Maths-

General

Easy

Question

# IfÂ Â andÂ Â thenÂ

- 1
- 3
- 4
- 5

Hint:

### In the question we will simplify the given equations using some trigonometric identities.

## The correct answer is: 4

on multiplying equation 1 and 2,

### Related Questions to study

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### The maximum value ofÂ Â under the restrictionsÂ Â andÂ Â is

Given,Â

### The maximum value ofÂ Â under the restrictionsÂ Â andÂ Â is

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Given,Â

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### LetÂ Â be two sets. Then

Given two setsÂ

### LetÂ Â be two sets. Then

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Given two setsÂ

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### IfÂ Â thenÂ Â is

### IfÂ Â thenÂ Â is

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physics-

### A long cylindrical tube carries a highly polished piston and has a side opening. A tuning fork of frequency n is sounded at the sound heard by the listener charges if the piston is moves in or out. At a particular position of the piston is moved through a distance of 9 cm, the intensity of sound becomes minimum, if the speed of sound is 360 m/s, the value of n is

When piston moves a distance, path difference change by 2 xs.

the path difference between maxima and consecutive minima=

âˆ´

Or

Î»=4x=4Ã—9 cm=36cm=0.36m

the path difference between maxima and consecutive minima=

âˆ´

Or

Î»=4x=4Ã—9 cm=36cm=0.36m

### A long cylindrical tube carries a highly polished piston and has a side opening. A tuning fork of frequency n is sounded at the sound heard by the listener charges if the piston is moves in or out. At a particular position of the piston is moved through a distance of 9 cm, the intensity of sound becomes minimum, if the speed of sound is 360 m/s, the value of n is

physics-General

When piston moves a distance, path difference change by 2 xs.

the path difference between maxima and consecutive minima=

âˆ´

Or

Î»=4x=4Ã—9 cm=36cm=0.36m

the path difference between maxima and consecutive minima=

âˆ´

Or

Î»=4x=4Ã—9 cm=36cm=0.36m

Maths-

The value ofÂ

The value ofÂ

Maths-General

Maths-

For any real , the maximum value ofÂ Â is

As cos is a decreasing function and sin is an increasing function so the functionÂ Â is maximum whenÂ =.

So, the maximum value ofÂ Â =Â

So, the maximum value ofÂ Â =Â

For any real , the maximum value ofÂ Â is

Maths-General

As cos is a decreasing function and sin is an increasing function so the functionÂ Â is maximum whenÂ =.

So, the maximum value ofÂ Â =Â

So, the maximum value ofÂ Â =Â

chemistry-

### Identify A and B

With H

_{2}/ pd / CaCO_{3}- is addition of â€˜Hâ€™ know is oxidizing agent### Identify A and B

chemistry-General

With H

_{2}/ pd / CaCO_{3}- is addition of â€˜Hâ€™ know is oxidizing agentMaths-

### IfÂ Â then sin (A-B) =

GIVEN-

$âˆš2ÂcosÂAÂ=ÂcosÂBÂ+Âcos_{3Â}B..... equation 1$

$cosÂAÂcosÂBÂ+Âcos_{3Â}âˆš2)$

# âˆš2Â sinÂ AÂ =Â sinÂ B -Â sin

âˆ´Â Â Â Â sinÂ AÂ = 1/âˆš2 (sinÂ B -Â sin

TO FIND-

Sin (A - B)

SOLUTION-

We know that-

Sin (A - B) = (Sin A * Cos B) - (Cos A * Sin B)

âˆ´Â Sin (A - B) = [1/âˆš2 (sinÂ B -Â sin

âˆ´Â Sin (A - B) =Â [1/âˆš2 * (sin B Cos B - sin

âˆ´Â Sin (A - B) =Â 1/âˆš2 sin B Cos B - 1/âˆš2 sin

âˆ´Â Sin (A - B) =Â 1/âˆš2 sin B Cos B - 1/âˆš2Â sin

âˆ´Â Sin (A - B) =Â 1/âˆš2 sin B Cos BÂ - 1/âˆš2Â sin

âˆ´Â Sin (A - B) = -1/âˆš2Â sin

âˆ´Â Sin (A - B) = -1/âˆš2Â sinÂ B Cos B (sin

Â Â Â Â Â Â Â Â Â Â =Â -1/2âˆš2Â sin2B

Now squaring and adding equation 1 and 2 we get

$âˆš2ÂcosÂAÂ=ÂcosÂBÂ+Âcos_{3Â}B..... equation 1$

$cosÂAÂcosÂBÂ+Âcos_{3Â}âˆš2)$

# âˆš2Â sinÂ AÂ =Â sinÂ B -Â sin^{3}Â B..... equation 2

âˆ´Â Â Â Â sinÂ AÂ = 1/âˆš2 (sinÂ B -Â sin^{3}Â B) ........ (Dividing both sides byÂ âˆš2)TO FIND-

