Trigonometric Functions Test 21

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Trigonometric Functions Test 21

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Weekly Quiz Competition

Question 1
1 / -0

Genral solution of the equation $$\theta -cosec\theta =\dfrac { 4 }{ 3 }$$ is :

A
$$\frac { 1 }{ 2 } \left( \left[ n\pi +{ \left( -1 \right) }^{ n }{ sin }^{ -1 }\left( 3/4 \right) \right] \right) n\epsilon Z$$

B
$$n\pi +{ \left( -1 \right) }^{ n }{ sin }^{ -1 }\left( 3/4 \right) n\epsilon Z$$

C
$$\frac { n\pi }{ 2 } +{ \left( -1 \right) }^{ n }{ sin }^{ -1 }\left( 3/4 \right) n\epsilon Z$$

D
none of these

Solution
Question 2
1 / -0

If $$\sin \alpha = a \sin \beta$$ and $$\cos \alpha = b \cos \beta$$ then $$\tan \alpha =$$

A
$$\pm \sqrt { \left( \frac { a ^ { 2 } b ^ { 2 } - a ^ { 2 } } { b ^ { 2 } - a ^ { 2 } b ^ { 2 } } \right) }$$

B
$$\pm \sqrt { \left( \frac { a ^ { 2 } - a ^ { 2 } b ^ { 2 } } { b ^ { 2 } +a ^ { 2 } b ^ { 2 } } \right) }$$

C
$$\pm \sqrt { \left( \frac { 1 - a ^ { 2 } } { 1 - b ^ { 2 } } \right) }$$

D
$$\pm \sqrt { \left( \frac { 1 + a ^ { 2 } } { 1 + b ^ { 2 } } \right. ) } $$

Solution
Question 3
1 / -0

If $$\tan \theta +\sec \theta =\sqrt {3}, 0< \theta < \pi$$ then $$\theta $$ is equal to

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

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

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

D
$$\dfrac {5\pi}{6}$$
Question 4
1 / -0

If $$\cosh x=\sec \theta$$, then $$\coth^{2}(x/2)=$$

A
$$\tan^{2}(\theta/2)$$

B
$$\tan^{2}\theta$$

C
$$\cot^{2}(\theta/2)$$

D
$$\cot^{2}\theta$$

Solution

Let $$t=\tanh^2 (x/2)$$

Given $$\cosh x=\text{sec}\theta$$

$$\implies \dfrac{1-\tanh^2 x}{1+\tanh^2 x}=\dfrac{1}{\cos \theta}$$

$$\implies \dfrac{1-t^2}{1+t^2}=\dfrac{1}{\cos \theta}$$

$$\implies (1-t^2)\cos \theta=1+t^2$$

$$\implies t^2(1+\cos \theta)=1-\cos \theta$$

$$\implies t^2=\dfrac{1-\cos \theta}{1+\cos \theta}=\dfrac{2\sin^2 \theta/2}{2\cos^2 \theta/2}=\tan^2 \theta/2$$

$$\coth^2(x/2)=\dfrac{1}{t^2}=\cot^2 \dfrac{\theta}{2}$$
Question 5
1 / -0

Choose the correct answer the following:
2) The principle solution of $$\sec{x}+2=0$$ is____

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

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

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

D
$$-\dfrac{2\pi}{3}$$
Question 6
1 / -0

$$\tan\left(\dfrac{\pi}{4}+x\right)+\tan\left(\dfrac{\pi}{4}-x\right)=2$$, then $$x=$$

