TRIGONOMETRY - 2 (Trigonometric Functions)
1. $\sin 47^{\circ}+\sin 61^{\circ}-\sin 11^{\circ}-\sin 25^{\circ}$ is equal to
(a) $\sin 36^{\circ}$
(b) $\cos 36^{\circ}$
(c) $\sin 7^{\circ}$
(d) $\cos 7^{\circ}$
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Solution:
$ \begin{aligned} & \sin 47+\sin 61^{\circ}-\sin 11^{\circ}-\sin 25^{\circ} \\ & \left(2 \sin 54^{\circ} \cos 7^{\circ}\right)-\left(2 \sin 18^{\circ} \cos 7^{\circ}\right) \\ & =2 \cos 7^{\circ}\left(\sin 54^{\circ}-\sin 18^{\circ}\right) \\ & =2 \cos 7^{\circ}\left(\cos 36^{\circ}-\sin 18^{\circ}\right) \\ & =2 \cos 7^{\circ}\left(\frac{\sqrt{5}+1}{4}-\left(\frac{\sqrt{5}-1}{4}\right)\right) \\ & =2 \cos 7^{\circ}\left(\frac{1}{2}\right) \\ & =\cos 7^{\circ} \end{aligned} $
Correct option is ’d ‘
2. If $\frac{\mathrm{x}}{\cos \theta}=\frac{\mathrm{y}}{\cos \left(\theta-\frac{2 \pi}{3}\right)}=\frac{\mathrm{z}}{\cos \left(\theta+\frac{2 \pi}{3}\right)}$, then $\mathrm{x}+\mathrm{y}+\mathrm{z}$ is equal to
(a) $1$
(b) $0$
(c) $-1$
(d) none of these
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Solution:
We have
$ \frac{\mathrm{x}}{\cos \theta}=\frac{\mathrm{y}}{\cos \left(\theta-\frac{2 \pi}{3}\right)}=\frac{\mathrm{z}}{\cos \left(\theta+\frac{2 \pi}{3}\right)} $
we know that each ratio is equal to $\frac{\text { sum of Numerator }}{\text { sum of denominator }}$
$\therefore \frac{\mathrm{x}}{\cos \theta}=\frac{\mathrm{y}}{\cos \left(\theta-\frac{2 \pi}{3}\right)}=\frac{\mathrm{z}}{\cos \left(\theta+\frac{2 \pi}{3}\right)}=\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}}{\cos \theta+\cos \left(\theta-\frac{2 \pi}{3}\right)+\cos \left(\theta+\frac{2 \pi}{3}\right)}$
now, $\quad \cos \theta+\cos \left(\theta-\frac{2 \pi}{3}\right)+\cos \left(\theta+\frac{2 \pi}{3}\right)$
$=\cos \theta+2 \cos \theta \cos \frac{2 \pi}{3} \quad(\cos (\mathrm{A}+\mathrm{B})+\cos (\mathrm{A}-\mathrm{B})=2 \cos \mathrm{A} \cos \mathrm{B})$
$=\cos \theta+2 \cos \theta\left(-\frac{1}{2}\right)$
$=0$
$\therefore \mathrm{x}+\mathrm{y}+\mathrm{z}=0$
Correct option is: ‘b’
3. The value of $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$ is
(a) $\frac{1}{2}$
(b) $\cos \frac{\pi}{8}$
(c) $\frac{1}{8}$
(d) $\frac{1+\sqrt{2}}{2 \sqrt{2}}$
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Solution:
$ \begin{aligned} & \cos \frac{7 \pi}{8}=\cos \left(\pi-\frac{\pi}{8}\right)=-\cos \frac{\pi}{8} \\ & \cos \frac{3 \pi}{8}=\cos \left(\frac{\pi}{2}-\frac{\pi}{8}\right)=\sin \frac{\pi}{8} \\ & \cos \frac{5 \pi}{8}=\cos \left(\frac{\pi}{2}+\frac{\pi}{8}\right)=-\sin \frac{\pi}{8} \\ & =\left(1+\cos \frac{\pi}{8}\right)\left(1+\sin \frac{\pi}{8}\right)\left(1-\sin \frac{\pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right) \\ & =\left(1-\cos ^{2} \frac{\pi}{8}\right)\left(1-\sin ^{2} \frac{\pi}{8}\right) \\ & =\sin ^{2} \frac{\pi}{8} \cos ^{2} \frac{\pi}{8} \\ & =\frac{\left(2 \sin \frac{\pi}{8} \cos \frac{\pi}{8}\right)^{2}}{4}=\frac{\left(\sin \frac{\pi}{4}\right)^{2}}{4}=\frac{1}{8} \end{aligned} $
