Rs Aggarwal 2018 for Class 10 Math Chapter 5 - Trigonometric Ratios

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Rs Aggarwal 2018 Solutions for Class 10 Math Chapter 5 Trigonometric Ratios are provided here with simple step-by-step explanations. These solutions for Trigonometric Ratios are extremely popular among Class 10 students for Math Trigonometric Ratios Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 10 Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

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 = $\frac{1}{\sqrt{7}}$ then prove that $(\frac{{cosec}^{2} θ + \sec^{2} θ}{{cosec}^{2} θ - \sec^{2} θ}) = \frac{4}{3} .$

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 = $\frac{17}{8}$ then prove 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:

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

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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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