosA)}]}^{2} + {[\frac{(cosecA + cosA) - (cosecA - cosA)}{2}]}^{2} - 1 = {[\frac{2}{cosecA + cosA + cosecA - cosA}]}^{2} + {[\frac{cosecA + cosA - cosecA + cosA}{2}]}^{2} - 1 = {[\frac{2}{2 cosecA}]}^{2} + {[\frac{2 cosA}{2}]}^{2} - 1$
$= {[\frac{1}{cosecA}]}^{2} + {[cosA]}^{2} - 1 = {[sinA]}^{2} + {[cosA]}^{2} - 1 = \sin^{2} A + \cos^{2} A - 1 = 1 - 1 = 0 = RHS$

Page No 548:

Answer:

$LHS = {(\frac{x - y}{x + y})}^{2} + {(\frac{x - y}{2})}^{2} = {[\frac{(cotA + cosA) - (cotA - cosA)}{(cotA + cosA) + (cotA - cosA)}]}^{2} + {[\frac{(cotA + cosA) - (cotA - cosA)}{2}]}^{2} = {[\frac{cotA + cosA - cotA + cosA}{cotA + cosA + cotA - cosA}]}^{2} + {[\frac{cotA + cosA - cotA + cosA}{2}]}^{2} = {[\frac{2 cosA}{2 cotA}]}^{2} + {[\frac{2 cosA}{2}]}^{2} = {[\frac{cosA}{(\frac{cosA}{sinA})}]}^{2} + {[cosA]}^{2} = {[\frac{sinA cosA}{cosA}]}^{2} + {[cosA]}^{2} = {[sinA]}^{2} + {[cosA]}^{2} = \sin^{2} A + \cos^{2} A = 1 = RHS$

Page No 548:

Answer:

In $∆ PQR, ∠ Q = 90 °$ ,

Using Pythagoras theorem, we get

$PQ = \sqrt{{PR}^{2} - {QR}^{2}} = \sqrt{{(x + 2)}^{2} - x^{2}} = \sqrt{x^{2} + 4 x + 4 - x^{2}} = \sqrt{4 (x + 1)} = 2 \sqrt{x + 1}$

Now,

$(i) (\sqrt{x + 1}) \cot ϕ = (\sqrt{x + 1}) \times \frac{QR}{PQ} = (\sqrt{x + 1}) \times \frac{x}{2 \sqrt{x + 1}} = \frac{x}{2}$

$(ii) (\sqrt{x^{3} + x^{2}}) \tan θ = (\sqrt{x^{2} (x + 1)}) \times \frac{QR}{PQ} = x \sqrt{(x + 1)} \times \frac{x}{2 \sqrt{x + 1}} = \frac{x^{2}}{2}$

$(iii) \cos θ = \frac{PQ}{PR} = \frac{2 \sqrt{x + 1}}{(x + 2)}$

Page No 548:

Answer:

$Given : \cot A + \frac{1}{\cot A} = 2 \cot A + \frac{1}{\cot A} = 2 Squaring both sides, we get \Rightarrow {(\cot A + \frac{1}{\cot A})}^{2} = 2^{2} \Rightarrow \cot^{2} A + {(\frac{1}{\cot A})}^{2} + 2 (\cot A) (\frac{1}{\cot A}) = 4 \Rightarrow \cot^{2} A + \frac{1}{\cot^{2} A} + 2 = 4 \Rightarrow \cot^{2} A + \frac{1}{\cot^{2} A} = 4 - 2 \Rightarrow \cot^{2} A + \frac{1}{\cot^{2} A} = 2 Hence, the value of (\cot^{2} A + \frac{1}{\cot^{2} A}) is 2 .$

Page No 548:

Answer:

Page No 548:

Answer:

In $∆$ ABC, $∠$ C = 90 $°$
sinA = $\frac{BC}{AB}$ and
sinB = $\frac{AC}{AB}$

As, sinA = sinB
$\Rightarrow$ $\frac{BC}{AB}$ = $\frac{AC}{AB}$
$\Rightarrow$ BC = AC
So, $∠$ A = $∠$ B (Angles opposite to equal sides are equal)

Page No 548:

Answer:

