Mathematics 3000 Secondary 4 Science Option Answers

• Textbook Solutions
• Class 10
• Math
• trigonometry

Mathematics_part_ _II_%28solutions%29 Solutions for Class 10 Math Chapter 6 Trigonometry are provided here with simple step-by-step explanations. These solutions for Trigonometry are extremely popular among Class 10 students for Math Trigonometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics_part_ _II_%28solutions%29 Book of Class 10 Math Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation's Mathematics_part_ _II_%28solutions%29 Solutions. All Mathematics_part_ _II_%28solutions%29 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 131:

Question 1:

If$\sin \theta =\frac{7}{25}$, find the values of cosθ and tan​θ.

Answer:

We have,

$$\begin{gathered} {\sin }^2\theta +{\cos }^2\theta =1 \\ \Rightarrow {\left(\frac{7}{25}\right)}^2+{\cos }^2\theta =1 \\ \Rightarrow {\cos }^2\theta =1-\frac{49}{625}=\frac{625-49}{625}=\frac{576}{625} \\ \Rightarrow \cos \theta =\sqrt{\frac{576}{625}}=\frac{24}{25} \end{gathered}$$

Now,

$$\begin{gathered} \tan \theta =\frac{\sin \theta }{\cos \theta } \\ \Rightarrow \tan \theta =\frac{{\displaystyle \frac{7}{25}}}{{\displaystyle \frac{24}{25}}} \\ \Rightarrow \tan \theta =\frac{{\displaystyle 7}}{{\displaystyle 24}} \end{gathered}$$

Thus, the values of cosθ and tanθ are

$\frac{24}{25}$

 and

$\frac{7}{24}$

, respectively.

Page No 131:

Question 2:

If $\mathrm{tan\theta }=\frac{3}{4}$, find the values of sec​θ and cos​θ

Answer:

We have,

$$\begin{gathered} {\sec }^2\theta =1+{\tan }^2\theta \\ \Rightarrow {\sec }^2\theta =1+{\left(\frac{3}{4}\right)}^2 \\ \Rightarrow {\sec }^2\theta =1+\frac{9}{16}=\frac{16+9}{16}=\frac{25}{16} \\ \Rightarrow \sec \theta =\sqrt{\frac{25}{16}}=\frac{5}{4} \end{gathered}$$

Now,

$$\begin{gathered} \cos \theta =\frac{1}{\sec \theta } \\ \Rightarrow \cos \theta =\frac{1}{\left({\displaystyle \frac{5}{4}}\right)} \\ \Rightarrow \cos \theta =\frac{4}{5} \end{gathered}$$

Thus, the values of sec​θ and cos​θ are

$\frac{5}{4}$

 and

$\frac{4}{5}$

, respectively.

Page No 131:

Question 3:

If $\mathrm{cot\theta }=\frac{40}{9}$ , find the values of cosecθ and sinθ.

Answer:

We have,

$$\begin{gathered} {\mathrm{cosec}}^2\theta =1+{\cot }^2\theta \\ \Rightarrow {\mathrm{cosec}}^2\theta =1+{\left(\frac{40}{9}\right)}^2 \\ \Rightarrow {\mathrm{cosec}}^2\theta =1+\frac{1600}{81}=\frac{1681}{81} \\ \Rightarrow \mathrm{cosec}\theta =\sqrt{\frac{1681}{81}}=\frac{41}{9} \end{gathered}$$

Now,

$$\begin{gathered} \sin \theta =\frac{1}{\mathrm{cosec}\theta } \\ \Rightarrow \sin \theta =\frac{1}{{\displaystyle \frac{41}{9}}} \\ \Rightarrow \sin \theta =\frac{9}{{\displaystyle 41}} \end{gathered}$$

Thus, the values of cosecθ and sinθare

$\frac{41}{9}$

 and

$\frac{9}{41}$

, respectively.

Page No 131:

Question 4:

If 5 secθ – 12 cosecθ = 0, find the values of secθ, cosθ and sinθ.

Answer:

$$\begin{gathered} 5\sec \theta -12\mathrm{cosec}\theta =0 \\ \Rightarrow 5\sec \theta =12\mathrm{cosec}\theta \\ \Rightarrow \frac{5}{\cos \theta }=\frac{12}{\sin \theta } \\ \Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{12}{5} \\ \Rightarrow \tan \theta =\frac{12}{5} \end{gathered}$$

We have,

$$\begin{gathered} {\sec }^2\theta =1+{\tan }^2\theta \\ \Rightarrow {\sec }^2\theta =1+{\left(\frac{12}{5}\right)}^2 \\ \Rightarrow {\sec }^2\theta =1+\frac{144}{25}=\frac{169}{25} \\ \Rightarrow \sec \theta =\sqrt{\frac{169}{25}}=\frac{13}{5} \end{gathered}$$

Now,

$$\begin{gathered} \cos \theta =\frac{1}{\sec \theta } \\ \Rightarrow \cos \theta =\frac{1}{\left({\displaystyle \frac{13}{5}}\right)} \\ \Rightarrow \cos \theta =\frac{5}{13} \end{gathered}$$

Also,

$$\begin{gathered} \frac{\sin \theta }{\cos \theta }=\tan \theta \\ \Rightarrow \sin \theta =\tan \theta \times \cos \theta \\ \Rightarrow \sin \theta =\frac{12}{5}\times \frac{5}{13}=\frac{12}{13} \end{gathered}$$

Thus, the values of secθ, cosθ and sinθare

$\frac{13}{5}$

,

$\frac{5}{13}$

 and

$\frac{12}{13}$

, respectively.

