Rs Aggarwal 2018 for Class 10 Math Chapter 1 - Real Numbers

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Page No 288:

Question 1:

If sin $θ = \frac{\sqrt{3}}{2}$ , find the value of all T-ratios of θ.

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 288:

Question 2:

If cos $θ = \frac{7}{25}$ find the values of all T-ratios of θ.

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 288:

Question 3:

If tan $θ = \frac{15}{8}$ find the values of all T-ratios of θ.

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 288:

Question 4:

If cot θ = 2, find the value of all T-ratios of θ.

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 288:

Question 5:

If cosec θ = $\sqrt{10}$ , the find the values of all T-ratios of θ.

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 288:

Question 6:

If $\sin θ = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$ , find the values of all T-ratios of $θ$ .

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 288:

Question 7:

If $15 \cot A = 8$ , find the values of sinA and secA.

Answer:

We have,

$15 \cot A = 8 \Rightarrow \cot A = \frac{8}{15}$

As,

${cosec}^{2} A = 1 + \cot^{2} A = 1 + {(\frac{8}{15})}^{2} = 1 + \frac{64}{225} = \frac{225 + 64}{225}$
$\Rightarrow {cosec}^{2} A = \frac{289}{225} \Rightarrow cosec A = \sqrt{\frac{289}{225}} \Rightarrow cosec A = \frac{17}{15} \Rightarrow \frac{1}{\sin A} = \frac{17}{15} \Rightarrow \sin A = \frac{15}{17}$

Also,

$\cos^{2} A = 1 - \sin^{2} A = 1 - {(\frac{15}{17})}^{2} = 1 - \frac{225}{289} = \frac{289 - 225}{289}$
$\Rightarrow \cos^{2} A = \frac{64}{289} \Rightarrow \cos A = \sqrt{\frac{64}{289}} \Rightarrow \cos A = \frac{8}{17} \Rightarrow \frac{1}{\sec A} = \frac{8}{17} \Rightarrow \sec A = \frac{17}{8}$

Page No 288:

Question 8:

If $\sin A = \frac{9}{41}$ , find the values of cosA and tanA.

Answer:

We have $\sin A = \frac{9}{41}$ ,

As,

$\cos^{2} A = 1 - \sin^{2} A = 1 - {(\frac{9}{41})}^{2} = 1 - \frac{81}{1681} = \frac{1681 - 81}{1681}$
$\Rightarrow \cos^{2} A = \frac{1600}{1681} \Rightarrow \cos A = \sqrt{\frac{1600}{1681}} \Rightarrow \cos A = \frac{40}{41}$

Also,

$\tan A = \frac{\sin A}{\cos A} = \frac{(\frac{9}{41})}{(\frac{40}{41})} = \frac{9}{40}$

Page No 289:

Question 9:

If cos θ = 0.6, show that (5 sin θ − 3 tan θ) = 0.

Answer:

Let us consider a right $△$ ABC right angled at B.
Now, we know that cos $θ$ = 0.6 = $\frac{B C}{A C}$ = $\frac{3}{5}$

So, if BC = 3k, then AC = 5k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC²
⇒ AB² = (5k)² $-$ (3k)² = 25k² $-$ 9k²
⇒ AB² = 16k²
⇒ AB = 4k

Finding out the other T-ratios using their definitions, we get:
sin $θ$ = $\frac{A B}{A C} = \frac{4 k}{5 k} = \frac{4}{5}$

tan $θ$ = $\frac{A B}{B C} = \frac{4 k}{3 k} = \frac{4}{3}$

Substituting the values in the given expression, we get:
5 sin $θ$ $-$ 3 tan $θ$
$\Rightarrow 5 (\frac{4}{5}) - 3 (\frac{4}{3}) \Rightarrow 4 - 4 = 0 = RHS$
i.e., LHS = RHS

Hence proved.