In $∆ ABC, ∠ C = 90 °$ ,

$tanA = \frac{BC}{AC} and tanB = \frac{AC}{BC}$

$As, tanA = tanB$
$\Rightarrow \frac{BC}{AC} = \frac{AC}{BC} \Rightarrow {BC}^{2} = {AC}^{2} \Rightarrow BC = AC So, ∠ A = ∠ B (Angles opposite to equal sides are equal)$

Page No 555:

Answer:

$Given : \tan θ = \frac{8}{15} Since, \tan θ = \frac{P}{B} \Rightarrow P = 8 and B = 15 Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow 8^{2} + {(15)}^{2} = H^{2} \Rightarrow H^{2} = 64 + 225 \Rightarrow H^{2} = 289 \Rightarrow H = 17 Therefore, cosec θ = \frac{H}{P} = \frac{17}{8}$

Hence, the correct option is (c).

Page No 555:

Answer:

$Given : \tan θ = \frac{\sqrt{3}}{1} Since, \tan θ = \frac{P}{B} \Rightarrow P = \sqrt{3} and B = 1 Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow {(\sqrt{3})}^{2} + 1^{2} = H^{2} \Rightarrow H^{2} = 3 + 1 \Rightarrow H^{2} = 4 \Rightarrow H = 2 Therefore, \sec θ = \frac{H}{B} = \frac{2}{1} = 2$

Hence, the correct option is (a).

Page No 555:

Answer:

$Given : cosec θ = \frac{\sqrt{10}}{1} Since, cosec θ = \frac{H}{P} \Rightarrow P = 1 and H = \sqrt{10} Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow 1^{2} + B^{2} = {(\sqrt{10})}^{2} \Rightarrow B^{2} = 10 - 1 \Rightarrow B^{2} = 9 \Rightarrow B = 3 Therefore, \sec θ = \frac{H}{B} = \frac{\sqrt{10}}{3}$

Hence, the correct option is (d).

Page No 555:

Answer:

$Given : \sec θ = \frac{25}{7} Since, \sec θ = \frac{H}{B} \Rightarrow B = 7 and H = 25 Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow P^{2} + {(7)}^{2} = {(25)}^{2} \Rightarrow P^{2} = 625 - 49 \Rightarrow P^{2} = 576 \Rightarrow P = 24 Therefore, \sin θ = \frac{P}{H} = \frac{24}{25}$

Hence, the correct option is (b).

Page No 555:

Answer:

$Given : \sin θ = \frac{1}{2} Since, \sin θ = \frac{P}{H} \Rightarrow P = 1 and H = 2 Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow 1^{2} + B^{2} = 2^{2} \Rightarrow B^{2} = 4 - 1 \Rightarrow B^{2} = 3 \Rightarrow B = \sqrt{3} Therefore, \cot θ = \frac{B}{P} = \frac{\sqrt{3}}{1}$

Hence, the correct option is (c).

Page No 555:

Answer:

$Given : \cos θ = \frac{4}{5} Since, \cos θ = \frac{B}{H} \Rightarrow B = 4 and H = 5 Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow P^{2} + 4^{2} = 5^{2} \Rightarrow P^{2} = 25 - 16 \Rightarrow P^{2} = 9 \Rightarrow P = 3 Therefore, \tan θ = \frac{P}{B} = \frac{3}{4}$

Hence, the correct option is (a).

Page No 556:

Answer:

$Given : \tan θ = \frac{4}{3} Since, \tan θ = \frac{P}{B} \Rightarrow P = 4 and B = 3 Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow 4^{2} + 3^{2} = H^{2} \Rightarrow H^{2} = 16 + 9 \Rightarrow H^{2} = 25 \Rightarrow H = 5 Therefore, \sin θ = \frac{P}{H} = \frac{4}{5} \cos θ = \frac{B}{H} = \frac{3}{5} Now, \sin θ + \cos θ = \frac{4}{5} + \frac{3}{5} = \frac{7}{5}$

Hence, the correct option is (c).