A
$$n\pi\pm\dfrac{\pi}{4}$$

B
$$2n\pi\pm\dfrac{\pi}{4}$$

C
$$n\pi$$

D
$$(2n+1)\dfrac{\pi}{4}$$

Solution

Given $$\tan \bigg(\dfrac{\pi}{4}+x\bigg)+\tan \bigg(\dfrac{\pi}{4}-x\bigg)=2$$

$$\implies \dfrac{\tan \dfrac{\pi}{4}+\tan x}{1-\tan \dfrac{\pi}{4}\tan x}+\dfrac{\tan \dfrac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4}\tan x}=2$$

$$\implies \dfrac{1+\tan x}{1-\tan x}+\dfrac{1-\tan x}{1+\tan x}=2$$

$$\implies \dfrac{(1+\tan x)^2+(1-\tan x)^2}{(1-\tan x)(1+\tan x)}=2$$

$$\implies \dfrac{2(1+\tan^2 x)}{1-\tan^2 x}=2$$ $$(\because (a+b)^2+(a-b)^2=2(a^2+b^2)$$ and $$(a+b)(a-b)=a^2-b^2)$$

$$\implies \dfrac{1+\tan^2 x}{1-\tan^2 x}=1$$

$$\implies 1-\tan^2 x=1+\tan^2 x$$

$$\implies \tan^2 x=0=\tan^2 0$$

$$\implies x=n\pi$$
Question 7
1 / -0

One of principal solution of $$\sqrt 3 \sec x = - 2$$ is equal to

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

B
$$5\dfrac{\pi }{3}$$

C
$$5\dfrac{\pi }{4}$$

D
none

Solution
Question 8
1 / -0

If $$\alpha +\beta =\dfrac { 5\pi }{ 4 } $$, then value of $$\dfrac { \cot { \alpha } \cdot\cot { \beta } }{ \left( 1+\cot { \alpha } \right) \left( 1+\cot { \beta } \right) }$$ is (wherever defined)-

A
$$1$$

B
$$\frac{3}{2}$$

C
$$\frac{1}{2}$$

D
$$\frac{5}{4}$$

Solution

Given $$\alpha+\beta=\dfrac{5\pi}{4}$$
$$\implies \cot (\alpha+\beta)=\cot \dfrac{5\pi}{4}$$
$$\implies \dfrac{\cot \alpha\cot \beta-1}{\cot \alpha+\cot \beta}=1$$
$$\implies \cot \alpha\cot \beta-1=\cot \alpha+\cot \beta$$
$$\implies \cot \alpha\cot \beta=1+\cot \alpha+\cot \beta$$
$$\implies 2\cot \alpha\cot \beta=1+\cot \alpha+\cot \beta+\cot \alpha\cot \beta$$
$$\implies 2\cot \alpha\cot \beta=(1+\cot \alpha)(1+\cot \beta)$$
$$\implies \dfrac{\cot \alpha\cot \beta}{(1+\cot \alpha)(1+\cot \beta)}=\dfrac{1}{2}$$
Question 9
1 / -0

$$cos[tan^{-1}(sin(cot^{-1}x))]$$ is equal to-

A
$$\sqrt{\frac{x^{2}+2}{x^{2}+3}}$$

B
$$\sqrt{\frac{x^{2}+2}{x^{2}+1}}$$

C
$$\sqrt{\frac{x^{2}+1}{x^{2}+2}}$$

D
$$\sqrt{\frac{x^{2}-1}{x^{2}+1}}$$

Solution

Let $$\text{cot}^{-1} x=y\implies x=\cot y\implies \sin y=\dfrac{1}{\sqrt{1+x^2}}$$
Let $$\text{tan}^{-1} \dfrac{1}{\sqrt{x^2+1}}=z\implies \tan z=\dfrac{1}{\sqrt{1+x^2}} $$
So $$\cos z=\dfrac{1}{\sqrt{1+(\frac{1}{\sqrt{x^2+1}})^2}}=\dfrac{1}{\sqrt{1+\frac{1}{x^2+1}}}=\dfrac{1}{\sqrt{\frac{x^2+2}{x^2+1}}}=\sqrt{\dfrac{x^2+1}{x^2+2}}$$
So $$\cos [\text{tan}^{-1}\{\sin (\text{cot}^{-1} x)\}]=\cos [\text{tan}^{-1}\{\sin y\}]=\cos [\text{tan}^{-1}\frac{1}{\sqrt{1+x^2}}]=\cos z=\sqrt{\dfrac{x^2+2}{x^2+1}}$$
Question 10
1 / -0