Page No 131:

Question 5:

If tanθ = 1 them, find the values of $\frac{\mathrm{sin\theta }+\mathrm{cos\theta }}{\mathrm{sec\theta }+\mathrm{cosec\theta }}$ .

Answer:

tanθ = 1

We know that, tan45º = 1

∴θ= 45º

Now,

$$\begin{gathered} \sin 45°=\frac{1}{\sqrt{2}} \\ \cos 45°=\frac{1}{\sqrt{2}} \\ \sec 45°=\sqrt{2} \\ \mathrm{cosec}45°=\sqrt{2} \end{gathered}$$
$\therefore \frac{\sin \theta +\cos \theta }{\sec \theta +\mathrm{cosec}\theta }=\frac{\sin 45°+\cos 45°}{\sec 45°+\mathrm{cosec}45°}=\frac{{\displaystyle \frac{1}{\sqrt{2}}}+{\displaystyle \frac{1}{\sqrt{2}}}}{\sqrt{2}+\sqrt{2}}=\frac{{\displaystyle \frac{2}{\sqrt{2}}}}{2\sqrt{2}}=\frac{1}{2}$

Page No 131:

Question 6:

Prave that:

(1)

$\frac{{\sin }^2\mathrm{\theta }}{\mathrm{cos\theta }}+\mathrm{cos\theta }=\mathrm{sec\theta }$

(2)

${\cos }^2\mathrm{\theta }\left(1+{\tan }^2\mathrm{\theta }\right)=1$

(3)

$\sqrt{\frac{1-\mathrm{sin\theta }}{1+\mathrm{sin\theta }}}=\mathrm{sec\theta }-\mathrm{tan\theta }$

(4)

$\left(\mathrm{sec\theta }-\mathrm{cos\theta }\right)\left(\mathrm{cot\theta }+\mathrm{tan\theta }\right)=\mathrm{tan\theta }\mathrm{sec\theta }$

(5)

$\mathrm{cot\theta }+\mathrm{tan\theta }=\mathrm{cosec\theta }\mathrm{sec\theta }$

(6)

$\frac{1}{\mathrm{sec\theta }-\mathrm{tan\theta }}=\mathrm{sec\theta }+\mathrm{tan\theta }$

(7)

${\sec }^4\mathrm{\theta }-{\cos }^4\mathrm{\theta }=1-2{\cos }^2\mathrm{\theta }$

(8)

$\mathrm{sec\theta }+\mathrm{tan\theta }=\frac{\cos {\displaystyle \mathrm{\theta }}}{1-\mathrm{sin\theta }}$

(9) If

$t\mathrm{an\theta }+\frac{1}{\mathrm{tan\theta }}=2$

, then show that

${\tan }^2\mathrm{\theta }+\frac{1}{{\tan }^2\mathrm{\theta }}=2$

(10)

$\frac{\mathrm{tanA}}{{\left(1+{\tan }^2\mathrm{A}\right)}^2}+\frac{\mathrm{cotA}}{{\left(1+{\cot }^2\mathrm{A}\right)}^2}=\sin \mathrm{A}\cos \mathrm{A}$

(11)

${\sec }^4\mathrm{A}\left(1-{\sin }^4\mathrm{A}\right)-2{\tan }^2\mathrm{A}=1$

(12)

$\frac{\mathrm{tan\theta }}{\mathrm{sec\theta }-1}=\frac{\mathrm{tan\theta }+\mathrm{sec\theta }+1}{\mathrm{tan\theta }+\mathrm{sec\theta }-1}$

Answer:

(1)

$$\begin{gathered} \frac{{\sin }^2\theta }{\cos \theta }+\cos \theta \\ =\frac{{\displaystyle {\sin }^2}{\displaystyle \theta }{\displaystyle +}{\displaystyle {{\displaystyle \cos }}^2}{\displaystyle \theta }}{\cos {\displaystyle \theta }} \\ =\frac{1}{\cos {\displaystyle \theta }} \\ =\sec \theta \end{gathered}$$

(2)

$$\begin{gathered} {\cos }^2\theta \left(1+{\tan }^2\theta \right) \\ ={\cos }^2\theta \times {\sec }^2\theta \\ ={\cos }^2\theta \times \frac{1}{{\cos }^2\theta } \\ =1 \end{gathered}$$

(3)

$$\begin{gathered} \sqrt{\frac{1-\sin \theta }{1+\sin \theta }} \\ =\sqrt{\frac{1-\sin \theta }{1+\sin \theta }\times \frac{1-\sin \theta }{1-\sin \theta }} \\ =\sqrt{\frac{{\left(1-\sin \theta \right)}^2}{1-{\sin }^2\theta }} \\ =\sqrt{\frac{{\left(1-\sin \theta \right)}^2}{{\cos }^2\theta }}\left({\cos }^2\theta +{\sin }^2\theta =1\right) \end{gathered}$$
$$\begin{gathered} =\frac{1-\sin \theta }{\cos \theta } \\ =\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta } \\ =\sec \theta -\tan \theta \end{gathered}$$