Page No 289:

Question 10:

If cosec θ = 2, show that $(cotθ + \frac{sinθ}{1 + cosθ}) = 2 .$

Answer:

Let us consider a right $△$ ABC, right angled at B and $∠ C = θ$ .
Now, it is given that cosec $θ$ = 2.
Also, sin $θ$ = $\frac{1}{\cos e c θ} = \frac{1}{2} = \frac{A B}{A C}$

So, if AB = k, then AC = 2k, where k is a positive number.
Using Pythagoras theorem, we have:
⇒ AC² = AB² + BC²
⇒ BC² = AC² $-$ AB²
⇒ BC² = (2k)² $-$ (k)²
⇒ BC² = 3k²
⇒ BC = $\sqrt{3}$ k
Finding out the other T-ratios using their definitions, we get:
cos $θ$ = $\frac{B C}{A C} = \frac{\sqrt{3} k}{2 k} = \frac{\sqrt{3}}{2}$
tan $θ$ = $\frac{A B}{B C} = \frac{k}{\sqrt{3} k} = \frac{1}{\sqrt{3}}$
cot $θ$ = $\frac{1}{\tan θ} = \sqrt{3}$
Substituting these values in the given expression, we get:
$c o t θ + \frac{\sin θ}{1 + \cos θ} = \sqrt{3} + \frac{(\frac{1}{2})}{1 + \frac{\sqrt{3}}{2}} = \sqrt{3} + \frac{\frac{1}{2}}{\frac{2 + \sqrt{3}}{2}}$

$= \sqrt{3} + \frac{1}{2 + \sqrt{3}} = \frac{\sqrt{3} (2 + \sqrt{3}) + 1}{2 + \sqrt{3}} = \frac{2 \sqrt{3} + 3 + 1}{2 + \sqrt{3}} = \frac{2 (2 + \sqrt{3})}{2 + \sqrt{3}} = 2$
i.e., LHS = RHS

Hence proved.

Page No 289:

Question 11:

If tan θ = $\frac{1}{\sqrt{7}}$ , show that $\frac{({cosec}^{2} θ - \sec^{2} θ)}{({cosec}^{2} θ + \sec^{2} θ)} = \frac{3}{4} .$

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 289:

Question 12:

If tan θ = $\frac{20}{21}$ , show that $\frac{(1 - sinθ + cosθ)}{(1 + sinθ + cosθ)} = \frac{3}{7} .$

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 289:

Question 13:

If $\sec θ = \frac{5}{4}$ , show that $\frac{(\sin θ - 2 \cos θ)}{(\tan θ - \cot θ)} = \frac{12}{7}$ .

Answer:

We have,

$\sec θ = \frac{5}{4} \Rightarrow \frac{1}{\cos θ} = \frac{5}{4} \Rightarrow \cos θ = \frac{4}{5}$

Also,

$\sin^{2} θ = 1 - \cos^{2} θ = 1 - {(\frac{4}{5})}^{2} = 1 - \frac{16}{25} = \frac{9}{25} \Rightarrow \sin θ = \frac{3}{5}$

Now,

$LHS = \frac{(\sin θ - 2 \cos θ)}{(\tan θ - \cot θ)} = \frac{(\sin θ - 2 \cos θ)}{(\frac{\sin θ}{\cos θ} - \frac{\cos θ}{\sin θ})} = \frac{(\sin θ - 2 \cos θ)}{(\frac{\sin^{2} θ - \cos^{2} θ}{\sin θ \cos θ})} = \frac{\sin θ \cos θ (\sin θ - 2 \cos θ)}{(\sin^{2} θ - \cos^{2} θ)}$
$= \frac{\frac{3}{5} \times \frac{4}{5} (\frac{3}{5} - 2 \times \frac{4}{5})}{{(\frac{3}{5})}^{2} - {(\frac{4}{5})}^{2}} = \frac{\frac{12}{25} (\frac{3}{5} - \frac{8}{5})}{(\frac{9}{25} - \frac{16}{25})} = \frac{\frac{12}{25} \times (\frac{- 5}{5})}{(\frac{- 7}{25})} = \frac{12}{7} = RHS$

Page No 289:

Question 14:

If $\cot θ = \frac{3}{4}$ , show that $\sqrt{\frac{\sec θ - cosec θ}{\sec θ + cosec θ}} = \frac{1}{\sqrt{7}}$ .

Answer:

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

Page No 289:

Question 15:

If $\sin θ = \frac{3}{4}$ , show that $\sqrt{\frac{{cosec}^{2} θ - \cot^{2} θ}{\sec^{2} θ - 1}} = \frac{\sqrt{7}}{3}$ .

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 289:

Question 16:

If $\sin θ = \frac{a}{b}$ , show that $(\sec θ + \tan θ) = \sqrt{\frac{b + a}{b - a}}$ .

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 289:

Question 17:

If $\cos θ = \frac{3}{5}$ , show that $\frac{(\sin θ - \cot θ)}{2 \tan θ} = \frac{3}{160}$ .

