Exercise 3.12 - Chapter 3 - Trigonometry - 11th Maths Guide Samacheer Kalvi Solutions

You can Download the Exercise 3.12 - Chapter 3 - Trigonometry - 11th Maths Guide Samacheer Kalvi Solutions with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

Ex 3.12
Choose the correct answer or most suitable answer:
Question 1.
$
\frac{1}{\cos 80^{\circ}}-\frac{\sqrt{3}}{\sin 80^{\circ}}=
$
(a) $\sqrt{2}$
(b) $\sqrt{3}$
(c) 2
(d) 4
Solution:
(d) 4
Hint:
$
\begin{aligned}
& \frac{1}{\cos 80^{\circ}}-\frac{\sqrt{3}}{\sin 80^{\circ}}=\frac{\frac{1}{2}}{\frac{1}{2} \times \cos 80^{\circ}}-\frac{\frac{\sqrt{3}}{2}}{\sin 80^{\circ} \times \frac{1}{2}} \\
& =\frac{\frac{1}{2} \sin 80^{\circ}-\frac{\sqrt{3}}{2} \cos 80^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}}=\frac{\sin 80^{\circ} \cos 60^{\circ}-\cos 80^{\circ} \sin 60^{\circ}}{\frac{1}{2}\left(\frac{2 \sin 80^{\circ} \cos 80^{\circ}}{2}\right)} \\
& =\frac{\sin 20^{\circ}}{\frac{1}{4}\left[\sin 160^{\circ}\right]}=\frac{4 \sin 20^{\circ}}{\sin \left(180^{\circ}-20^{\circ}\right)}=\frac{4 \sin 20^{\circ}}{\sin 20^{\circ}}=4 \\
&
\end{aligned}
$
Question 2.
If $\cos 28^{\circ}+\sin 28^{\circ}=k^3$, then $\cos 17^{\circ}$ is equal to
(a) $\frac{k^3}{\sqrt{2}}$
(b) $-\frac{k^3}{\sqrt{2}}$
(c) $\pm \frac{k^3}{\sqrt{2}}$
(d) $-\frac{k^3}{\sqrt{3}}$
Solution:

(a) $\frac{k^3}{\sqrt{2}}$
Hint:
$
\begin{aligned}
& \cos 28^{\circ}+\sin 28^{\circ}=k^3 \\
& \text { (i.e.) } \cos 28^{\circ}+\sin \left(90^{\circ}-62^{\circ}\right)=k^3 \\
& \Rightarrow \quad \cos 28^{\circ}+\cos 62^{\circ}=k^3 \\
& 2 \cos \frac{90}{2} \cos \frac{34}{2}=k^3 \\
& 2 \cos 45^{\circ} \cos 17^{\circ}=k^3 \\
& 2 \times \frac{1}{\sqrt{2}} \cos 17^{\circ}=k^3 \\
& \Rightarrow \cos 17^{\circ}=k^3 \times \frac{\sqrt{2}}{2}=\frac{k^3}{\sqrt{2}} \\
&
\end{aligned}
$
Question 3.
The maximum value of $4 \sin ^2 x+3 \cos ^2 x+\sin \frac{x}{2}+\cos \frac{x}{2}$ is
(a) $4+\sqrt{2}$
(b) $3+\sqrt{2}$
(c) 9
(d) 4
Solution:
(a) $4+\sqrt{2}$
Hint:
The max value of $\sin x$ is 1 at $x=\frac{\pi}{2}$ and the max value of $\cos x$ is 1 at $x=0$
The value at $x=0$ is $3+1=4$
The value at $x=\frac{\pi}{2}$ is $4+0+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=4+\frac{2}{\sqrt{2}}=4 \sqrt{2}$
So the maximum value is $4+\sqrt{2}$