(4)

$$\begin{gathered} \left(\sec \theta -\cos \theta \right)\left(\cot \theta +\tan \theta \right) \\ =\left(\frac{1}{\cos \theta }-\cos \theta \right)\left(\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta }\right) \\ =\left(\frac{1-{\cos }^2\theta }{\cos \theta }\right)\left(\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }\right) \\ =\frac{{\sin }^2\theta }{\cos \theta }\times \frac{1}{\sin \theta \cos \theta }\left({\sin }^2\theta +{\cos }^2\theta =1\right) \\ =\frac{\sin \theta }{\cos \theta }\times \frac{1}{\cos \theta } \\ =\tan \theta \sec \theta \end{gathered}$$

(5)

$$\begin{gathered} \cot \theta +\tan \theta \\ =\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta } \\ =\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta } \\ =\frac{1}{\sin \theta \cos \theta }\left({\sin }^2\theta +{\cos }^2\theta =1\right) \\ =\frac{1}{\sin \theta }\times \frac{1}{\cos \theta } \\ =\mathrm{cosec}\theta \sec \theta \end{gathered}$$

(6)

$$\begin{gathered} \frac{1}{\sec \theta -\tan \theta } \\ =\frac{{\displaystyle 1}}{{\displaystyle \sec \theta -\tan \theta }}\times \frac{\sec \theta +\tan \theta }{\sec \theta +\tan \theta } \\ =\frac{\sec \theta +\tan \theta }{{\sec }^2\theta -{\tan }^2\theta }\left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\sec \theta +\tan \theta \left(1+{\tan }^2\theta ={\sec }^2\theta \right) \end{gathered}$$

(7)
Disclaimer: There is printing mistake in the question. The correct question should be

${\sin }^4\mathrm{\theta }-{\cos }^4\mathrm{\theta }=1-2{\cos }^2\mathrm{\theta }$

. The solution has been provided accordingly.

$$\begin{gathered} {\sin }^4\theta -{\cos }^4\theta \\ ={\left({\sin }^2\theta \right)}^2-{\left({\cos }^2\theta \right)}^2 \\ =\left({\sin }^2\theta -{\cos }^2\theta \right)\left({\sin }^2\theta +{\cos }^2\theta \right)\left[a^2-b^2=\left(a+b\right)\left(a-b\right)\right] \\ ={\sin }^2\theta -{\cos }^2\theta \left({\sin }^2\theta +{\cos }^2\theta =1\right) \\ =1-{\cos }^2\theta -{\cos }^2\theta \\ =1-2{\cos }^2\theta \end{gathered}$$

(8)

$$\begin{gathered} \sec \theta +\tan \theta \\ =\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta } \\ =\frac{1+\sin \theta }{\cos \theta } \\ =\frac{\cos \theta \left(1+\sin \theta \right)}{{\cos }^2\theta } \\ =\frac{\cos \theta \left(1+\sin \theta \right)}{1-{\sin }^2\theta }\left({\sin }^2\theta +{\cos }^2\theta =1\right) \end{gathered}$$
$$\begin{gathered} =\frac{\cos {\displaystyle \theta }{\displaystyle \left(1+\sin \theta \right)}}{\left(1+\sin \theta \right){\displaystyle \left(1-\sin \theta \right)}} \\ =\frac{{\displaystyle \cos \theta }}{{\displaystyle 1-\sin \theta }} \end{gathered}$$

(9)

$\tan \theta +\frac{1}{\tan \theta }=2$

Squaring on both sides, we get

$$\begin{gathered} {\left(\tan \theta +\frac{1}{\tan \theta }\right)}^2=2^2 \\ \Rightarrow {\tan }^2\theta +\frac{1}{{\tan }^2\theta }+2\times \tan \theta \times \frac{1}{\tan \theta }=4 \\ \Rightarrow {\tan }^2\theta +\frac{1}{{\tan }^2\theta }+2=4 \\ \Rightarrow {\tan }^2\theta +\frac{1}{{\tan }^2\theta }=4-2=2 \end{gathered}$$

(10)

$$\begin{gathered} \frac{\tan A}{{\left(1+{\tan }^{2}A\right)}^2}+\frac{\cot A}{{\displaystyle {\left(1+{\cot }^{2}A\right)}^2}} \\ =\frac{{\displaystyle \tan A}}{{\displaystyle {\left({\sec }^{2}A\right)}^2}}+\frac{{\displaystyle \cot A}}{{\displaystyle {\left({\mathrm{cosec}}^{2}A\right)}^2}}\left(1+{\tan }^2\theta ={\sec }^2\theta \mathrm{and}1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta \right) \\ =\frac{\sin {\displaystyle A}}{\cos {\displaystyle A}}\times {\cos }^{4}A+\frac{\cos {\displaystyle A}}{\sin {\displaystyle A}}\times {\sin }^{4}A\left(\cos \theta =\frac{1}{\sec \theta }\mathrm{and}\sin \theta =\frac{1}{\cos {\displaystyle \mathrm{ec}}{\displaystyle \theta }}\right) \\ =\sin A{\cos }^{3}A+\cos A{\sin }^{3}A \end{gathered}$$
$$\begin{gathered} =\sin A\cos A\left({\cos }^{2}A+{\sin }^{2}A\right) \\ =\sin A\cos A\left({\cos }^2\theta +{\sin }^2\theta =1\right) \end{gathered}$$