Correct option is: ‘c’
4. If $\mathrm{x} _{1}, \mathrm{x} _{2} \ldots \mathrm{x} _{\mathrm{n}}$ are in AP whose common difference is $\alpha$, then the value of $\sin \alpha\left(\operatorname{\sec x} _{1} \operatorname{\sec x} _{2}+\operatorname{\sec x} _{2} \sec _{3}+\ldots \ldots . .+\operatorname{\sec x} _{n-1} \sec _{n}\right)$ is
(a) $\frac{\sin (\mathrm{n}-1) \alpha}{\cos \mathrm{x} _{1} \cos \mathrm{x} _{\mathrm{n}}}$
(b) $\frac{\sin n \alpha}{\cos x _{1} \cos x _{n}}$
(c) $\sin (\mathrm{n}-1) \alpha \cos \mathrm{x} _{1} \cos \mathrm{x} _{\mathrm{n}}$
(d) $\sin \alpha \cos x _{1} \cos x _{n}$
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Solution:
We have
$\sin \alpha \sec x _{1} \sec _{2}+\sin \alpha \sec x _{2} \sec x _{3}+\ldots \ldots \ldots \ldots+\sin \alpha \sec x _{n-1} \sec x _{n}$
$=\frac{\sin \left(\mathrm{x} _{2}-\mathrm{x} _{1}\right)}{\cos \mathrm{x} _{1} \cos \mathrm{x} _{2}}+\frac{\sin \left(\mathrm{x} _{3}-\mathrm{x} _{2}\right)}{\cos \mathrm{x} _{2} \cos \mathrm{x} _{3}}+\ldots \ldots \ldots+\frac{\sin \left(\mathrm{x} _{\mathrm{n}}-\mathrm{x} _{\mathrm{n}-1}\right)}{\cos \mathrm{x} _{\mathrm{n}-1} \cos \mathrm{x} _{\mathrm{n}}}$
$=\frac{\sin x _{2} \cos x _{1}-\cos x _{2} \sin x _{1}}{\cos x _{1} \cos x _{2}}+\frac{\sin x _{3} \cos x _{2}-\cos x _{3} \sin x _{2}}{\cos x _{2} \cos x _{3}}+$
$+\frac{\sin x _{n} \cos x _{n-1}-\cos x _{n} \sin x _{n-1}}{\cos x _{n-1} \cos x _{n}}$
$=\tan x _{2}-\tan x _{1}+\tan x _{3}-\tan x _{2}+\ldots \ldots .+\tan x _{n}-\tan x _{n-1}$
$=\tan x _{n}-\tan x _{1}$
$=\frac{\sin x _{n} \cos x _{1}-\cos x _{n} \sin x _{1}}{\cos x _{n} \cos x _{1}}$
$=\frac{\sin \left(\mathrm{x} _{\mathrm{n}}-\mathrm{x} _{1}\right)}{\cos \mathrm{x} _{\mathrm{n}} \cos \mathrm{x} _{1}}$
$=\frac{\sin (\mathrm{n}-1) \alpha}{\cos \mathrm{x} _{\mathrm{n}} \cos \mathrm{x} _{1}} \quad\left(\because \mathrm{x} _{\mathrm{n}}=\mathrm{x} _{1}+(\mathrm{n}-1) \alpha \quad \mathrm{x} _{\mathrm{n}}-\mathrm{x} _{1}=(\mathrm{n}-1) \alpha\right)$
$\therefore$ correct option is ‘a’
5. Let $f(\mathrm{n})=2 \operatorname{cosnx} \forall \mathrm{n} \in \mathrm{N}$, then $f(1) f(\mathrm{n}+1)-f(\mathrm{n})$ is equal to
(a) $\mathrm{f}(\mathrm{n}+3)$
(b) $\mathrm{f}(\mathrm{n}+2)$
(c) $\mathrm{f}(\mathrm{n}+1) \mathrm{f}(2)$
(d) $\mathrm{f}(\mathrm{n}+2) \mathrm{f}(2)$
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Solution:
We have $\mathrm{f}(\mathrm{n})=2 \operatorname{cosnx} \forall \mathrm{n} \in \mathrm{N}$
$\mathrm{f}(1)=2 \cos \mathrm{x}$