Page No 556:

Answer:

$Given : \tan θ + \cot θ = 5 \tan θ + \cot θ = 5 Squaring both sides, we get \Rightarrow {(\tan θ + \cot θ)}^{2} = 5^{2} \Rightarrow \tan^{2} θ + \cot^{2} θ + 2 (\cot θ) (\tan θ) = 25 \Rightarrow \tan^{2} θ + \cot^{2} θ + 2 (\frac{1}{\tan θ}) (\tan θ) = 25 (∵ \cot θ = \frac{1}{\tan θ}) \Rightarrow \tan^{2} θ + \cot^{2} θ + 2 = 25 \Rightarrow \tan^{2} θ + \cot^{2} θ = 23 Hence, the correct option is (d) .$

Page No 556:

Answer:

$Given : \cos θ + \sec θ = \frac{5}{2} \cos θ + \sec θ = \frac{5}{2} Squaring both sides, we get \Rightarrow {(\cos θ + \sec θ)}^{2} = {(\frac{5}{2})}^{2} \Rightarrow \cos^{2} θ + \sec^{2} θ + 2 (\cos θ) (\sec θ) = \frac{25}{4} \Rightarrow \cos^{2} θ + \sec^{2} θ + 2 (\cos θ) (\frac{1}{\cos θ}) = \frac{25}{4} (∵ \sec θ = \frac{1}{\cos θ}) \Rightarrow \cos^{2} θ + \sec^{2} θ + 2 = \frac{25}{4} \Rightarrow \cos^{2} θ + \sec^{2} θ = \frac{25}{4} - 2 \Rightarrow \cos^{2} θ + \sec^{2} θ = \frac{25 - 8}{4} \Rightarrow \cos^{2} θ + \sec^{2} θ = \frac{17}{4} Hence, the correct option is (a) .$

Page No 556:

Answer:

$Given : 4 \tan θ = 3 \Rightarrow \tan θ = \frac{3}{4} Since, \tan θ = \frac{P}{B} \Rightarrow P = 3 and B = 4 Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow 3^{2} + 4^{2} = H^{2} \Rightarrow H^{2} = 9 + 16 \Rightarrow H^{2} = 25 \Rightarrow H = 5 Therefore, \sin θ = \frac{P}{H} = \frac{3}{5} \cos θ = \frac{B}{H} = \frac{4}{5} \cos^{2} θ - \sin^{2} θ = {(\frac{4}{5})}^{2} - {(\frac{3}{5})}^{2} = \frac{16}{25} - \frac{9}{25} = \frac{16 - 9}{25} = \frac{7}{25} Hence, the correct option is (b) .$

Page No 556:

Answer:

$Given : 4 \cot θ = 3 \Rightarrow \cot θ = \frac{3}{4} Since, \cot θ = \frac{B}{P} \Rightarrow P = 4 and B = 3 Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow 4^{2} + 3^{2} = H^{2} \Rightarrow H^{2} = 16 + 9 \Rightarrow H^{2} = 25 \Rightarrow H = 5 Therefore, \sin θ = \frac{P}{H} = \frac{4}{5} \cos θ = \frac{B}{H} = \frac{3}{5} (\frac{\sin θ - \cos θ}{\sin θ + \cos θ}) = (\frac{\frac{4}{5} - \frac{3}{5}}{\frac{4}{5} + \frac{3}{5}}) = (\frac{\frac{4 - 3}{5}}{\frac{4 + 3}{5}}) = \frac{1}{7} Hence, the correct option is (c) .$

Page No 556:

Answer:

$Given : 3 \cos θ = 2 \Rightarrow \cos θ = \frac{2}{3} Since, \cos θ = \frac{B}{H} \Rightarrow B = 2 and H = 3 Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow P^{2} + 2^{2} = 3^{2} \Rightarrow P^{2} = 9 - 4 \Rightarrow P^{2} = 5 \Rightarrow P = \sqrt{5} Therefore, \sec θ = \frac{H}{B} = \frac{3}{2} \tan θ = \frac{P}{B} = \frac{\sqrt{5}}{2} Now, (2 \sec^{2} θ + 2 \tan^{2} θ - 7) = (2 {(\frac{3}{2})}^{2} + 2 {(\frac{\sqrt{5}}{2})}^{2} - 7) = (2 (\frac{9}{4}) + 2 (\frac{5}{4}) - 7) = (\frac{9}{2} + \frac{5}{2} - 7) = (\frac{9 + 5}{2} - 7) = (\frac{14}{2} - 7) = (7 - 7) = 0 Hence, the correct option is (a) .$

Page No 556:

Answer:

$Given : \sec θ + \tan θ + 1 = 0 \sec θ + \tan θ + 1 = 0 \Rightarrow \sec θ + \tan θ = - 1 Multiplying and dividing LHS by \sec θ - \tan θ, we get \Rightarrow (\sec θ + \tan θ) \times (\frac{\sec θ - \tan θ}{\sec θ - \tan θ}) = - 1 \Rightarrow (\frac{\sec^{2} θ - \tan^{2} θ}{\sec θ - \tan θ}) = - 1 \Rightarrow (\frac{1 + \tan^{2} θ - \tan^{2} θ}{\sec θ - \tan θ}) = - 1 (∵ \sec^{2} θ = 1 + \tan^{2} θ) \Rightarrow (\frac{1}{\sec θ - \tan θ}) = - 1 \Rightarrow (\sec θ - \tan θ) = - 1 Hence, the correct option is (b) .$

Page No 556:

Answer:

$Given : \cos A + \cos^{2} A = 1 \cos A + \cos^{2} A = 1 \Rightarrow \cos A = 1 - \cos^{2} A \Rightarrow \cos A = \sin^{2} A (∵ \sin^{2} A + \cos^{2} A = 1) Now, \sin^{2} A + \sin^{4} A = \sin^{2} A + {(\sin^{2} A)}^{2} = \cos A + {(\cos A)}^{2} = \cos A + \cos^{2} A = 1 Hence, the correct option is (c) .$

Page No 556:

Answer:

$Given : \sin θ = \frac{\sqrt{3}}{2} Since, \sin θ = \frac{P}{H} \Rightarrow P = \sqrt{3} and H = 2 Using Pythagoras theorem, P^{2} + B^{2} = H^{2} \Rightarrow {(\sqrt{3})}^{2} + B^{2} = 2^{2} \Rightarrow B^{2} = 4 - 3 \Rightarrow B^{2} = 1 \Rightarrow B = 1 Therefore, cosec θ = \frac{H}{P} = \frac{2}{\sqrt{3}} \cot θ = \frac{B}{P} = \frac{1}{\sqrt{3}} Now, (cosec θ + \cot θ) = (\frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}}) = (\frac{2 + 1}{\sqrt{3}}) = (\frac{3}{\sqrt{3}}) = \sqrt{3} Hence, the correct option is (d) .$

Page No 556:

Answer:

$Given : \sqrt{3} \tan θ = 3 \sin θ \sqrt{3} \tan θ = 3 \sin θ \Rightarrow \sqrt{3} \frac{\sin θ}{\cos θ} = 3 \sin θ \Rightarrow \frac{\sqrt{3} \sin θ}{\cos θ} - 3 \sin θ = 0 \Rightarrow \frac{\sqrt{3} \sin θ - 3 \sin θ \cos θ}{\cos θ} = 0 \Rightarrow \sqrt{3} \sin θ - 3 \sin θ \cos θ = 0 \Rightarrow \sqrt{3} \sin θ (1 - \sqrt{3} \cos θ) = 0 \Rightarrow \sin θ = 0 or 1 - \sqrt{3} \cos θ = 0 \Rightarrow \sin θ = 0 or \cos θ = \frac{1}{\sqrt{3}} \Rightarrow \sin θ = 0 or \cos^{2} θ = \frac{1}{3} \Rightarrow \sin θ = 0 or 1 - \cos^{2} θ = 1 - \frac{1}{3} \Rightarrow \sin θ = 0 or \sin^{2} θ = \frac{2}{3} \Rightarrow \sin θ = 0 or \sin θ = \sqrt{\frac{2}{3}} For \sin θ = 0, \Rightarrow \sin^{2} θ = 0 \Rightarrow 1 - \sin^{2} θ = 1 - 0 \Rightarrow \cos^{2} θ = 1 Thus, \sin^{2} θ - \cos^{2} θ = - 1 For \sin θ = \sqrt{\frac{2}{3}}, \Rightarrow \sin^{2} θ = \frac{2}{3} \Rightarrow 1 - \sin^{2} θ = 1 - \frac{2}{3} \Rightarrow \cos^{2} θ = \frac{1}{3} Thus, \sin^{2} θ - \cos^{2} θ = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$

Hence, the correct option is (a).

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