(11)

$$\begin{gathered} {\sec }^{4}A\left(1-{\sin }^{4}A\right)-2{\tan }^{2}A \\ ={\sec }^{4}A-{\sec }^{4}A{\sin }^{4}A-2{\tan }^{2}A \\ ={\left(1+{\tan }^{2}A\right)}^2-\frac{{\sin }^{4}A}{{\cos }^{4}A}-2{\tan }^{2}A\left({\sec }^2\theta =1+{\tan }^2\theta \mathrm{and}\sec \theta =\frac{1}{\cos \theta }\right) \\ =1+{\tan }^{4}A+2{\tan }^{2}A-{\tan }^{4}A-2{\tan }^{2}A\left[{\left(a+b\right)}^2=a^2+b^2+2ab\right] \\ =1 \end{gathered}$$

(12)

$$\begin{gathered} \frac{\tan \theta +\sec \theta +1}{\tan \theta +\sec \theta -1} \\ =\frac{{\displaystyle \tan \theta +\sec \theta +1}}{{\displaystyle \tan \theta +\sec \theta -1}}\times \frac{{\displaystyle \tan \theta +\sec \theta +1}}{{\displaystyle \tan \theta +\sec \theta +1}} \\ =\frac{{\displaystyle {\tan }^2\theta +{\sec }^2\theta +1+2\tan \theta +2\sec \theta +2\sec \theta \tan \theta }}{{\displaystyle {\tan }^2\theta +{\sec }^2\theta +2\tan \theta \sec \theta -1}} \\ =\frac{{\displaystyle 2{\sec }^2\theta +2\tan \theta +2\sec \theta +2\sec \theta \tan \theta }}{{\displaystyle 2{\tan }^2\theta +2\tan \theta \sec \theta }}\left(1+{\tan }^2\theta ={\sec }^2\theta \right) \\ =\frac{{\displaystyle 2\left(\tan \theta +\sec \theta \right)+2\sec \theta \left(\tan \theta +\sec \theta \right)}}{{\displaystyle 2\tan \theta \left(\tan \theta +\sec \theta \right)}} \end{gathered}$$
$$\begin{gathered} =\frac{2\left(\tan \theta +\sec \theta \right)\left(\sec \theta +1\right)}{2\tan \theta \left(\tan \theta +\sec \theta \right)} \\ =\frac{\sec \theta +1}{\tan \theta } \\ =\frac{\sec \theta +1}{\tan \theta }\times \frac{\sec \theta -1}{\sec \theta -1} \\ =\frac{{\sec }^2\theta -1}{\tan \theta \left(\sec \theta -1\right)}\left[\left(a-b\right)\left(a+b\right)=a^2-b^2\right] \\ =\frac{{\tan }^2\theta }{\tan \theta \left(\sec \theta -1\right)} \\ =\frac{\tan \theta }{\sec \theta -1} \end{gathered}$$

Page No 137:

Question 1:

A person is standing at a distance of 80 m from a church looking at its top. The angle of elevation is of 45°. Find the height of the church.

Answer:

Let AB be the church and C be the position of the person from the church.

Suppose the height of the church beh m.

In right ∆ABC,

$$\begin{gathered} \tan 45°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow 1=\frac{h}{80} \\ \Rightarrow h=80\mathrm{m} \end{gathered}$$

Thus, the height of the church is 80 m.

Page No 137:

Question 2:

From the top of a lighthouse, an observer looking at a ship makes angle of depression of 60°. If the height of the lighthouse is 90 metre, then find how far the ship is from the lighthouse. $\left(\sqrt{3}=1.73\right)$

Answer:

Let AB be the lighthouse and C be the position of the ship from the lighthouse.

Suppose the distance of the ship from the lighthouse bex m.

In right ∆ABC,

$$\begin{gathered} \tan 60°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \sqrt{3}=\frac{90}{x} \\ \Rightarrow x=\frac{90}{\sqrt{3}}=30\sqrt{3} \\ \Rightarrow x=30\times 1.73=51.9\mathrm{m} \end{gathered}$$

Thus, the ship is 51.9 m away from the lighthouse.

Page No 137:

Question 3:

Two buildings are facing each other on a road of width 12 metre. From the top of the first building, which is 10 metre high, the angle of elevation of the top of the second is found to be 60°. What is the height of the second building ?

Answer:

Let AB and CD be the two building standing on the road.

Suppose the height of the second building beh m.

Here, CD = 10 m, BD = 12 m and CE ⊥ AB.

