Exercise 6.5 - Chapter 6 Trigonometry 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\operatorname{Ex} 6.5$
Multiple choice questions.
Question 1.
The value of $\sin ^2 \theta+\frac{1}{1+\tan ^2 \theta}$ is equal to
(1) $\tan ^2 \theta$
(2) 1
(3) $\cot ^2 \theta$
(4) 0
Solution:
(2) 1
Hint:
$
\begin{aligned}
& \sin ^2 \theta+\frac{1}{1+\frac{\sin ^2 \theta}{\cos ^2 \theta}} \\
& =\sin ^2 \theta+\frac{1}{\frac{\cos ^2 \theta+\sin ^2 \theta}{\cos ^2 \theta}}=\sin ^2 \theta+\frac{\cos ^2 \theta}{1} \\
& =1
\end{aligned}
$
Question 2 .
$\tan \theta \operatorname{cosec}^2 \theta-\tan \theta$ is equal to
(1) $\sec \theta$
(2) $\cot ^2 \theta$
(3) $\sin \theta$
(4) $\cot \theta$

Solution:
(4) $\cot \theta$
Hint:
$
\begin{aligned}
& \frac{\sin \theta}{\cos \theta} \times \frac{1}{\sin ^2 \theta}-\frac{\sin \theta}{\cos \theta} \\
& =\frac{1}{\sin \theta \cos \theta}-\frac{\sin \theta}{\cos \theta} \\
& =\frac{1-\sin ^2 \theta}{\sin \theta \cos \theta}=\frac{\cos ^2 \theta}{\sin \theta \cos \theta}=\cot \theta
\end{aligned}
$
Question 3.
If $(\sin \alpha+\operatorname{cosec} \alpha)^2+(\cos \alpha+\sec \alpha)^2=k+\tan ^2 \alpha+\cot ^2 \alpha$, then the value of $k$ is equal to (1) 9
(2) 7
(3) 5
(4) 3
Solution:
(2) 7
Hint:
$
\begin{aligned}
& (\sin \alpha+\operatorname{cosec} \alpha)^2+(\cos \alpha+\sec \alpha)^2=\mathrm{k}+\tan ^2 \alpha+\cot ^2 \alpha \\
& \sin ^2 \alpha+\operatorname{cosec}^2 \alpha+2 \sin \alpha \operatorname{cosec} \alpha+\cos ^2 \alpha+\sec ^2 \alpha+2 \cos \alpha \sec \alpha=\mathrm{k}+\tan ^2 \alpha+\cot ^2 \alpha \\
& \sin ^2 \alpha+\cos ^2 \alpha+\operatorname{cosec}^2 \alpha+\sec ^2 \alpha+2 \sin \alpha \times \frac{1}{\sin \alpha}+2 \cos \alpha \times \frac{1}{\cos \alpha}=\mathrm{k}+\tan ^2 \alpha+\cot ^2 \alpha \\
& 1+1+\cot ^2 \alpha+1+\tan ^2 \alpha+2+2=\mathrm{k}+\tan ^2 \alpha+\cot ^2 \alpha \\
& 7+\cot ^2 \alpha+\tan ^2 \alpha=\mathrm{k}+\tan ^2 \alpha+\cot ^2 \alpha \\
& \therefore \mathrm{k}=7
\end{aligned}
$
Question 4.
If $\sin \theta+\cos \theta=a$ and $\sec \theta+\operatorname{cosec} \theta=b$, then the value of $b\left(a^2-1\right)$ is equal to
(1) $2 a$
(2) $3 a$
(3) 0
(4) $2 a b$
Solution:
(1) $2 a$
$
a=\sin \theta+\cos \theta
$

$
\begin{aligned}
\mathrm{b} & =\sec \theta+\operatorname{cosec} \theta \\
& b\left(a^2-1\right)=\sec \theta+\operatorname{cosec} \theta\left[(\sin \theta+\cos \theta)^2-1\right] \\
& =\sec \theta+\operatorname{cosec} \theta \\
& =(\sec \theta+\operatorname{cosec} \theta)[2 \sin \theta \cos \theta] \\
& =2 \sin \theta \cos \theta \cdot \frac{1}{\cos \theta}+2 \sin \theta \cos \theta \times \sin \theta \\
& =2 \sin \theta+2 \cos \theta \\
& =2(\sin \theta+\cos \theta) \\
& =2 a
\end{aligned}
$
Question 5.
If $3 \mathrm{x}=\sec \theta$ and $\frac{5}{x}=\tan \theta$, then $x^2-\frac{1}{x^2}$ is
(1) 25
(2) $\frac{1}{25}$
(3) 5
(4) 1