$\mathrm{f}(\mathrm{n}+1)=2 \cos (\mathrm{n}+1) \mathrm{x}$
$f(1) f(n+1)=4 \cos x \cos (n+1) x$
$\mathrm{f}(1) \mathrm{f}(\mathrm{n}+1)-\mathrm{f}(\mathrm{n})=4 \cos \mathrm{x} \cos (\mathrm{n}+1) \mathrm{x}-2 \cos n \mathrm{x}$
$=2[2 \cos x \cos (n+1) x-\cos n x]$
$=2[\cos (n+2) x+\cos n x-\cos n x]$
$=2 \cos (\mathrm{n}+2) \mathrm{x}$
$=\mathrm{f}(\mathrm{n}+2)$
Option ‘b’ is correct
6. The ratio of the greatest value of $2-\cos x+\sin ^{2} x$ to its least value is
(a) $1 / 4$
(b) $9 / 4$
(c) $13 / 4$
(d) None of these
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Solution:
$2-\cos x+\sin ^{2} x$
$=2-\cos x+1-\cos ^{2} x$
$=-\left(\cos ^{2} x+\cos x\right)+3$
$=-\left(\cos x+\frac{1}{2}\right)^{2}+\frac{1}{4}+3$
$=-\left(\cos x+\frac{1}{2}\right)^{2}+\frac{13}{4}$
$ =\frac{13}{4}-\left(\cos x+\frac{1}{2}\right)^{2} $
Maximum value $\frac{13}{4}$ occurs at $\cos x=-\frac{1}{2}$
Minimum value occurs at $\cos x=1$
$ \begin{aligned} & \therefore \frac{13}{4}-\left(1+\frac{1}{2}\right)^{2} \\ & \frac{13}{4}-\frac{9}{4}=\frac{4}{4}=1 \end{aligned} $
$\therefore$ The ratio of greatest to the least is $\frac{13}{4}$
Option ‘c’ is correct
7. If $\sin (120-\alpha)=\sin (120-\beta), 0<\alpha, \beta<\pi$, then find the relation between $\alpha$ and $\beta$.
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Solution:
$ \begin{aligned} & \text { If } \sin A=\sin B \\ & \text { Then } A=B \text { or } A=\pi-B \\ & \text { Here, } \sin \left(120^{\circ} \alpha\right)=\sin \left(120^{\circ}-\beta\right) \\ & \therefore 120^{\circ} \alpha=120^{\circ}-\beta \text { or } 120^{\circ} \alpha=\pi-\left(120^{\circ} \beta\right) \\ & \alpha=\beta \text { or } 120^{\circ}-\alpha=60^{\circ}+\beta \alpha+\beta=60^{\circ} \end{aligned} $
8. If $x, y, z$ are in $A P$, then $\frac{\sin x-\sin z}{\cos z-\cos x}$ is equal to
(a) $\tan y$
(b) $\cot y$
(c) $\sin y$
(d) $\cos y$
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Solution:
We have $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in $\mathrm{AP}$
$\therefore \mathrm{x}+\mathrm{z}=2 \mathrm{y}$
Here, $\frac{\sin x-\sin z}{\cos z-\cos x}=\frac{2 \cos \frac{x+z}{2} \sin \left(\frac{x-z}{2}\right)}{2 \sin \left(\frac{x+z}{2}\right) \sin \left(\frac{x-z}{2}\right)}$
$ \begin{aligned} & =\frac{\cos y}{\sin y} \\ & =\operatorname{\cot y} \end{aligned} $
Correct option is ‘b’
Practice questions
1. It $\cos \theta=\frac{a \cos \phi+b}{a+b \cos \phi}$, then $\tan \theta / 2$ equals
(a). $\sqrt{\frac{a-b}{a+b}} \tan \frac{\phi}{2}$
(b). $\sqrt{\frac{a+b}{a-b}} \cos \frac{\phi}{2}$
(c). $\sqrt{\frac{a-b}{a+b}} \sin \frac{\phi}{2}$
(d). none
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Answer: (a)2. If $a \cos 2 \theta+b \sin 2 \theta=c$ has $\alpha$ and $\beta$ as its solution, then $\tan \alpha+\tan \beta$ equals