AE = AB − EB = (h − 10) m            (EB = CD)

CE = BD = 12 m

In right ∆AEC,

$$\begin{gathered} \tan 60°=\frac{\mathrm{AE}}{\mathrm{CE}} \\ \Rightarrow \sqrt{3}=\frac{h-10}{12} \\ \Rightarrow h-10=12\sqrt{3} \\ \Rightarrow h=\left(10+12\sqrt{3}\right)\mathrm{m} \end{gathered}$$

Thus, the height of the second building is

$\left(10+12\sqrt{3}\right)$

 m.

Page No 137:

Question 4:

Two poles of heights 18 metre and 7 metre are erected on a ground. The length of the wire fastened at their tops in 22 metre. Find the angle made by the wire with the horizontal.

Answer:

Let AB and CD be the two poles standing on the ground.

Suppose the angle made by the wire with the horizontal beθ.

Here, AB = 18 m and CD = 7 m.

Length of the wire fastened at their tops = AC = 22 m

AE = AB − EB = 18 − 7 = 11 m            (EB = CD)

In right ∆AEC,

$$\begin{gathered} \sin \theta =\frac{\mathrm{AE}}{\mathrm{AC}} \\ \Rightarrow \sin \theta =\frac{11}{22}=\frac{1}{2} \\ \Rightarrow \sin \theta =\frac{1}{2}=\sin 30° \\ \Rightarrow \theta =30° \end{gathered}$$

Thus, the angle made by the wire with the horizontal is 30º.

Page No 137:

Question 5:

A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60° with the horizontal. Find the height of the tree.

Answer:

Let AC be the original height of the tree. Suppose BD be the broken part of the tree which is rested at D from the base of the tree.

Here, CD = 20 m and ∠BDC = 60º.

In right ∆BCD,

$$\begin{gathered} \tan 60°=\frac{\mathrm{BC}}{\mathrm{CD}} \\ \Rightarrow \sqrt{3}=\frac{\mathrm{BC}}{20} \\ \Rightarrow \mathrm{BC}=20\sqrt{3}\mathrm{m}.....\left(1\right) \end{gathered}$$

Also,

$$\begin{gathered} \cos 60°=\frac{\mathrm{CD}}{\mathrm{BD}} \\ \Rightarrow \frac{1}{2}=\frac{20}{\mathrm{BD}} \\ \Rightarrow \mathrm{BD}=40\mathrm{m}.....\left(2\right) \end{gathered}$$

∴ Height of the tree = AB + BC = BD + BC =

$\left(40+20\sqrt{3}\right)\mathrm{m}$

               [Using (1) and (2)]

Thus, the height of the tree is

$\left(40+20\sqrt{3}\right)\mathrm{m}$

.

Page No 137:

Question 6:

A kite is flying at a height of 60 m above the ground. The string attached to the kite is tied at the ground. It makes an angle of 60° with the ground. Assuming that the string is straight, find the length of the string. $\left(\sqrt{3}=1.73\right)$

Answer:

Let AB be the height of kite above the ground and C be the position of the string attached to the kite which is tied at the ground.

Suppose the length of the string bex m.

Here, AB = 60 m and ∠ACB = 60º

In right ∆ABC,

$$\begin{gathered} \sin 60°=\frac{\mathrm{AB}}{\mathrm{AC}} \\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{60}{x} \\ \Rightarrow x=\frac{120}{\sqrt{3}}=40\sqrt{3} \\ \Rightarrow x=40\times 1.73=69.2\mathrm{m} \end{gathered}$$

Thus, the length of the string is 69.2 m.

Page No 138:

Question 1:

Choose the correct alternative answer for the following questions.

(2) cosec 45^° =?

(A)

$\frac{1}{2}$

(B)

$\sqrt{2}$

(C)

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

(D)

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

(A) cot²

$\mathrm{\theta }$

(B) cosec²

$\mathrm{\theta }$

(C) sec²

$\mathrm{\theta }$

(D) tan²

$\mathrm{\theta }$

(4) When we see at a higher level , from the horizontal line,angle formed is ....... .

(A) angle of elevation.
(B) angle of depression.
(C) 0
(D) straight angle.

Answer:

(1)

$$\begin{gathered} \sin \theta \mathrm{cosec}\theta \\ =\sin \theta \times \frac{1}{\sin \theta } \\ =1 \end{gathered}$$

Hence, the correct answer is option (A).

(2)

$\mathrm{cosec}45°=\sqrt{2}$

Hence, the correct answer is option (B).

(3)

$1+{\tan }^2\theta ={\sec }^2\theta$

Hence, the correct answer is option (C).

(4)
When we see at a higher level, from the horizontal line, angle formed is

angle of elevation

.

Hence, the correct answer is option (A).

Page No 138:

Question 2:

If $\mathrm{sin\theta }=\frac{11}{61}$, find the values of cosθ using trigonometric identity.

Answer:

We have,

$$\begin{gathered} {\sin }^2\theta +{\cos }^2\theta =1 \\ \Rightarrow {\cos }^2\theta =1-{\sin }^2\theta \\ \Rightarrow {\cos }^2\theta =1-{\left(\frac{11}{61}\right)}^2 \\ \Rightarrow {\cos }^2\theta =1-\frac{121}{3721}=\frac{3721-121}{3721}=\frac{3600}{3721} \end{gathered}$$
$\Rightarrow \cos \theta =\sqrt{{\left(\frac{60}{61}\right)}^2}=\frac{60}{61}$

Thus, the value of cosθis

$\frac{60}{61}$

.