Solution:
$
\begin{aligned}
5 x & =\sec \theta, \frac{5}{x}=\tan \theta \\
x^2-\frac{1}{x^2} & =\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right) \\
& =\left(\frac{\sec \theta}{5}+\frac{\tan \theta}{5}\right)\left(\frac{\sec \theta}{5}-\frac{\tan \theta}{5}\right) \\
& =\left(\frac{\sec \theta+\tan ^2}{5}\right)\left(\frac{\sec \theta-\tan \theta}{5}\right) \\
& =\frac{\sec ^2 \theta-\tan ^2 \theta}{25}=\frac{\sec ^2 \theta-\left(\sec ^2 \theta-1\right)}{25} \\
& =\frac{\sec ^2 \theta-\sec ^2 \theta+1}{25} \\
& =\frac{1}{25}
\end{aligned}
$
Question 6.
If $\sin \theta=\cos \theta$, then $2 \tan ^2 \theta+\sin ^2 \theta-1$ is equal to
(1) $\frac{-3}{2}$
(2) $\frac{3}{2}$
(3) $\frac{2}{3}$
(4) $\frac{-2}{3}$
Solution:
(2) $\frac{3}{2}$

Hint:

$
\begin{aligned}
\text { If } \sin \theta & =\cos \theta \\
\Rightarrow \quad \theta & =\frac{\pi}{4} \text { or } 45^{\circ} \\
2 \tan ^2 \theta+\sin ^2 \theta-1 & =2(1)+\frac{1}{2}-1 \\
& =\frac{3}{2}
\end{aligned}
$
Question 7.
If $\mathrm{x}=\mathrm{a} \tan \theta$ and $\mathrm{y}=\mathrm{b} \sec \theta$ then
(1) $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$
(2) $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
(3) $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
(4) $\frac{x^2}{a^2}-\frac{y^2}{b^2}=0$
Solution:
(1) $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$

Hint:
$
\begin{aligned}
x=a \tan \theta & \Rightarrow \frac{x}{a}=\tan \theta \\
y= & b \sec \theta \Rightarrow \frac{y}{b}=\sec \theta \\
\frac{x^2}{a^2}-\frac{y^2}{b^2} & =\tan ^2 \theta-\sec ^2 \theta \\
& =\sec ^2 \theta-1-\sec ^2 \theta \\
\Rightarrow \frac{x^2}{a^2}-\frac{y^2}{b^2} & =-1 \\
\Rightarrow \frac{y^2}{b^2}-\frac{x^2}{a^2} & =1
\end{aligned}
$
Question 8.
$(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)$ is equal to
(1) 0
(2) 1
(3) 2
(4) $-1$
Solution:
(3) 2

Hint:
$
\begin{aligned}
& (1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta) \\
& =\left(1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)\left(1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right) \\
& =\left(\frac{\cos \theta+\sin \theta+1}{\cos \theta}\right)\left(\frac{\sin \theta+\cos \theta-1}{\sin \theta}\right) \\
& =\frac{((\cos \theta+\sin \theta)+1)((\sin \theta+\cos \theta)-1)}{\sin \theta \cos \theta} \\
& =\frac{(\cos \theta+\sin \theta)^2-1^2}{\sin \theta \cos \theta} \\
& =\frac{\cos ^2 \theta+2 \cos \theta \sin \theta+\sin ^2 \theta-1}{\sin \theta \cos \theta} \\
& =\frac{2 \cos ^2 \theta \sin \theta}{\cos \theta \sin \theta}=2
\end{aligned}
$
Question 9.
$a \cot \theta+b \operatorname{cosec} \theta=p$ and $b \cot \theta+a \operatorname{cosec} \theta=q$ then $p^2-q^2$ is equal to
(1) $a^2-b^2$
(2) $b^2-a^2$
(3) $a^2+b^2$
(4) $b-a$
Answer:
(2) $b^2-a^2$
Hint:
$
\begin{aligned}
& \mathrm{p}^2-\mathrm{q}^2=(\mathrm{p}+\mathrm{q})(\mathrm{p}-\mathrm{q}) \\
& =(\mathrm{a} \cot \theta+\mathrm{b} \operatorname{cosec} \theta+\mathrm{b} \cot \theta+\mathrm{a} \operatorname{cosec} \theta)(\mathrm{a} \cot \theta+\mathrm{b} \operatorname{cosec} \theta-\mathrm{b} \cot \theta-\mathrm{a} \operatorname{cosec} \theta) \\
& =[\cot \theta(\mathrm{a}+\mathrm{b})+\operatorname{cosec} \theta(\mathrm{a}+\mathrm{b})][\cot \theta(\mathrm{a}-\mathrm{b})+\operatorname{cosec} \theta(\mathrm{b}-\mathrm{a})] \\
& =(\mathrm{a}+\mathrm{b})[\cot \theta+\operatorname{cosec} \theta](\mathrm{a}-\mathrm{b})[\operatorname{cosec} \theta(\mathrm{a}-\mathrm{b})] \\
& =(\mathrm{a}+\mathrm{b})[\cot \theta+\operatorname{cosec} \theta](\mathrm{a}-\mathrm{b})[\cot \theta-\operatorname{cosec} \theta] \\
& =(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\left(\cot ^2 \theta-\operatorname{cosec}^2 \theta\right) \\
& =\left(\mathrm{a}^2-\mathrm{b}^2\right)(-1)=-\left(\mathrm{a}^2-\mathrm{b}^2\right) \\
& \mathrm{p}^2-\mathrm{q}^2=\mathrm{b}^2-\mathrm{a}^2
\end{aligned}
$