(a). $\frac{2 a}{b+c}$
(b). $\frac{2 b}{c+a}$
(c). $\frac{2 c}{a+b}$
(d). none
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Answer: (b, c)3. The value of $\frac{1}{\sin 10^{0}}-\frac{\sqrt{3}}{\cos 10^{\circ}}$ equals
(a). 1
(b). 4
(c). 2
(d). 0
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Answer: (b)4. $\cos ^{4} \frac{\pi}{8}+\cos ^{4} \frac{3 \pi}{8}+\cos ^{4} \frac{5 \pi}{8}+\cos ^{4} \frac{7 \pi}{8}$ equals
(a). $\frac{1}{2}$
(b). $1$
(c). $\frac{3}{2}$
(d). $2$
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Answer: (c)5. $\sin \frac{\pi}{n}+\sin \frac{3 \pi}{n}+\sin \frac{5 \pi}{n}+…….n$ terms equals
(a). $1$
(b). $0$
(c). $\frac{\mathrm{n}}{2}$
(d). none
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Answer: (b)6. If $x=\frac{\sqrt{1-\sin 4 \theta}+1}{\sqrt{1+\sin 4 \theta}-1}$ then one of the values of $x$ is
(a). $-\tan \theta$
(b). $\cot \theta$
(c). $\tan (\pi / 4+\theta)$
(d). $-\cot (\pi / 4+\theta)$
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Answer: (a, b, c, d)7. The value of $\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}$ is
(a). $\frac{1}{2}$
(b). $1$
(c). $-\frac{1}{2}$
(d). $\frac{1}{8}$
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Answer: (c)8. $\sin 6^{\circ}-\sin 66^{\circ}+\sin 78^{\circ}-\sin 42^{\circ}$ is
(a). $-1$
(b). $-\frac{1}{2}$
(c). $\frac{1}{2}$
(d). $1$
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Answer: (b)9. $\cos \frac{\pi}{65} \cos \frac{2 \pi}{65} \cos \frac{4 \pi}{65} \cos \frac{8 \pi}{65} \cos \frac{16 \pi}{65} \cos \frac{32 \pi}{65}$ equals
(a). $\frac{1}{8}$
(b). $\frac{1}{16}$
(c). $\frac{1}{32}$
(d). $\frac{1}{64}$
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Answer: (d)10. $\sin 36^{\circ} \sin 72^{\circ} \sin 108^{\circ} \sin 144^{\circ}$ equals
(a). $\frac{5}{16}$
(b). $\frac{3}{16}$
(c). $\frac{1}{16}$
(d). none
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Answer: (a)11. Passage
If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ be the angles of a triangle, then
(a). $\quad \sum \sin 2 \mathrm{~A}=4 \sin \mathrm{A} \sin \mathrm{B} \sin \mathrm{C}$
(b). $\quad \sum \cos 2 \mathrm{~A}=1-4 \cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C}$
(c). $\quad \sum \sin \mathrm{A}=4 \cos \mathrm{A} / 2 \cos \mathrm{B} / 2 \cos \mathrm{C} / 2$
(d). $\quad \sum \cos \mathrm{A}=-1+4 \sin \mathrm{A} / 2 \sin \mathrm{B} / 2 \sin \mathrm{C} / 2$
(e). $\quad \sum \tan \mathrm{A}=\tan \mathrm{A} \tan \mathrm{B} \tan \mathrm{C}$ ie., $\mathrm{S} _{1}=\mathrm{S} _{3}$
(f). $\quad \sum \tan \mathrm{A} / 2 \tan \mathrm{B} / 2=1$ or $\sum \cot \mathrm{A} / 2=\cot \mathrm{A} / 2 \cot \mathrm{B} / 2 \cot \mathrm{C} / 2$
Answer the following questions based upon above passage.