Page No 138:

Question 3:

If tanθ = 2, find the values of other trigonometric ratios.

Answer:

tanθ = 2         (Given)

We have,

$$\begin{gathered} {\sec }^2\theta =1+{\tan }^2\theta \\ \Rightarrow \sec \theta =\sqrt{1+{\tan }^2\theta } \\ \Rightarrow \sec \theta =\sqrt{1+2^2} \\ \Rightarrow \sec \theta =\sqrt{1+4}=\sqrt{5} \end{gathered}$$
$\therefore \cos \theta =\frac{1}{\sec \theta }=\frac{1}{\sqrt{5}}$

Now,

$$\begin{gathered} \tan \theta =\frac{\sin \theta }{\cos \theta } \\ \Rightarrow \sin \theta =\tan \theta \times \cos \theta \\ \Rightarrow \sin \theta =2\times \frac{1}{\sqrt{5}}=\frac{2}{\sqrt{5}} \\ \therefore \mathrm{cosec}\theta =\frac{1}{\sin \theta }=\frac{1}{{\displaystyle \frac{2}{\sqrt{5}}}}=\frac{\sqrt{5}}{2} \end{gathered}$$

Also,

$\cot \theta =\frac{1}{\tan \theta }=\frac{1}{2}$

Page No 138:

Question 4:

If $\mathrm{sec\theta }=\frac{13}{12}$ , find the values of other trigonometric ratios.

Answer:

$\sec \theta =\frac{13}{12}$

              (Given)

$\therefore \cos \theta =\frac{1}{\sec \theta }=\frac{1}{{\displaystyle \frac{13}{12}}}=\frac{12}{13}$

We have,

$$\begin{gathered} 1+{\tan }^2\theta ={\sec }^2\theta \\ \Rightarrow \tan \theta =\sqrt{{\sec }^2\theta -1} \\ \Rightarrow \tan \theta =\sqrt{{\left(\frac{13}{12}\right)}^2-1} \\ \Rightarrow \tan \theta =\sqrt{\frac{169}{144}-1} \end{gathered}$$
$$\begin{gathered} \Rightarrow \tan \theta =\sqrt{\frac{169-144}{144}}=\sqrt{\frac{25}{144}} \\ \Rightarrow \tan \theta =\frac{5}{12} \\ \therefore \cot \theta =\frac{1}{\tan \theta }=\frac{1}{{\displaystyle \frac{5}{12}}}=\frac{12}{5} \end{gathered}$$

Now,

$$\begin{gathered} \tan \theta =\frac{\sin \theta }{\cos \theta } \\ \Rightarrow \sin \theta =\tan \theta \times \cos \theta \\ \Rightarrow \sin \theta =\frac{5}{12}\times \frac{12}{13}=\frac{5}{13} \\ \therefore \mathrm{cosec}\theta =\frac{1}{\sin \theta }=\frac{1}{{\displaystyle \frac{5}{13}}}=\frac{13}{5} \end{gathered}$$

Page No 138:

Question 5:

Prove the following.

(1) sec θ (1 – sin θ ) (secθ + tanθ) = 1

(2) (secθ + tanθ) (1 – sinθ) = cosθ

(3) sec² θ + cosec² θ = sec² θ × cosec² θ

(4) cot² θ – tan² θ = cosec² θ – sec² θ

(5) tan⁴ θ + tan² θ = sec⁴ θ – sec² θ

(6)

$\frac{1}{1-\sin \theta }+\frac{{\displaystyle 1}}{{\displaystyle 1+\sin \theta }}=2{\sec }^2\theta$

(7) sec⁶ x – tan⁶ x = 1 + 3sec² x × tan² x

(8)

$\frac{\tan \theta }{sec\theta +1}=\frac{{\displaystyle sec\theta -1}}{{\displaystyle \tan \theta }}$

(9)

$\frac{{\tan }^3\theta -1}{\tan \theta -1}={\sec }^2\theta +\tan \theta$

(10)

$\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{1}{\sin \theta -\tan \theta }$

Answer:

(1)

$$\begin{gathered} \sec \theta \left(1-\sin \theta \right)\left(\sec \theta +\tan \theta \right) \\ =\left(\sec \theta -\sec \theta \sin \theta \right)\left(\sec \theta +\tan \theta \right) \\ =\left(\sec \theta -\frac{\sin \theta }{\cos \theta }\right)\left(\sec \theta +\tan \theta \right) \\ =\left(\sec \theta -\tan \theta \right)\left(\sec \theta +\tan \theta \right) \\ ={\sec }^2\theta -{\tan }^2\theta \\ =1\left(1+{\tan }^2\theta ={\sec }^2\theta \right) \end{gathered}$$

(2)

$$\begin{gathered} \left(\sec \theta +\tan \theta \right)\left(1-\sin \theta \right) \\ =\left(\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }\right)\left(1-\sin \theta \right) \\ =\left(\frac{1+\sin \theta }{\cos \theta }\right)\left(1-\sin \theta \right) \\ =\frac{1-{\sin }^2\theta }{\cos \theta } \\ =\frac{{\cos }^2\theta }{\cos \theta }\left({\sin }^2\theta +{\cos }^2\theta =1\right) \\ =\cos \theta \end{gathered}$$