Answer:

$LHS = \frac{(\sin θ - \cot θ)}{2 \tan θ} = \frac{(\sin θ - \frac{\cos θ}{\sin θ})}{2 (\frac{\sin θ}{\cos θ})} = \frac{(\frac{\sin^{2} θ - \cos θ}{\sin θ})}{(\frac{2 \sin θ}{\cos θ})} = \frac{\cos θ (\sin^{2} θ - \cos θ)}{2 \sin^{2} θ} = \frac{\cos θ (1 - \cos^{2} θ - \cos θ)}{2 (1 - \cos^{2} θ)}$
$= \frac{\frac{3}{5} [1 - {(\frac{3}{5})}^{2} - \frac{3}{5}]}{2 [1 - {(\frac{3}{5})}^{2}]} = \frac{\frac{3}{5} (\frac{1}{1} - \frac{9}{25} - \frac{3}{5})}{2 (1 - \frac{9}{25})} = \frac{\frac{3}{5} (\frac{25 - 9 - 15}{25})}{2 (\frac{25 - 9}{25})} = \frac{\frac{3}{5} (\frac{1}{25})}{2 (\frac{16}{25})}$
$= \frac{3}{5 \times 2 \times 16} = \frac{3}{160} = RHS$

Page No 289:

Question 18:

If tan θ = $\frac{4}{3}$ , show that (sin θ + cos θ) = $\frac{7}{5}$ .

Answer:

Let us consider a right $△$ ABC, right angled at B and $∠ C = θ$ .
Now, we know that tan $θ$ = $\frac{A B}{B C}$ = $\frac{4}{3}$ .

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC² = (4k)² + (3k)²
⇒ AC² = 16k²+ 9k² = 25k²
⇒ AC = 5k
Finding out the values of sin $θ$ and cos $θ$ using their definitions, we have:
sin $θ$ = $\frac{A B}{A C} = \frac{4 k}{5 k} = \frac{4}{5}$
cos $θ$ = $\frac{B C}{A C} = \frac{3 k}{5 k} = \frac{3}{5}$
Substituting these values in the given expression, we get:
(sin $θ$ + cos $θ$ ) = $(\frac{4}{5} + \frac{3}{5}) = (\frac{7}{5}) = RHS$
i.e., LHS = RHS

Hence proved.

Page No 289:

Question 19:

If tan θ = $\frac{a}{b}$ , show that $(\frac{a sinθ - b cosθ}{a sinθ + b cosθ}) = \frac{(a^{2} - b^{2})}{(a^{2} + b^{2})} .$

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 289:

Question 20:

If 3 tan θ = 4, show that $\frac{4 cosθ - sinθ}{2 cosθ + sinθ} = \frac{4}{5} .$

Answer:

Let us consider a right $△$ ABC right angled at B and $∠ C = θ$ .
We know that tan $θ$ = $\frac{A B}{B C}$ = $\frac{4}{3}$

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AC² = 16k² + 9k²
⇒ AC² = 25k²
⇒ AC = 5k

Now, we have:

sin $θ$ = $\frac{A B}{A C} = \frac{4 k}{5 k} = \frac{4}{5}$

cos $θ$ = $\frac{B C}{A C} = \frac{3 k}{5 k} = \frac{3}{5}$

Substituting these values in the given expression, we get:

$\frac{4 \cos θ - \sin θ}{2 \cos θ + \sin θ} = \frac{4 (\frac{3}{5}) - \frac{4}{5}}{2 (\frac{3}{5}) + \frac{4}{5}} = \frac{\frac{12}{5} - \frac{4}{5}}{\frac{6}{5} + \frac{4}{5}} = \frac{\frac{12 - 4}{5}}{\frac{6 + 4}{5}} = \frac{8}{10} = \frac{4}{5} = RHS$
i.e., LHS = RHS

Hence proved.

Page No 289:

Question 21:

If 3 cot θ = 2, show that $(\frac{4 sinθ - 3 cosθ}{2 sinθ + 6 cosθ}) = \frac{1}{3} .$

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 289:

Question 22:

If $3 \cot θ = 4$ , show that $\frac{(1 - \tan^{2} θ)}{(1 + \tan^{2} θ)} = (\cos^{2} θ - \sin^{2} θ)$ .