Question 4.
$
\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)=
$
Solution:
(a) $\frac{1}{8}$
Hint:
$
\begin{aligned}
\text { LHS } & =\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \pi-\frac{3 \pi}{8}\right)\left(1+\cos \pi-\frac{\pi}{8}\right) \\
& =\left[\left(1+\cos \frac{\pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{3 \pi}{8}\right)\right] \\
& =\left(1-\cos ^2 \frac{\pi}{8}\right)\left(1-\cos ^2 \frac{3 \pi}{8}\right)=\sin ^2 \frac{\pi}{8} \sin ^2 \frac{3 \pi}{8} \\
& =\left(\sin \frac{\pi}{8} \sin \frac{3 \pi}{8}\right)^2=\left\{\frac{1}{2}\left[2 \sin \frac{\pi}{8} \sin \frac{3 \pi}{8}\right]\right\}^2 \\
& =\frac{1}{4}[\cos (\mathrm{A}-\mathrm{B})-\cos (\mathrm{A}+\mathrm{B})]^2 \\
& =\frac{1}{4}\left[\cos \frac{\pi}{4}-\cos \frac{\pi}{2}\right]^2=\frac{1}{4}\left(\frac{1}{\sqrt{2}}\right)^2=\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}
\end{aligned}
$

Question 5 .
If $\pi<2 \theta<\frac{3 \pi}{2}$, then $\sqrt{2+\sqrt{2+2 \cos 4 \theta}}$ equals to
Solution:
(c) $2 \cos \theta$
Hint:
$
\begin{aligned}
2+2 \cos 4 \theta & =2(1+\cos 4 \theta)=2\left(2 \cos ^2 2 \theta\right)=4 \cos ^2 2 \theta \\
\therefore \sqrt{4 \cos ^2 2 \theta} & =2 \cos 2 \theta \\
\sqrt{2+2 \cos 2 \theta} & =\sqrt{2(1+\cos 2 \theta)}=\sqrt{2 \times 2 \cos ^2 \theta} \\
& =\sqrt{4 \cos ^2 \theta}=2 \cos \theta
\end{aligned}
$
Here $\theta$ is in II Quadrant
Question 6.
If $\tan 40^{\circ}=\lambda$, then $\frac{\tan 140^{\circ}-\tan 130^{\circ}}{1+\tan 140^{\circ} \tan 130^{\circ}}=$
(a) $\frac{1-\lambda^2}{\lambda}$
(b) $\frac{1+\lambda^2}{\lambda}$
(c) $\frac{1+\lambda^2}{2 \lambda}$
(d) $\frac{1-\lambda^2}{2 \lambda}$

Solution:
(d) (d) $\frac{1-\lambda^2}{2 \lambda}$
Hint:
Given $\tan 40^{\circ}=\lambda$
$
\begin{aligned}
\text { Now } \frac{\tan 140^{\circ}-\tan 130^{\circ}}{1+\tan 140^{\circ} \tan 130^{\circ}} & =\tan \left(140^{\circ}-130^{\circ}\right)=\tan 10^{\circ} \\
& =\frac{1}{\cot 10^{\circ}}=\frac{1}{\cot \left(90^{\circ}-80^{\circ}\right)}=\frac{1}{\tan 80^{\circ}} \\
\text { Now } \tan 80^{\circ} & =\tan 2(40) \\
& =\frac{2 \tan 40^{\circ}}{1-\tan ^2 40^{\circ}}=\frac{2 \lambda}{1-\lambda^2} \\
\therefore \frac{1}{\tan 80^{\circ}} & =\frac{1-\lambda^2}{2 \lambda}
\end{aligned}
$
Question 7.
$\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+\ldots .+\cos 179^{\circ}=$
(a) 0
(b) 1
(c) -1
(d) 89
Solution:
(a) 0
Hint:
$
\begin{aligned}
& \text { LHS }=\left(\cos 10+\cos 179^{\circ}\right)+\left(\cos 2^{\circ} \div \cos 178^{\circ}\right)+\ldots \ldots+\cos \left(89^{\circ}+\cos 91^{\circ}\right)+\cos 90^{\circ} \\
& \cos 179^{\circ}=\cos \left(180^{\circ}-1\right)=-\cos 1^{\circ} \\
& \cos 178^{\circ}=\cos \left(180^{\circ}-2\right)=-\cos 2^{\circ} \\
& \text { So }\left(\cos 1^{\circ}-\cos 1^{\circ}\right)+\left(\cos 2^{\circ}-\cos 2^{\circ}\right)+\left(\cos 89^{\circ}-\cos 89^{\circ}\right)+\cos 90^{\circ} \\
& =0+0 \ldots+0+0=0 .
\end{aligned}
$
Question 8.
Let $f_k(x)=\frac{1}{k}\left[\sin ^k x+\cos ^k x\right]$ where $x \in \mathrm{R}$ and $k \geq 1$. then $f_4(x)-f_6(x)=$
(a) $\frac{1}{4}$
(b) $\frac{1}{12}$
(c) $\frac{1}{6}$
(d) $\frac{1}{3}$
Solution:
(b) $\frac{1}{12}$