(3)

$$\begin{gathered} {\sec }^2\theta +{\mathrm{cosec}}^2\theta \\ =\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta } \\ =\frac{{\sin }^2\theta +{\cos }^2\theta }{{\cos }^2\theta {\sin }^2\theta } \\ =\frac{1}{{\cos }^2\theta {\sin }^2\theta } \\ =\frac{1}{{\cos }^2\theta }\times \frac{1}{{\sin }^2\theta } \\ ={\sec }^2\theta {\mathrm{cosec}}^2\theta \end{gathered}$$

(4)

$$\begin{gathered} {\cot }^2\theta -{\tan }^2\theta \\ =\left({\mathrm{cosec}}^2\theta -1\right)-\left({\sec }^2\theta -1\right)\left(1+{\tan }^2\theta ={\sec }^2\theta \&1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta \right) \\ ={\mathrm{cosec}}^2\theta -1-{\sec }^2\theta +1 \\ ={\mathrm{cosec}}^2\theta -{\sec }^2\theta \end{gathered}$$

(5)

$$\begin{gathered} {\tan }^4\theta +{\tan }^2\theta \\ ={\left({\sec }^2\theta -1\right)}^2+\left({\sec }^2\theta -1\right)\left(1+{\tan }^2\theta ={\sec }^2\theta \right) \\ ={\sec }^4\theta -2{\sec }^2\theta +1+{\sec }^2\theta -1\left[{\left(a-b\right)}^2=a^2-2ab+b^2\right] \\ ={\sec }^4\theta -{\sec }^2\theta \end{gathered}$$

(6)

$$\begin{gathered} \frac{1}{1-\sin \theta }+\frac{1}{1+\sin \theta } \\ =\frac{{\displaystyle 1+\sin \theta +1-\sin \theta }}{{\displaystyle \left(1-\sin \theta \right)\left(1+\sin \theta \right)}} \\ =\frac{2}{1-{\sin }^2\theta }\left[\left(a-b\right)\left(a+b\right)=a^2-b^2\right] \\ =\frac{{\displaystyle 2}}{{\displaystyle {\cos }^2\theta }}\left({\sin }^2\theta +{\cos }^2\theta =1\right) \\ =2{\sec }^2\theta \end{gathered}$$

(7)
We have,

${\sec }^{2}x-{\tan }^{2}x=1$

Cubing on both sides, we get

$$\begin{gathered} {\left({\sec }^{2}x-{\tan }^{2}x\right)}^3=1^3 \\ \Rightarrow {\left({\sec }^{2}x\right)}^3-{\left({\tan }^{2}x\right)}^3-3\times {\sec }^{2}x\times {\tan }^{2}x\times \left({\sec }^{2}x-{\tan }^{2}x\right)=1\left[{\left(a-b\right)}^3=a^3-b^3-3ab\left(a-b\right)\right] \\ \Rightarrow {\sec }^{6}x-{\tan }^{6}x-3{\sec }^{2}x{\tan }^{2}x=1 \\ \Rightarrow {\sec }^{6}x-{\tan }^{6}x=1+3{\sec }^{2}x{\tan }^{2}x \end{gathered}$$

(8)

$$\begin{gathered} \frac{\tan \theta }{\sec \theta +1} \\ =\frac{{\displaystyle \tan \theta }}{{\displaystyle \sec \theta +1}}\times \frac{\sec \theta -1}{\sec \theta -1} \\ =\frac{\tan {\displaystyle \theta }{\displaystyle \left(\sec \theta -1\right)}}{{\sec }^2\theta -1} \\ =\frac{{\displaystyle \tan \theta \left(\sec \theta -1\right)}}{{\tan }^2\theta }\left(1+{\tan }^2\theta ={\sec }^2\theta \right) \\ =\frac{{\displaystyle \sec \theta -1}}{\tan \theta } \end{gathered}$$

(9)

$$\begin{gathered} \frac{{\tan }^3\theta -1}{\tan \theta -1} \\ =\frac{{\displaystyle \left(\tan \theta -1\right)\left({\tan }^2\theta +\tan \theta \times 1+1\right)}}{{\displaystyle \tan \theta -1}}\left[a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\right] \\ ={\tan }^2\theta +\tan \theta +1 \\ ={\sec }^2\theta +\tan \theta \left(1+{\tan }^2\theta ={\sec }^2\theta \right) \end{gathered}$$

(10)

$\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}$
$=\frac{{\displaystyle \frac{\sin \theta -\cos \theta +1}{\cos \theta }}}{{\displaystyle \frac{\sin \theta +\cos \theta -1}{\cos \theta }}}$

             (Dividing numerator and denominator by cosθ)

