Trigonometric Functions Test -4

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Trigonometric Functions Test -4

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Weekly Quiz Competition

Question 1
1 / -0

If $$ X\sin { \left( { 90 }^{ \circ }-\theta \right) \cot { \left( { 90 }^{ \circ }-\theta \right) } } =\cos { \left( { 9 }0^{ \circ }-\theta \right) } $$, then x =

A
0

B
1

C
-1

D
2

Solution

$$x\sin(90-\theta)\cot(90^o-\theta)=\cos(90-\theta)$$
$$\Rightarrow x\cos \theta\tan\theta =\sin\theta$$
$$\Rightarrow x=1$$.
Question 2
1 / -0

The value of $$\operatorname { \sec } ^ { 2 } A \operatorname { \tan } ^ { 2 } B - \tan ^ { 2 } A \operatorname { \sec } ^ { 2 } B $$ is

A
$$\operatorname { \tan } ^ { 2 } B - \tan ^ { 2 } A$$

B
$$\operatorname { \sec } ^ { 2 } B + \operatorname { \sec } ^ { 2 } A$$

C
$$\tan ^ { 2 } B - \operatorname { \sec } ^ { 2 } A$$

D
$$\operatorname { \sec } ^ { 2 } B - \tan ^ { 2 } A$$

Solution

$$\left( 1+{ \tan }^{ 2 }A \right) { \tan }^{ 2 }B-{ \tan }^{ 2 }A\left( 1+{ \tan }^{ 2 }B \right) $$
$$={ \tan }^{ 2 }B+{ \tan }^{ 2 }A{ \tan }^{ 2 }B-{ \tan }^{ 2 }A-{ \tan }^{ 2 }A{ \tan }^{ 2 }B$$
$$={ tan }^{ 2 }B-{ tan }^{ 2 }A\quad \left[ A \right] $$
Question 3
1 / -0

$$\dfrac{\cos(90-A)\sin(90-A)}{\tan (90-A)}$$=

A
$$\sin^{2}A$$

B
$$\cos^{2}A$$

C
$$\sin A$$

D
$$1$$

Solution

$$ = \dfrac{cos\,(90-A)\,sin\,(90-A)}{tan(90-A)}$$
$$ = \dfrac{sin\,A \times cos\,A}{\dfrac{cos\,A}{sin\,A}} = \dfrac{sin^{2}A\times cos\,A}{cos\,A}$$
$$ = sin^{2}A$$
Question 4
1 / -0

Which of the following is the principal value of $$\cos { ^{ -1 }\left( \dfrac { -1 }{ 2 } \right) } $$?

A
$$\dfrac { \pi }{ 3 } $$

B
$$\dfrac { \pi }{ 6 } $$

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

D
$$\dfrac { 3\pi }{ 3 } $$

Solution

we know $$ cos^{-1}(-x) = \pi -cos^{-1}x.$$

$$ cos^{-1}(-\dfrac{1}{2})\\ = \pi -cos^{-1}(\dfrac{1}{2})$$
$$ = \pi -cos^{-1}(cos\dfrac{\pi }{3}).$$
$$ = \pi -\dfrac{\pi }{3}.$$
$$ = \dfrac{2\pi }{3} $$
$$ \therefore $$ principal value of $$ cos^{-1}(-\dfrac{1}{2}) $$ is $$ \dfrac{2\pi }{3}.$$
Question 5
1 / -0

If $$cos x=b$$, then for what b do the roots of the equation form an AP ?

A
$$-1$$

B
$$\cfrac { 1 }{ 2 } $$

C
$$\cfrac { \sqrt { 3 } }{ 2 } $$

D
None of these

Solution

$$ \lim_{x\rightarrow 0} \frac{log(1+x^{3})}{sin^{3}x} = \lim_{x\rightarrow 0} \frac{log(1+x^{3})}{x^{3}} \frac{x^{3}}{sin^{3}x}$$
$$ = \lim_{x\rightarrow 0} \frac{d}{dx} \frac{log(1+x^{3})}{\frac{d}{dx}}\times 1 $$
$$ = \lim_{x\rightarrow 0} \frac{\frac{1}{1+x^{3}}(3x)^{2}}{3x^{2}}\times 1$$
$$ =1\times 1 = 1 $$
Question 6
1 / -0

What is the value of $$\cfrac{\tan{A}-\sin{A}}{\sin^3{A}}$$?

A
$$\cfrac{\sec{A}}{1-\cos{A}}$$

B
$$\cfrac{\sec{A}}{1+\cos^2{A}}$$

C
$$\cfrac{\sec{A}}{1+\cos{A}}$$

D
None of these

Solution

$$\dfrac{\tan A-\sin A}{\sin^3 A}=\dfrac{\dfrac{\sin A}{\cos A}-\sin A}{\sin^3 A}=\dfrac{\dfrac{1-\cos A}{\cos A}}{\sin^2 A}=\dfrac{\text{sec} A(1-\cos A)}{1-\cos ^2 A}$$
$$=\dfrac{\text{sec}A(1-\cos A)}{(1-\cos A)(1+\cos A)}$$
$$=\dfrac{\text{sec}A}{1+\cos A}$$
Question 7
1 / -0

The range of the function $$f(x)=\left (\cos ^2x+4\sec ^2x\right )$$ is

A
$$[4,\infty )$$

B
$$[0,\infty )$$

C
$$[5,\infty )$$

D
$$(0,\infty )$$

Solution

As we know that
$$AM\ge GM$$
$$\implies \dfrac{\cos^2 x+\text{sec}^2 x}{2}\ge \sqrt{(\cos^2 x)(4\text{sec}^2 x)}$$
$$\implies \dfrac{f(x)}{2}\ge \sqrt{4}$$
$$\implies f(x)\ge 4$$
So the range of $$f(x)$$ is $$[4,\infty)$$
Question 8
1 / -0

