Trigonometric Functions Test 15

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Trigonometric Functions Test 15

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Question 1
1 / -0

If $$\alpha cos^23\theta +\beta cos^4\theta= 16 cos^6\theta + 9 cos^2\theta$$ is an identity then-

A
$$\alpha = 1, \beta = 18$$

B
$$\alpha = 1, \beta = 24$$

C
$$\alpha = 3, \beta = 24$$

D
$$\alpha = 4, \beta = 2$$

Solution

$$\alpha\cos^2 3\theta+\beta\cos^4\theta=16\cos^6\theta+9\cos^2\theta$$

$$\Rightarrow$$ $$\alpha(4\cos^3\theta-3\cos\theta)^2+\beta\cos^4\theta=16\cos^6\theta+9\cos^2\theta$$

$$\Rightarrow$$ $$\alpha(16\cos^6\theta+9\cos^2\theta-24\cos^4\theta)+\beta\cos^4\theta=16\cos^6\theta+9\cos^2\theta$$

$$\Rightarrow$$ $$\alpha(16\cos^6\theta+9\cos^2\theta)-24\alpha\cos^4\theta+\beta\cos^4\theta=(16\cos^6\theta+9\cos^2\theta)$$

$$\Rightarrow$$ $$\alpha(16\cos^6\theta+9\cos^2\theta)-\cos^4\theta(24\alpha-\beta)=(16\cos^6\theta+9\cos^2\theta)$$

$$\Rightarrow$$ $$\alpha(16\cos^6\theta+9\cos^2\theta)-\cos^4\theta(24\alpha-\beta)=(16\cos^6\theta+9\cos^2\theta)+\cos^4\theta\times 0$$

Now, we are going to compare,
$$\alpha(16\cos^6+9\cos^2\theta)=16\cos^6+9\cos^2\theta$$
$$\therefore$$ $$\alpha=1$$

Again,
$$\Rightarrow$$ $$-\cos^4\theta(24\alpha-\beta)=\cos^4\theta\times 0$$
$$\Rightarrow$$ $$24(1)-\beta=0$$ [ Since, $$\alpha=1$$ ]
$$\Rightarrow$$ $$\beta=24$$

$$\therefore$$ $$\alpha=1$$ and $$\beta=24$$
Question 2
1 / -0

If $$\sin \left( {\pi \cos x} \right) = \cos \left( {\pi \sin x} \right)$$, then $$\sin 2x = $$

A
$$ -\dfrac{3}{4}$$

B
$$ -\dfrac{4}{3}$$

C
$$ \dfrac{1}{3}$$

D
none of these

Solution

Given $$\sin(\pi\cos x)=\cos(\pi\sin x)$$
$$\sin(\pi cosx)=\sin(\dfrac{\pi}{2}-\pi\sin x)$$
$$\pi \cos x=\dfrac{\pi}{2}-\pi \sin x $$
$$\cos x+\sin x=\dfrac{1}{2}$$
Squaring on both sides
$$\sin^2x+\cos^2x+2\sin x \cos x= \dfrac{1}{4}$$
$$1+2\sin x\cos x=\dfrac{1}{4}$$
$$\sin 2x=-\dfrac{3}{4}$$
Question 3
1 / -0

If $$x \in (\pi, 2\pi)$$ and $$\cos x + \sin x = \dfrac{1}{2}$$, then the value of $$\tan x$$ is

A
$$\dfrac{4 - \sqrt{7}}{3}$$

B
$$\dfrac{\sqrt{7} - 4}{3}$$

C
$$\dfrac{-4 + \sqrt{7}}{3}$$

D
$$-\left(\dfrac{4 + \sqrt{7}}{3}\right)$$

Solution

$$\cos x+\sin x=\dfrac{1}{2}$$ $$x\in (\pi, 2\pi)$$
divide by $$\cos x$$
$$1+\dfrac{\sin x}{\cos x}=\dfrac{1}{2\cos x}$$
$$\Rightarrow 1+\tan x=\dfrac{1}{2}\sec x$$ [we know $$\sec^2x-\tan^2x=1$$]
$$\Rightarrow 1+\tan x=\dfrac{1}{2}\sqrt{1+\tan^2x}$$
Let $$\tan x=t$$
$$1+t=\dfrac{\sqrt{1+t^2}}{2}$$
$$(2(1+t))^2=1+t^2$$
$$\Rightarrow 3t^2+8t+3=0$$
$$\therefore t=\dfrac{-8\pm \sqrt{8^2-4(3)(3)}}{2\times 3}$$
$$\Rightarrow \dfrac{-8\pm 2\sqrt{7}}{6}$$
$$\Rightarrow t=\dfrac{-4\pm \sqrt{7}}{3}$$
$$\therefore \tan x=\dfrac{-4\pm \sqrt{7}}{3}$$ but $$x\in (\pi, 2\pi)$$
$$\therefore \tan x=\dfrac{-4+\sqrt{7}}{3}$$.
Question 4
1 / -0