If $$sin A=\dfrac{3}{5},\tan B=\dfrac{1}{2}$$ and $$\dfrac{\pi}{2}<A<\pi<B<\dfrac{3\pi}{2}$$, then the value of $$8\tan A-\sqrt{5}sec B=$$

A
$$5/2$$

B
$$-5/2$$

C
$$-7/2$$

D
$$7/2$$

Solution

Given $$\dfrac{\pi}{2}<A<\pi,\implies \tan A<0$$
$$\sin A=\dfrac{3}{5}\implies \tan A=-\dfrac{3}{\sqrt{5^2-3^2}}=-\dfrac{3}{4}$$
$$\pi<B<\dfrac{3\pi}{2}\implies \text{sec} B<0$$
$$\tan B=\dfrac{1}{2}\implies \text{sec} B=-\sqrt{1+\dfrac{1}{4}}=-\dfrac{\sqrt{5}}{2}$$
So $$8\tan A-\sqrt{5}\text{sec}B=8\times \bigg(-\dfrac{3}{4}\bigg)-\sqrt{5}\times \bigg(-\dfrac{\sqrt{5}}{2}\bigg)=-6+\dfrac{5}{2}=-\dfrac{7}{2}$$

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Trigonometric Functions Test 21

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Trigonometric Functions Test 21

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SHARING IS CARING

Question 1

$$\frac { 1 }{ 2 } \left( \left[ n\pi +{ \left( -1 \right) }^{ n }{ sin }^{ -1 }\left( 3/4 \right) \right] \right) n\epsilon Z$$

$$n\pi +{ \left( -1 \right) }^{ n }{ sin }^{ -1 }\left( 3/4 \right) n\epsilon Z$$

$$\frac { n\pi }{ 2 } +{ \left( -1 \right) }^{ n }{ sin }^{ -1 }\left( 3/4 \right) n\epsilon Z$$

none of these

Solution

Question 2

$$\pm \sqrt { \left( \frac { a ^ { 2 } b ^ { 2 } - a ^ { 2 } } { b ^ { 2 } - a ^ { 2 } b ^ { 2 } } \right) }$$

$$\pm \sqrt { \left( \frac { a ^ { 2 } - a ^ { 2 } b ^ { 2 } } { b ^ { 2 } +a ^ { 2 } b ^ { 2 } } \right) }$$

$$\pm \sqrt { \left( \frac { 1 - a ^ { 2 } } { 1 - b ^ { 2 } } \right) }$$

$$\pm \sqrt { \left( \frac { 1 + a ^ { 2 } } { 1 + b ^ { 2 } } \right. ) } $$

Solution

Question 3

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

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

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

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

Question 4

$$\tan^{2}(\theta/2)$$

$$\tan^{2}\theta$$

$$\cot^{2}(\theta/2)$$

$$\cot^{2}\theta$$

Solution

Question 5

$$\dfrac{4\pi}{3}$$

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

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

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

Question 6

$$n\pi\pm\dfrac{\pi}{4}$$

$$2n\pi\pm\dfrac{\pi}{4}$$

$$n\pi$$

$$(2n+1)\dfrac{\pi}{4}$$

Solution

Question 7

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

$$5\dfrac{\pi }{3}$$

$$5\dfrac{\pi }{4}$$

none

Solution

Question 8

$$1$$

$$\frac{3}{2}$$

$$\frac{1}{2}$$

$$\frac{5}{4}$$

Solution

Question 9

$$\sqrt{\frac{x^{2}+2}{x^{2}+3}}$$

$$\sqrt{\frac{x^{2}+2}{x^{2}+1}}$$

$$\sqrt{\frac{x^{2}+1}{x^{2}+2}}$$

$$\sqrt{\frac{x^{2}-1}{x^{2}+1}}$$

Solution

Question 10

$$5/2$$

$$-5/2$$

$$-7/2$$

$$7/2$$

Solution

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