Hint:
$
\begin{aligned}
f_k(x)= & \frac{1}{k}\left[\sin ^k x+\cos ^k x\right] \Rightarrow f_4(x)=\frac{1}{4}\left(\sin ^4 x+\cos ^4 x\right) \\
f_4(x)-f_6(x)= & \frac{1}{4}\left[\sin ^4 x+\cos ^4 x\right]-\frac{1}{6}\left[\sin ^6 x+\cos ^6 x\right] \\
= & \frac{1}{4}\left[\left(\sin ^2 x+\cos ^2 x\right)-2 \sin ^2 x \cos ^2 x\right] \\
& \quad-\frac{1}{6}\left[\left(\sin ^2 x+\cos ^2 x\right)^3-3 \sin ^2 x \cos ^2 x\right] \\
= & \frac{1}{4}\left(1-2 \sin ^2 x \cos ^2 x\right)-\frac{1}{6}\left(1-3 \sin ^2 x \cos ^2 x\right) \\
= & \frac{1}{4}-\frac{1}{2} \sin ^2 x \cos ^2 x-\frac{1}{6}+\frac{1}{2} \sin ^2 x \cos ^2 x \\
= & \frac{1}{4}-\frac{1}{6}=\frac{3-2}{12}=\frac{1}{12}
\end{aligned}
$
Question 9.
Which of the following is not true?
(a) $\sin \theta=-\frac{3}{4}$
(b) $\cos \theta=-1$
(c) $\tan \theta=25$
(d) $\sec \theta=\frac{1}{4}$
Solution:
(d) $\sec \theta=\frac{1}{4}$
Hint:
$\sec \theta=\frac{1}{4} \Rightarrow \cos \theta=4$, which is not true.

Question 10 .
$\cos 2 \theta \cos 2 \phi+\sin ^2(\theta-\phi)-\sin ^2(\theta+\phi)$ is equal to
(a) $\sin 2(\theta+\phi)$
(b) $\cos 2(\theta+\phi)$
(c) $\sin 2(\theta-\phi)$
(d) $\cos 2(\theta-\phi)$
Solution:
(b) $\cos 2(\theta+\phi)$
Hint.
Given $\cos 2 \theta \cdot \cos 2 \phi+\sin ^2(\theta-\phi)-\sin ^2(\theta+\phi)$
$=\cos 2 \theta \cos 2 \phi+\sin (\theta-\phi+\theta+\phi) \sin (\theta-\phi-\theta-\phi)$
$=\cos 2 \theta \cos 2 \phi+\sin 2 \theta \sin (-2 \phi)$
$=\cos 2 \theta \cos 2 \phi-\sin 2 \theta \sin (2 \phi)$
$=\cos (2 \theta+2 \phi)=\cos 2(\theta+\phi)$

Question 11.
$\frac{\sin (A-B)}{\cos A \cos B}+\frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}$ is
(a) $\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C}$
(b) 1
(c) 0
(d) $\cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}$
Solution:
(c) 0
Hint:
$
\begin{aligned}
& \frac{\sin (A-B)}{\cos A \cos B}=\frac{\sin A \cos B-\cos A \sin B}{\cos A \cos B} \\
& =\tan A-\tan B \\
& \text { Similarly } \frac{\sin (B-C)}{\cos B \cos C}=\tan B-\tan C \\
& \text { and } \left.\quad \frac{\sin (C-A)}{\cos C \cos A}=\tan C-\tan A \text { (adding we get } 0\right)
\end{aligned}
$
Question 12.
$\cos p \theta+\cos q \theta=0$ and if $p \neq q$, then 0 is equal to (n is any integer) ......
(a) $\frac{\pi(3 n+1)}{p-q}$
(b) $\frac{\pi(2 n+1)}{p \pm q}$
(c) $\frac{\pi(n \pm 1)}{p \pm q}$
(d) $\frac{\pi(n+2)}{p+q}$
Solution:
(b) $\frac{\pi(2 n+1)}{p \pm q}$
Hint:

$
\begin{aligned}
& \cos p \theta+\cos q \theta=0 \\
& 2 \cos \left(\frac{p+q}{2}\right) \theta \cos \left(\frac{p-q}{2}\right) \theta=0
\end{aligned}
$
as $2 \neq 0, \cos \left(\frac{p+q}{2}\right) \theta=0$ (or) $\cos \left(\frac{p-q}{2}\right) \theta=0$

Question 13.
If $\tan \alpha$ and $\tan \beta$ are the roots $\mathrm{x}^2+\mathrm{ax}+\mathrm{b}=0$, then $\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}$ is equal to
Solution:
(b) $\frac{a}{b}$
Hint:
$
\tan ^2 \mathrm{x}+\mathrm{a} \tan \mathrm{x}+\mathrm{b}=0
$
$\alpha$ and $\beta$ are the roots of the equation
$
\begin{aligned}
& \Rightarrow \tan ^2 \alpha+\mathrm{a} \tan \alpha+\mathrm{b}=0 \ldots \ldots \text { (1) } \\
& \tan ^2 \beta+\mathrm{a} \tan \beta+\mathrm{b}=0 \ldots \ldots \text { (2) } \\
& (1)-(2) \Rightarrow \tan ^2 \alpha-\tan ^2 \beta+\mathrm{a}(\tan \alpha-\tan \beta)=0 \\
& (\tan \alpha-\tan \beta)(\tan \alpha+\tan \beta)+\mathrm{a}(\tan \alpha-\tan \beta)=0 \\
& \Rightarrow \tan \alpha+\tan \beta=-\mathrm{a} \ldots . \text { (A) }
\end{aligned}
$

$
\begin{aligned}
& \text { (1) } \times \tan \beta-(2) \times \tan \alpha \\
& \Rightarrow \tan ^2 \alpha \tan \beta-\tan ^2 \beta \tan \alpha+b(\tan \beta-\tan \alpha)=0 \\
& \tan \alpha \tan \beta(\tan \alpha-\tan \beta)+b(\tan \beta-\tan \alpha)=0 \\
& \Rightarrow \quad \tan \alpha \tan \beta-b=0 \\
& \Rightarrow \quad \tan \alpha \tan \beta=b \\
& \text { Now } \frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}=\frac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\sin \alpha \sin \beta} \\
& \div \text { by } \cos \alpha \cos \beta \\
& \text { we get } \frac{\tan \alpha+\tan \beta}{\tan \alpha \tan \beta}=-\frac{a}{b} \\
&
\end{aligned}
$
Question 14.
In a triangle $\mathrm{ABC}, \sin ^2 \mathrm{~A}+\sin ^2 \mathrm{~B}+\sin ^2 \mathrm{C}=2$, then the triangle is
(a) equilateral triangle
(b) isosceles triangle
(c) right triangle
(d) scalene triangle
Solution:
(c) right triangle
Hint.
On simplifying we get
$
\begin{aligned}
& \sin ^2 \mathrm{~A}+\sin ^2 \mathrm{~B}+\sin ^2 \mathrm{C}=2+2 \cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C} \\
& =2 \text { (given) } \\
& \Rightarrow \cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C}=0 \\
& \cos \mathrm{A} \text { (or) } \cos \mathrm{B} \text { (or) } \cos \mathrm{C}=0 \\
& \Rightarrow \mathrm{A} \text { (or) } \mathrm{B} \text { (or) } \mathrm{C}=\pi / 2 \\
& \Rightarrow \mathrm{ABC} \text { (is a right angled triangle). }
\end{aligned}
$
Question 15 .
If $\mathrm{f}(\theta)=|\sin \theta|+|\cos \theta|, \theta \in \mathrm{R}$, then $\mathrm{f}(\theta)$ is in the interval
(a) $[0,2]$
(b) $[1, \sqrt{2}]$
(c) $[1,2]$
(d) $[0,1]$
Solution:
(b) $[1, \sqrt{2}]$
Hint:

$
\text { at } \begin{aligned}
\theta=0, f(0)=1 & =|\sin \theta|+|\cos \theta| \\
\text { at } \theta=\pi, f(\pi) & =|\sin \pi|+|\cos \pi| \\
& =0+1=1 \\
\text { at } \theta=\frac{\pi}{6}, f\left(\frac{\pi}{6}\right) & =\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}=1.36 \\
\text { at } \theta=\frac{\pi}{4}, f\left(\frac{\pi}{4}\right) & =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}=1.414 \\
\text { at } \theta=\frac{\pi}{3}, f\left(\frac{\pi}{3}\right) & =\frac{\sqrt{3}}{2}+\frac{1}{2}=1.36
\end{aligned}
$
So the interval is $[1, \sqrt{2}]$
Question 16.
$\frac{\cos 6 x+6 \cos 4 x+15 \cos 2 x+10}{\cos 5 x+5 \cos 3 x+10 \cos x}$ is equal to
Solution:
$
\begin{aligned}
& \text { Nr: } \cos 6 x+\cos 4 x+5 \cos 4 x+5 \cos 2 x+10 \cos 2 x+10 \\
& =2 \cos 5 x \cos x+5(2 \cos 3 x \cos x)+10\left(2 \cos ^2 x\right) \\
& =2 \cos x[\cos 5 x+5 \cos 3 x+10 \cos x] \\
& \therefore \frac{\mathrm{Nr}}{\mathrm{Dr}}=\frac{2 \cos x[\cos 5 x+5 \cos 3 x+10 \cos x]}{\cos 5 x+5 \cos 3 x+10 \cos x}=\cos 2 x
\end{aligned}
$
Question 17.
The triangle of maximum area with constant perimeter $12 \mathrm{~m}$
(a) is an equilateral triangle with side $4 \mathrm{~m}$
(b) is an isosceles triangle with sides $2 \mathrm{~m}, 5 \mathrm{~m}, 5 \mathrm{~m}$
(c) is a triangle with sides $3 \mathrm{~m}, 4 \mathrm{~m}, 5 \mathrm{~m}$
(d) does not exists
Solution:
(a) is an equilateral triangle with side $4 \mathrm{~m}$
Hint.
A triangle will have a max area (with a given perimeter) when it is an equilateral triangle.

Question 18 .
A wheel is spinning at 2 radians/second. How many seconds will it take to make 10 complete rotations?
(a) $10 \pi$ seconds
(b) $20 \pi$ seconds
(c) $5 \pi$ seconds
(d) $15 \pi$ seconds
Solution:
(a) $10 \pi$ seconds
Hint.
1 rotation makes $2 \pi^{\mathrm{c}}$
Distance travelled in 1 second $=2$ radians
So time taken to complete 10 rotations $=6 \times 2 \pi=20 \pi^{\mathrm{c}}$ $=\frac{20 \pi}{2}=10 \pi$ seconds
Question 19.
If $\sin \alpha+\cos \alpha=b$, then $\sin 2 \alpha$ is equal to
(a) $b^2-1$, if $b \leq \sqrt{2}$
(b) $b^2-1$, if $b>\sqrt{2}$
(c) $b^2-1$, if $b \geq 1$
(d) $b^2-1$, if $b \geq \sqrt{2}$
Solution:
(b) $\mathrm{b}^2-1$, if $\mathrm{b}>\sqrt{2}$
Hint:
$
\begin{aligned}
& \sin \alpha+\cos \beta=\mathrm{b} \\
& (\sin \alpha+\cos \beta)^2=\mathrm{b}^2 \\
& \sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cos \alpha=\mathrm{b}^2 \\
& \sin ^2 \alpha=\mathrm{b}^2-1
\end{aligned}
$

Question 20.
In a $\triangle \mathrm{ABC},\left(\right.$ i) $\sin \frac{\mathrm{A}}{2} \sin \frac{\mathrm{B}}{2} \sin \frac{\mathrm{C}}{2}>0$ (ii) $\sin \mathrm{A} \sin \mathrm{B} \sin \mathrm{C}>0$ then
(a) Both (i) and (ii) are true
(b) Only (i) is true
(c) Only (ii) is true
(d) Neither (i) nor (ii) is true
Solution:
(a) Both (i) and (ii) are true
Hint.
When $\mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ}$
$\frac{\mathrm{A}}{2}+\frac{\mathrm{B}}{2}+\frac{\mathrm{C}}{2}=\frac{180^{\circ}}{2}=90^{\circ} \Rightarrow$ each is an acute angle.
So $\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}>0$
When $\mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ}$ each angle will be lesser than $180^{\circ}$
So $\sin A, \sin B, \sin C>0$
$\Rightarrow \sin \mathrm{A} \sin \mathrm{B} \sin C>0$
So both (i) and (ii) are true