If $$\dfrac{{\left( {1 - \cos A} \right)}}{2} = x$$ then find the value of x is

A
$${\cos ^2}\left( {\dfrac{A}{2}} \right)$$

B
$$\sqrt {\sin \left( {\dfrac{A}{2}} \right)} $$

C
$$\sqrt {\cos \left( {\dfrac{A}{2}} \right)} $$

D
$${\sin ^2}\left( {\dfrac{A}{2}} \right)$$

Solution

$$\implies\cfrac { 1-\cos A }{ 2 } =x$$
$$\implies\cfrac { 2{ \sin }^{ 2 }\cfrac { A }{ 2 } }{ 2 } =x$$
$$\implies x={ \sin }^{ 2 }\cfrac { A }{ 2 } $$
Question 5
1 / -0

If $$x=\dfrac{{2\left( {\sin {1^0} + \sin {2^0} + \sin {3^0} + ....... + \sin {{89}^0}} \right)}}{{2\left( {\cos {1^0} + \cos {2^0} + .............\cos {{44}^0}} \right) + 1}}$$ , then the value of $${\log_x}2$$ is equal

A
$$0$$

B
$$\dfrac { 1 }{ 2 } $$

C
$$1$$

D
$$2$$

Solution
Question 6
1 / -0

A flag staff on the top of the tower $$80\ meter$$ high, subtends an angle $$\tan^{-1}\left(\dfrac{1}{9}\right)$$ at point on the ground $$100\ meters$$ away from the foot of the tower. Find the height of the flag-staff.

A
$$20\ m$$

B
$$30\ m$$

C
$$25\ m$$

D
$$35\ m$$

Solution

$$\rightarrow$$ Let, height of the flag stuff be $$x$$.
$$\rightarrow$$ Angle subtended by the tower.
$$\therefore \theta = \tan^{-1} \left( \dfrac{80}{100} \right)= 38.7 $$ degree.
$$\rightarrow $$ Angle subtended by the flag stuff
$$\therefore \tan^{-1} (7/9)= 6.3$$ degree.
$$\rightarrow $$ total angle subtended $$= 38.7 + 6.3$$
$$= 45$$ degree.
$$\rightarrow $$ As shown in triangle
$$\tan 45= \dfrac{x +80}{100}$$
$$\therefore 1 =\dfrac{x+80}{100} $$
$$\therefore x= 100-80$$
$$\therefore x= 20$$
Question 7
1 / -0

Total number of solution of the equation $$3x+2\tan x=\dfrac {5\pi}{2}$$ in $$x\ \epsilon [0,2\pi]$$ is equal to

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Given equation $$3x+2\tan x=\dfrac{5\pi}{2}$$
$$2\tan x =5\dfrac{\pi}{2}-3x$$
$$\tan x=\dfrac{5\pi}{4}-\dfrac{3x}{2}$$
$$y=\tan x=\dfrac{5\pi}{4}-\dfrac{3x}{2}$$
Using graph we find point intersection of
$$y=\tan x;y=\dfrac{5\pi}{4}-\dfrac{3x}{2}\left(m=-\dfrac{3}{2};C=\dfrac{5\pi}{4}\right)$$
$$x-intercept :y=0;$$
$$\dfrac{3x}{2}=\dfrac{5\pi}{4}\left(x=\dfrac{5\pi}{6},y=0\right)$$
$$y-intercept\, x=0;$$
$$y=\dfrac{5\pi}{4}\left(x=0;y=\dfrac{5\pi}{4}\right)$$
Given region of graph $$[0,2\pi]$$
Number of solution = Number of intersections
$$='3'$$
$$\boxed{Number\, of \, solutions=3}$$
Question 8
1 / -0

$$\sin^{-1}x+\sin^{-1}\dfrac{1}{x}+\cos^{-1}x+\cos^{-1}\dfrac{1}{x}, x\notin \pm 1$$ is equal to?

A
$$\pi$$

B
$$\dfrac{\pi}{2}$$

C
$$\dfrac{3\pi}{2}$$

D
None of these

Solution
Question 9
1 / -0

Find $$\dfrac{\sin^2A-\sin^2B}{\sin A\cos A-\sin B\cos B}=?$$.