i. In a triangle $A B C \frac{\sin A+\sin B+\sin C}{\sin A+\sin B-\sin C}$ equals
(a). $\tan \mathrm{A} / 2 \cot \mathrm{B} / 2$
(b). $\cot \mathrm{A} / 2 \tan \mathrm{B} / 2$
(c). $\cot \mathrm{A} / 2 \cot \mathrm{B} / 2$
(d). $\tan \mathrm{A} / 2 \tan \mathrm{B} / 2$
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Answer: (c)ii. $\sin ^{2} \mathrm{~A}+\sin ^{2} \mathrm{~B}+\sin ^{2} \mathrm{C}-2 \cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C}$ equals
(a). 1
(b). 2
(c). 3
(d). 4
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Answer: (b)iii. In a $\triangle \mathrm{ABC}$, whose angles are acute and positive such that $\mathrm{A}+\mathrm{B}+\mathrm{C}=\pi$ and $\cot \mathrm{A} / 2 \cot \mathrm{B} / 2 \cot \mathrm{C} / 2=\mathrm{K}$ then
(a). $\mathrm{K} \leq 3$
(b). $\mathrm{K} \leq 3 \sqrt{3}$
(c). $\mathrm{K} \geq 3 \sqrt{3}$
(d). none
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Answer: (c)iv. $\tan \mathrm{A}, \tan \mathrm{B}, \tan C$ are the roots of the cubic equation $\mathrm{x}^{3}-7 \mathrm{x}^{2}+11 \mathrm{x}-7=0$ then $\mathrm{A}+\mathrm{B}+\mathrm{C}$ equals
(a). $\frac{\pi}{2}$
(b). $\pi$
(c). $0$
(d). none
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Answer: (b)12. Matchning type question :-
| Column I | Column II |
|---|---|
| a. $\sin (\mathrm{B}+\mathrm{C}-\mathrm{A})+\sin (\mathrm{C}+\mathrm{A}-\mathrm{B})+$ $\sin (\mathrm{A}+\mathrm{B}-\mathrm{C})=$ | p. $1-2\cos\frac{A}{2}\cos \frac{B}{2}\cos\frac{C}{2}$ |
| b. $\sin ^{2} \frac{\mathrm{A}}{2}+\sin ^{2} \frac{\mathrm{B}}{2}-\sin ^{2} \frac{\mathrm{C}}{2}=$ | q. $1$ |
| c. If $\tan ^{2} \frac{\mathrm{A}}{2}+\tan ^{2} \frac{\mathrm{B}}{2}+\tan ^{2} \frac{\mathrm{C}}{2}=\mathrm{K}$ | r. $4 \sin \mathrm{A} \sin \mathrm{B} \sin \mathrm{C}$ |
then $\mathrm{K} \geq$
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Answer: a - r, b - p, c - qAssertion and reason type questions
$\mathrm{A}$ : Both $\mathrm{A}$ and $\mathrm{R}$ are individually true and $\mathrm{R}$ is the correct explaination of $\mathrm{A}$
$\mathrm{B}$ : Both $\mathrm{A}$ and $\mathrm{R}$ are individually true and $\mathrm{R}$ is not the correct explaination of $\mathrm{A}$
$\mathrm{C}: \mathrm{A}$ is true but $\mathrm{R}$ is false
$\mathrm{D}: \mathrm{A}$ is false but $\mathrm{R}$ is true
13. Assertion (A) : $\tan \theta+2 \tan 2 \theta+4 \tan 4 \theta+8 \tan 8 \theta+16 \cot 16 \theta=\cot \theta$
Reason $(\mathrm{R}): \cot \theta-\tan \theta=2 \cot 2 \theta$
(a). A
(b). B
(c). C
(d). D
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Answer: (a)14. Assertion (A): $\sec ^{2} \theta=\frac{4 x y}{(x+y)^{2}} \Rightarrow x=y$
Reason $(\mathrm{R}): \sec \theta \geq 1$
(a). A
(b). B
(c). C
(d). D
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Answer: (a)15. Assertion (A) : If A, B, C, D be the angles of a cyclic quadrilateral then $\cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}+\cos \mathrm{D}=0$
Reason $(\mathrm{R}): \sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C}+\sin \mathrm{D}=0$
(a). A
(b). B
(c). C
(d). D
BETA