Question 10.
If the ratio of the height of a tower and the length of its shadow is $\sqrt{3}: 1$, then the angle of elevation of the sun has measure.
(1) $45^{\circ}$
(2) $30^{\circ}$
(3) $90^{\circ}$
(4) $60^{\circ}$
Solution:
(4) $60^{\circ}$
Hint:

Question 11.
The electric pole subtends an angle of $30^{\circ}$ at a point on the same level as its foot. At a second point ' $b$ ' metres above the first, the depression of the foot of the tower is $60^{\circ}$. The height of the tower (in metres) is equal to
(1) $\sqrt{3} b$
(2) $\frac{b}{3}$
(3) $\frac{b}{2}$
(4) $\frac{b}{\sqrt{3}}$
Solution:
(2) $\frac{b}{3}$
Hint:

$
\begin{aligned}
& A B \text { - Pole } \\
& \text { In } \Delta \mathrm{BEC} \\
& \tan 60^{\circ}=\frac{\mathrm{BE}}{\mathrm{EC}}=\sqrt{3} \\
& \Rightarrow \frac{b}{\mathrm{EC}}=\sqrt{3} \\
& \mathrm{EC}=\frac{b}{\sqrt{3}} \\
& \mathrm{EC}=\mathrm{BD}=\frac{b}{\sqrt{3}} \\
& \text { In } \Delta \mathrm{ABD}, \tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{1}{\sqrt{3}} \\
& \Rightarrow \quad \frac{h}{\frac{b}{\sqrt{3}}}=\frac{1}{\sqrt{3}} \\
& h=\frac{\sqrt{3}}{\sqrt{3}}=\frac{b}{\sqrt{3}} \times \frac{1}{\sqrt{3}}=\frac{b}{3} \\
&
\end{aligned}
$
Question 12.
A tower is $60 \mathrm{~m}$ height. Its shadow is $\mathrm{x}$ metres shorter when the sun's altitude is $45^{\circ}$ than when it has been $30^{\circ}$, then $\mathrm{x}$ is equal to
(1) $41.92 \mathrm{~m}$
(2) $43.92 \mathrm{~m}$
(3) $43 \mathrm{~m}$
(4) $45.6 \mathrm{~m}^{\circ}$
Solution:
(2) $43.92 \mathrm{~m}$

Hint:
$
\begin{gathered}
\tan 45^{\circ} \quad=1=\frac{60}{y} \\
y=60 \mathrm{~m} \\
\tan 30^{\circ}=\frac{60}{x+y}=\frac{1}{\sqrt{3}} \\
60+x=60 \sqrt{3} \\
x=60 \sqrt{3}-60 \\
x=60(\sqrt{3}-1) \\
=43.92 \mathrm{~m}
\end{gathered}
$
Question 13.
The angle of depression of the top and bottom of $20 \mathrm{~m}$ tall building from the top of a multistoried building are $30^{\circ}$ and $60^{\circ}$ respectively. The height of the multistoried building and the distance between two buildings (in metres) is
(1) $20,10 \sqrt{3}$
(2) $30,5 \sqrt{3}$
(3) 20,10
(4) $30,10 \sqrt{3}$
Solution:
(4) $30,10 \sqrt{3}$
Hint:

$\begin{aligned}
& \tan 30^{\circ}=\frac{x}{y}=\frac{1}{\sqrt{3}} \\
& \sqrt{3} x=y \\
& x=\frac{y}{\sqrt{3}} \\
& \tan 60^{\circ}=\frac{20+x}{y}=\sqrt{3} \\
& 20+x=y \sqrt{3} \\
& x=y \sqrt{3}-20 \\
&(1)=(2) \Rightarrow \frac{y}{\sqrt{3}}=y \sqrt{3}-20 \\
& y=y \sqrt{3} \sqrt{3}-20 \sqrt{3} \\
& y=3 y-20 \sqrt{3} \\
& \not 2 y=1020 \sqrt{3} \\
&=17.32 \\
& x=\frac{17.32 \times 100}{17.332 \times 100}=\frac{17320}{1732} \\
&=10 \mathrm{~m} \\
& \therefore \text { Height of tower }=20+10=30 \mathrm{~m} \\
& \text { distance }= 17.32 \mathrm{~m}=10 \sqrt{3}
\end{aligned}$

Question 14.
Two persons are standing ' $x$ ' metres apart from each other and the height of the first person is double that of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the shorter person (in metres) is
(1) $\sqrt{2} x$
(2) $\frac{x}{2 \sqrt{2}}$
(3) $\frac{x}{\sqrt{2}}$
(4) $2 x$
Solution:
(2) $\frac{x}{2 \sqrt{2}}$
Hint:

$
\begin{aligned}
& \text { In } \triangle \text { AEB, } \tan \theta=\frac{2 h}{\frac{x}{2}}=\frac{4 h}{x} \\
& \text { In } \Delta \text { CED }, \tan (90-\theta)=\frac{h}{x}=\frac{2 h}{x} \\
& \tan (90-\theta)=\cot \theta=\frac{2 h}{x}, \tan \theta=\frac{x}{2 h}
\end{aligned}
$
Equation (1) and (2)
$
\begin{aligned}
\frac{4 h}{x} & =\frac{x}{2 h} \Rightarrow 8 h^2=x^2 \\
\therefore \quad h^2 & =\frac{x^2}{8} \Rightarrow \quad h=\frac{x}{\sqrt{8}} \\
h & =\frac{x}{2 \sqrt{2}}
\end{aligned}
$
Question 15.
The angle of elevation of a cloud from a point $\mathrm{h}$ metres above a lake is $\beta$. The angle of depression of its reflection in the lake is $45^{\circ}$. The height of location of the cloud from the lake is

(1) $\frac{h(1+\tan \beta)}{1-\tan \beta}$
(2) $\frac{h(1-\tan \beta)}{1+\tan \beta}$
(3) $h \tan \left(45^{\circ}-\beta\right)$
(4) none of these
Solution:
(1) $\frac{h(1+\tan \beta)}{1-\tan \beta}$
Hint:

$\begin{aligned}
& \text { In } \mathrm{CPM}, \tan \beta=\frac{x}{\mathrm{AM}}=\frac{x}{\mathrm{AB}} \\
& \Rightarrow \quad \mathrm{AB}=x \cot \beta \text {. } \\
& \text { In } \triangle \mathrm{PMC}^{\prime} \\
& \tan 45^{\circ}=\frac{x+2 h}{\mathrm{PM}} \\
& =\frac{x+2 h}{\mathrm{AB}} \\
& \mathrm{AB}=(x+2 h) \cot 45^{\circ} \\
& \text { From (1) \& (2) } \\
& \Rightarrow \quad x \cot \beta=(x+2 h) \cot 45^{\circ} \\
& \Rightarrow x\left(\frac{1}{\tan \beta}-\frac{1}{\tan 45^{\circ}}\right)=\frac{2 h}{\tan 45^{\circ}} \\
& \Rightarrow x\left(\frac{\tan 45-\tan \beta}{\tan \beta \tan 45}\right)=\frac{2 h}{\tan 45^{\circ}} \\
& \Rightarrow \quad x=\frac{2 h \tan \beta}{1-\tan \beta} \\
& \mathrm{CB}=x+h=\frac{2 h \tan \beta}{1-\tan \beta}+h \\
& =\frac{2 h \tan \beta}{1-\tan \beta}+h(1-\tan \beta) \\
& =\frac{h+h \tan \beta}{1-\tan \beta} \\
& =\frac{h(1+\tan \beta)}{1-\tan \beta}=0 \\
&
\end{aligned}$