Exercise 3.12 - Chapter 3 Trigonometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

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