$=\frac{{\displaystyle \frac{\sin \theta }{\cos \theta }}-{\displaystyle \frac{\cos \theta }{\cos \theta }}+{\displaystyle \frac{1}{\cos \theta }}}{{\displaystyle \frac{\sin \theta }{\cos \theta }}+{\displaystyle \frac{\cos \theta }{\cos \theta }}-{\displaystyle \frac{1}{\cos \theta }}}$
$$\begin{gathered} =\frac{\tan \theta -1+\sec \theta }{\tan \theta +1-\sec \theta } \\ =\frac{\tan \theta -1+\sec \theta }{\tan \theta +\left({\sec }^2\theta -{\tan }^2\theta \right)-\sec \theta }\left(1+{\tan }^2\theta ={\sec }^2\theta \right) \\ =\frac{\tan \theta -1+\sec \theta }{\left(\sec \theta -\tan \theta \right)\left(\sec \theta +\tan \theta \right)-\left(\sec \theta -\tan \theta \right)}\left[a^2-b^2=\left(a-b\right)\left(a+b\right)\right] \\ =\frac{\tan \theta -1+\sec \theta }{\left(\sec \theta -\tan \theta \right)\left(\sec \theta +\tan \theta -1\right)} \\ =\frac{1}{\sec \theta -\tan \theta } \end{gathered}$$

Page No 139:

Question 6:

A boy standing at a distance of 48 meters from a building observes the top of the building and makes an angle of elevation of 30°. Find the height of the building.

Answer:

Let AB be the building and C be the position of the boy from the building.

Suppose the height of the building beh m.

Here, BC = 48 m and ∠ACB = 30º.

In right ∆ABC,

$$\begin{gathered} \tan 30°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{48} \\ \Rightarrow h=\frac{48}{\sqrt{3}}=16\sqrt{3}\mathrm{m} \end{gathered}$$

Thus, the height of the building is

$16\sqrt{3}$

m.

Page No 139:

Question 7:

From the top of the light house, an observer looks at a ship and finds the angle of depression to be 30°. If the height of the light-house is 100 meters, then find how far the ship is from the light-house.

Answer:

Let AB be the lighthouse and C be the position of the ship from the lighthouse.

Suppose the distance of the ship from the lighthouse bex m.

Here, AB = 100 m and ∠ACB = 30º.

In right ∆ABC,

$$\begin{gathered} \tan 30°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{x} \\ \Rightarrow x=100\sqrt{3}\mathrm{m} \end{gathered}$$

Thus, the ship is

$100\sqrt{3}$

 m away from the lighthouse.

Page No 139:

Question 8:

Two buildings are in front of each other on a road of width 15 meters. From the top of the first building, having a height of 12 meter, the angle of elevation of the top of the second building is 30°.What is the height of the second building?

Answer:

Let AB and CD be the two building standing on the road.

Suppose the height of the second building beh m.

Here, AB = 12 m, BD = 15 m, ∠CAE = 30º and AE ⊥ CD.

CE = CD − ED = (h − 12) m            (ED = AB)

AE = BD = 15 m

In right ∆AEC,

$$\begin{gathered} \tan 30°=\frac{\mathrm{CE}}{\mathrm{AE}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h-12}{15} \\ \Rightarrow h-12=\frac{15}{\sqrt{3}}=5\sqrt{3} \\ \Rightarrow h=\left(12+5\sqrt{3}\right)\mathrm{m} \end{gathered}$$

Thus, the height of the second building is

$\left(12+5\sqrt{3}\right)$

 m.

Page No 139:

Question 9:

A ladder on the platform of a fire brigade van can be elevated at an angle of 70° to the maximum. The length of the ladder can be extended upto 20 m. If the platform is 2m above the ground, find the maximum height from the ground upto which the ladder can reach. (sin 70° = 0.94)

Answer:

Let AB be the platform of the fire brigade van and AD be the ladder.

Suppose the maximum height from the ground upto which the ladder can reach beh m.

Here, AB = 2 m, AD = 20 m, ∠DAE = 70º and AE ⊥ CD.

DE = CD − CE = (h− 2) m             (CE = AB)

In right ∆AED,

$$\begin{gathered} \sin 70°=\frac{\mathrm{DE}}{\mathrm{AD}} \\ \Rightarrow 0.94=\frac{h-2}{20} \\ \Rightarrow h-2=20\times 0.94=18.8 \\ \Rightarrow h=18.8+2=20.8\mathrm{m} \end{gathered}$$

Thus, the maximum height from the ground upto which the ladder can reach is 20.8 m.

Page No 139:

Question 10:

While landing at an airport, a pilot made an angle of depression of 20°. Average speed of the plane was 200 km/hr. The plane reached the ground after 54 seconds. Find the height at which the plane was when it started landing. (sin 20° = 0.342)

Answer:

Let the plane was at a height ofh m when it started landing.

Average speed of the plane = 200 km/h =

$200\times \frac{5}{18}=\frac{500}{9}$

 m/s

Time taken by plane to reach the ground = 54 s

∴ Distance covered by the plane to reach the ground = Average speed of the plane × Time taken by plane to reach the ground

=

$\frac{500}{9}\times 54$

= 3000 m

Here, AC = 3000 m and ∠ACB = 20º

In right ∆ABC,

$$\begin{gathered} \sin 20°=\frac{\mathrm{AB}}{\mathrm{AC}} \\ \Rightarrow 0.342=\frac{h}{3000} \\ \Rightarrow h=0.342\times 3000=1026\mathrm{m} \end{gathered}$$

Thus, the plane was at the height of 1026 m when it started landing.

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