A
$$\cos (A+B)$$

B
$$\tan (A-B)$$

C
$$\cot (A+B)$$

D
$$\tan (A+B)$$

Solution

$$\cfrac { { \sin }^{ 2 }A-{ \sin }^{ 2 }B }{ \sin A\cos A-\sin B\cos B } =\cfrac { \cfrac { 1-\cos 2 A }{ 2 } -\cfrac { 1-\cos 2 B }{ 2 } }{ \cfrac { 1 }{ 2 } (\sin 2 A-\sin 2 B) } $$
$$=\cfrac { 1-\cos 2 A-1+\cos 2 B }{ \sin 2 A-\sin 2 B } $$
$$=\cfrac {- (\cos 2 A-\cos 2 B) }{ \sin 2 B-\sin 2 A } $$
$$=\cfrac { -\{ -2\sin(\cfrac { 2A+2B }{ 2 } )\sin(\cfrac { 2A-2B }{ 2 } )\} }{ \sin 2B-\sin 2A } $$
$$==\cfrac { -\{ -2\sin(\cfrac { 2A+2B }{ 2 } )\sin(\cfrac { 2A-2B }{ 2 } )\} }{ 2\sin(\cfrac { 2A-2B }{ 2 } )\cos(\cfrac { 2A+2B }{ 2 } ) } $$
$$=\cfrac { \sin(A+B) }{ \cos(A+B) } $$
$$=\tan(A+B)$$
Question 10
1 / -0

If $$\tan\theta=\dfrac{a}{b}$$ then $$\dfrac{a\sin\theta-b\cos\theta}{a\sin\theta +b\cos\theta}=$$

A
$$\dfrac{a^2b^2}{a^2+b^2}$$

B
$$\dfrac{ab}{a^2+b^2}$$

C
$$\dfrac{a^2b^2}{a+b}$$

D
None of these

Solution

$$\tan\theta =\cfrac { a }{ b } ,$$ then $$\cfrac { a \sin\theta -b \cos\theta }{ a \sin\theta +b \cos }=\cfrac { \cfrac { a \sin\theta }{ b \cos\theta } -1 }{ \cfrac { a \sin\theta }{ b \cos\theta } +1 } $$
$$=\cfrac { \cfrac { a }{ b } \tan\theta -1 }{ \cfrac { a }{ b } \tan\theta +1 } $$
$$=\cfrac { \cfrac { { a }^{ 2 } }{ { b }^{ 2 } } -1 }{ \cfrac { { a }^{ 2 } }{ { b }^{ 2 } } +1 } $$
$$=\cfrac { { a }^{ 2 }-{ b }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 } } $$

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Trigonometric Functions Test 15

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Trigonometric Functions Test 15

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SHARING IS CARING

Question 1

$$\alpha = 1, \beta = 18$$

$$\alpha = 1, \beta = 24$$

$$\alpha = 3, \beta = 24$$

$$\alpha = 4, \beta = 2$$

Solution

Question 2

$$ -\dfrac{3}{4}$$

$$ -\dfrac{4}{3}$$

$$ \dfrac{1}{3}$$

none of these

Solution

Question 3

$$\dfrac{4 - \sqrt{7}}{3}$$

$$\dfrac{\sqrt{7} - 4}{3}$$

$$\dfrac{-4 + \sqrt{7}}{3}$$

$$-\left(\dfrac{4 + \sqrt{7}}{3}\right)$$

Solution

Question 4

$${\cos ^2}\left( {\dfrac{A}{2}} \right)$$

$$\sqrt {\sin \left( {\dfrac{A}{2}} \right)} $$

$$\sqrt {\cos \left( {\dfrac{A}{2}} \right)} $$

$${\sin ^2}\left( {\dfrac{A}{2}} \right)$$

Solution

Question 5

$$0$$

$$\dfrac { 1 }{ 2 } $$

$$1$$

$$2$$

Solution

Question 6

$$20\ m$$

$$30\ m$$

$$25\ m$$

$$35\ m$$

Solution

Question 7

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 8

$$\pi$$

$$\dfrac{\pi}{2}$$

$$\dfrac{3\pi}{2}$$

None of these

Solution

Question 9

$$\cos (A+B)$$

$$\tan (A-B)$$

$$\cot (A+B)$$

$$\tan (A+B)$$

Solution

Question 10

$$\dfrac{a^2b^2}{a^2+b^2}$$

$$\dfrac{ab}{a^2+b^2}$$

$$\dfrac{a^2b^2}{a+b}$$

None of these

Solution

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