The principle solution of equation $$\cot x = - \sqrt 3 $$ is

A
$$\dfrac{\pi }{3}$$

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

C
$$\dfrac{\pi }{6}$$

D
$$\dfrac{5\pi }{6}$$

Solution

Given $$\cot x=-\sqrt{3}$$

$$\tan x=\dfrac{1}{\cot x}$$

$$\tan x=\dfrac{1}{-\sqrt{3}}$$

$$\tan x=\dfrac{-1}{\sqrt{3}}$$

We know that $$\tan 30^o=1/\sqrt{3}$$

We find the value of x where $$\tan$$ is negative

$$\tan$$ is negative in $$2$$ and $$4^{th}$$ quadrant.

Value in $$2^{nd}$$ Quadrant $$=180^o-30^o=150^o$$

Value in $$4^{th}$$ Quadrant $$=360^o-30^o=330^o$$

So, principal solution are
$$x=150^o$$ and $$x=330^o$$

$$x=150 \times \pi/180$$ and $$x=330\times \pi/180$$

$$x=5\pi/6$$ and $$x=11\pi/6$$

To find general solution
Let $$\tan x=\tan y$$ ………..$$(1)$$
$$\tan x=-1/\sqrt{3}$$ ………..$$(2)$$

From $$(1)$$ & $$(2)$$

$$\tan y=-1/\sqrt{3}$$
$$\tan y=\tan 5\pi/6$$
$$y=5\pi/6$$

Since $$\tan x=\tan y$$
General solution is $$x=n\pi +y$$ where $$n\in z$$.
Hence $$x=n\pi +5\pi /6$$ where $$n\in z$$.
Question 9
1 / -0

$$\displaystyle 1^c =?$$

A
$$\displaystyle 56^027'22''$$

B
$$\displaystyle 57^016'22''$$

C
$$\displaystyle 55^018'32''$$

D
$$\displaystyle 57^026'32''$$

Solution

$$\displaystyle \pi^c = 180^0 \\ \Rightarrow 1^c = \left(\dfrac{180}{\pi}\right)^0 = \left(180 \times \dfrac{7}{22}\right)^0 = \left(\dfrac{630}{11}\right)^0 = 57^016'22''$$ (app.).
Question 10
1 / -0

Let $$n$$ be a positive integer such that
$$\displaystyle \sin(\frac{\pi}{2^{n}})+\cos(\frac{\pi}{2^{n}})=\frac{\sqrt{n}}{2}$$

A
$$<4$$

B
$$n>8$$

C
$$4\leq n<8$$

D
$$6\leq n\leq 8$$

Solution

$$\left(\sin(\cfrac{\pi}{2^n})+\cos(\cfrac{\pi}{2^n})\right)^2=\dfrac{n}{2}$$

$$1+2\sin \cfrac{\pi}{2^n}\cos\cfrac{\pi}{2^n}=\dfrac{n}{4}$$......$$[\cos^{2}\theta +\sin^{2}\theta =1]$$

$$\displaystyle \Rightarrow 1+\sin \cfrac{\pi }{2^{n-1}}=\cfrac{n}{4}$$

$$\displaystyle \sin(\cfrac{\pi }{2^{n-1}})=\cfrac{n-4}{4}$$
AS $$n=+ve. \neq 1$$ and $$\sin\theta\le 1$$
$$\displaystyle 0 < \cfrac{n-4}{4} \leq 1$$
$$\therefore 4 < n \leq 8$$

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Trigonometric Functions Test -4

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Trigonometric Functions Test -4

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Question 1

0

1

-1

2

Solution

Question 2

$$\operatorname { \tan } ^ { 2 } B - \tan ^ { 2 } A$$

$$\operatorname { \sec } ^ { 2 } B + \operatorname { \sec } ^ { 2 } A$$

$$\tan ^ { 2 } B - \operatorname { \sec } ^ { 2 } A$$

$$\operatorname { \sec } ^ { 2 } B - \tan ^ { 2 } A$$

Solution

Question 3

$$\sin^{2}A$$

$$\cos^{2}A$$

$$\sin A$$

$$1$$

Solution

Question 4

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

$$\dfrac { \pi }{ 6 } $$

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

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

Solution

Question 5

$$-1$$

$$\cfrac { 1 }{ 2 } $$

$$\cfrac { \sqrt { 3 } }{ 2 } $$

None of these

Solution

Question 6

$$\cfrac{\sec{A}}{1-\cos{A}}$$

$$\cfrac{\sec{A}}{1+\cos^2{A}}$$

$$\cfrac{\sec{A}}{1+\cos{A}}$$

None of these

Solution

Question 7

$$[4,\infty )$$

$$[0,\infty )$$

$$[5,\infty )$$

$$(0,\infty )$$

Solution

Question 8

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

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

$$\dfrac{\pi }{6}$$

$$\dfrac{5\pi }{6}$$

Solution

Question 9

$$\displaystyle 56^027'22''$$

$$\displaystyle 57^016'22''$$

$$\displaystyle 55^018'32''$$

$$\displaystyle 57^026'32''$$

Solution

Question 10

$$<4$$

$$n>8$$

$$4\leq n<8$$

$$6\leq n\leq 8$$

Solution

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