Answer:

$LHS = \frac{(1 - \tan^{2} θ)}{(1 + \tan^{2} θ)} = \frac{(1 - \frac{1}{\cot^{2} θ})}{(1 + \frac{1}{\cot^{2} θ})} = \frac{(\frac{\cot^{2} θ - 1}{\cot^{2} θ})}{(\frac{\cot^{2} θ + 1}{\cot^{2} θ})} = \frac{\cot^{2} θ - 1}{\cot^{2} θ + 1} = \frac{{(\frac{4}{3})}^{2} - 1}{{(\frac{4}{3})}^{2} + 1} (As, 3 \cot θ = 4 or \cot θ = \frac{4}{3})$
$= \frac{\frac{16}{9} - 1}{\frac{16}{9} + 1} = \frac{(\frac{16 - 9}{9})}{(\frac{16 + 9}{9})} = \frac{(\frac{7}{9})}{(\frac{25}{9})} = \frac{7}{25}$
$RHS = (\cos^{2} θ - \sin^{2} θ) = \frac{(\cos^{2} θ - \sin^{2} θ)}{1} = \frac{(\frac{\cos^{2} θ - \sin^{2} θ}{\sin^{2} θ})}{(\frac{1}{\sin^{2} θ})} = \frac{(\frac{\cos^{2} θ}{\sin^{2} θ} - \frac{\sin^{2} θ}{\sin^{2} θ})}{{cosec}^{2} θ} = \frac{(\cot^{2} θ - 1)}{(\cot^{2} θ + 1)}$
$= \frac{[{(\frac{4}{3})}^{2} - 1]}{[{(\frac{4}{3})}^{2} + 1]} = \frac{(\frac{16}{9} - \frac{1}{1})}{(\frac{16}{9} + \frac{1}{1})} = \frac{(\frac{16 - 9}{9})}{(\frac{16 + 9}{9})} = \frac{(\frac{7}{9})}{(\frac{25}{9})} = \frac{7}{25}$

Since, LHS=RHS
Hence, verified.

Page No 289:

Question 23:

If sec θ = $\frac{17}{8}$ , verify that $\frac{3 - 4 \sin^{2} θ}{4 \cos^{2} θ - 3} = \frac{3 - \tan^{2} θ}{1 - 3 \tan^{2} θ}$ .

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 289:

Question 24:

In the adjoining figure, $∠ B = 90 °, ∠ BAC = θ °, BC = CD = 4 cm and AD = 10 cm . Find (i) \sin θ and (ii) \cos θ .$

Answer:

In $∆$ ABD,

Using Pythagoras theorem, we get

$AB = \sqrt{{AD}^{2} - {BD}^{2}} = \sqrt{10^{2} - 8^{2}} = \sqrt{100 - 64} = \sqrt{36} = 6 cm$

Again,

In $∆$ ABC,

Using Pythagoras therem, we get

$AC = \sqrt{{AB}^{2} + {BC}^{2}} = \sqrt{6^{2} + 4^{2}} = \sqrt{36 + 16} = \sqrt{52} = 2 \sqrt{13} cm$

Now,

$(i) \sin θ = \frac{BC}{AC} = \frac{4}{2 \sqrt{13}} = \frac{2}{\sqrt{13}} = \frac{2 \sqrt{13}}{13}$

$(ii) \cos θ = \frac{AB}{AC} = \frac{6}{2 \sqrt{13}} = \frac{3}{\sqrt{13}} = \frac{3 \sqrt{13}}{13}$

Page No 289:

Question 25:

In a ∆ABC, it is given that ∠B = 90°, AB = 24 cm and BC = 7 cm.
Figure

Find the value of
(i) sin A
(ii) cos A
(iii) sin C
(iv) cos C

Answer:

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC²= (24)² + (7)²
⇒ AC² = 576 + 49 = 625
⇒ AC = 25 cm
Now, for T-Ratios of ∠A, base = AB and perpendicular = BC
(i) sin A = $\frac{B C}{A C} = \frac{7}{25}$

(ii) cos A = $\frac{A B}{A C} = \frac{24}{25}$

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) sin C = $\frac{A B}{A C} = \frac{24}{25}$

(iv) cos C = $\frac{B C}{A C} = \frac{7}{25}$

Page No 290:

Question 26:

Given a ∆ABC, in which ∠C = 90°, ∠ABC = θ°, BC = 21 units, AB = 29 units.
Show that (cos²θ − sin²θ) = $\frac{41}{841} .$

Answer:

Using Pythagoras theorem, we get:
AB²=AC²+ BC²
⇒ AC² = AB² $-$ BC²
⇒ AC² = (29)² $-$ (21)²
⇒ AC² = 841 $-$ 441
⇒ AC²= 400
⇒ AC = $\sqrt{400}$ = 20 units

Now, sin $θ = \frac{A C}{A B} = \frac{20}{29}$ and cos $θ$ = $\frac{B C}{A B} = \frac{21}{29}$

cos² $θ$ $-$ sin² $θ$ = ${(\frac{21}{29})}^{2} - {(\frac{20}{29})}^{2} = \frac{441}{841} - \frac{400}{841} = \frac{41}{841}$

Hence Proved.

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Question 27:

In a ∆ABC, right-angled at B, it is given that AB = 12 cm and BC = 5 cm.

Find the value of
(i) cos A
(ii) cosec A
(iii) cos C
(iv) coses C.

Answer:

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC² = 12² + 5² = 144 + 25
⇒ AC² = 169
⇒ AC = 13 cm
Now, for T-Ratios of ∠A, base = AB and perpendicular = BC
(i) cos A = $\frac{A B}{A C} = \frac{12}{13}$
(ii) cosec A = $\frac{1}{\sin A} = \frac{A C}{B C} = \frac{13}{5}$

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) cos C = $\frac{B C}{A C} = \frac{5}{13}$
(iv) cosec C = $\frac{1}{\sin C} = \frac{A C}{A B} = \frac{13}{12}$

Page No 290:

Question 28:

If $\sin α = \frac{1}{2}$ , prove that $(3 \cos α - 4 \cos^{3} α) = 0$ .

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 290:

Question 29:

In a $∆$ ABC, $∠$ B = 90 $°$ and tanA = $\frac{1}{\sqrt{3}}$ . Prove that
(i) sinA $\cdot$ cosC + cosA $\cdot$ sinC = 1 (ii) cosA $\cdot$ cosC $-$ sinA $\cdot$ sinC = 0

Answer:

$In ∆ ABC, ∠ B = 90 °, As, tanA = \frac{1}{\sqrt{3}} \Rightarrow \frac{BC}{AB} = \frac{1}{\sqrt{3}} Let BC = x and AB = x \sqrt{3} Using Pythagoras theorem, we get AC = \sqrt{{AB}^{2} + {BC}^{2}} = \sqrt{{(x \sqrt{3})}^{2} + x^{2}} = \sqrt{3 x^{2} + x^{2}} = \sqrt{4 x^{2}} = 2 x$

Now,

$(i) LHS = sinA \cdot cosC + cosA \cdot sinC = \frac{BC}{AC} \cdot \frac{BC}{AC} + \frac{AB}{AC} \cdot \frac{AB}{AC} = {(\frac{BC}{AC})}^{2} + {(\frac{AB}{AC})}^{2} = {(\frac{x}{2 x})}^{2} + {(\frac{x \sqrt{3}}{2 x})}^{2} = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1 = RHS$

$(ii) LHS = cosA \cdot cosC - sinA \cdot sinC = \frac{AB}{AC} \cdot \frac{BC}{AC} - \frac{BC}{AC} \cdot \frac{AB}{AC} = \frac{x \sqrt{3}}{2 x} . \frac{x}{2 x} - \frac{x}{2 x} . \frac{x \sqrt{3}}{2 x} = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0 = RHS$

Page No 290:

Question 30:

If $∠$ A and $∠$ B are acute angles such that sinA = sinB, then prove that $∠$ A = $∠$ B.

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 290:

Question 31:

If $∠ A$ and $∠ B$ are acute angles such that tanA = tanB, the prove that $∠ A = ∠ B$ .

Answer:

In $∆ ABC, ∠ C = 90 °$ ,

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

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

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Question 32:

In a right $∆ ABC$ , right-angled at $B$ , if $tanA = 1$ , then verify that $2 sinA \cdot cosA = 1$ .

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 290:

Question 33:

In the figure of $∆ PQR, ∠ P = θ ° and ∠ R = ϕ °$ . Find
$(i) (\sqrt{x + 1}) \cot ϕ (ii) (\sqrt{x^{3} + x^{2}}) \tan θ (iii) \cos θ$

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 290:

Question 34:

If $x = cosecA + cosA$ and $y = cosecA - cosA$ , then prove that ${(\frac{2}{x + y})}^{2} + {(\frac{x - y}{2})}^{2} - 1 = 0$ .

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 290:

Question 35:

If $x = cotA + cosA$ and $y = cotA - cosA$ , prove that ${(\frac{x - y}{x + y})}^{2} + {(\frac{x - y}{2})}^{2} = 1$ .

Answer:

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

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