Sin (A - B)

SOLUTION-

We know that-

Sin (A - B) = (Sin A * Cos B) - (Cos A * Sin B)

âˆ´Â Sin (A - B) = [1/âˆš2 (sinÂ B -Â sin

^{3}Â B) * cos B] - [1/âˆš2 (cosÂ BÂ +Â cos^{3}Â B) * sin B] ...... (From Equations i & ii)âˆ´Â Sin (A - B) =Â [1/âˆš2 * (sin B Cos B - sin

^{3}B Cos B)] -Â [1/âˆš2 * (sin BÂ cosÂ BÂ +Â cos^{3}Â B * sin B)]âˆ´Â Sin (A - B) =Â 1/âˆš2 sin B Cos B - 1/âˆš2 sin

^{3}Â B Cos B - (1/âˆš2 sin BÂ cosÂ BÂ +Â 1/âˆš2Â cos^{3}Â B * sin B)âˆ´Â Sin (A - B) =Â 1/âˆš2 sin B Cos B - 1/âˆš2Â sin

^{3}Â B Cos B - 1/âˆš2 sin BÂ cosÂ BÂ -Â 1/âˆš2Â cos^{3}Â B * sin Bâˆ´Â Sin (A - B) =Â 1/âˆš2 sin B Cos BÂ - 1/âˆš2Â sin

^{3}Â B Cos B - 1/âˆš2 sin BÂ cosÂ BÂ -Â 1/âˆš2Â cos^{3}Â B * sin Bâˆ´Â Sin (A - B) = -1/âˆš2Â sin

^{3}Â B Cos B -Â 1/âˆš2Â cos^{3}Â B * sin Bâˆ´Â Sin (A - B) = -1/âˆš2Â sinÂ B Cos B (sin

^{2}Â B +cos^{2}B)Â Â Â Â Â Â Â Â Â Â =Â -1/2âˆš2Â sin2B

Now squaring and adding equation 1 and 2 we get

### IfÂ Â then sin (A-B) =

Maths-General

GIVEN-

$âˆš2ÂcosÂAÂ=ÂcosÂBÂ+Âcos_{3Â}B..... equation 1$

$cosÂAÂcosÂBÂ+Âcos_{3Â}âˆš2)$

# âˆš2Â sinÂ AÂ =Â sinÂ B -Â sin

âˆ´Â Â Â Â sinÂ AÂ = 1/âˆš2 (sinÂ B -Â sin

TO FIND-

Sin (A - B)

SOLUTION-

We know that-

Sin (A - B) = (Sin A * Cos B) - (Cos A * Sin B)

âˆ´Â Sin (A - B) = [1/âˆš2 (sinÂ B -Â sin

âˆ´Â Sin (A - B) =Â [1/âˆš2 * (sin B Cos B - sin

âˆ´Â Sin (A - B) =Â 1/âˆš2 sin B Cos B - 1/âˆš2 sin

âˆ´Â Sin (A - B) =Â 1/âˆš2 sin B Cos B - 1/âˆš2Â sin

âˆ´Â Sin (A - B) =Â 1/âˆš2 sin B Cos BÂ - 1/âˆš2Â sin

âˆ´Â Sin (A - B) = -1/âˆš2Â sin

âˆ´Â Sin (A - B) = -1/âˆš2Â sinÂ B Cos B (sin

Â Â Â Â Â Â Â Â Â Â =Â -1/2âˆš2Â sin2B

Now squaring and adding equation 1 and 2 we get

$âˆš2ÂcosÂAÂ=ÂcosÂBÂ+Âcos_{3Â}B..... equation 1$

$cosÂAÂcosÂBÂ+Âcos_{3Â}âˆš2)$

# âˆš2Â sinÂ AÂ =Â sinÂ B -Â sin^{3}Â B..... equation 2

âˆ´Â Â Â Â sinÂ AÂ = 1/âˆš2 (sinÂ B -Â sin^{3}Â B) ........ (Dividing both sides byÂ âˆš2)TO FIND-

Sin (A - B)

SOLUTION-

We know that-

Sin (A - B) = (Sin A * Cos B) - (Cos A * Sin B)

âˆ´Â Sin (A - B) = [1/âˆš2 (sinÂ B -Â sin

^{3}Â B) * cos B] - [1/âˆš2 (cosÂ BÂ +Â cos^{3}Â B) * sin B] ...... (From Equations i & ii)âˆ´Â Sin (A - B) =Â [1/âˆš2 * (sin B Cos B - sin

^{3}B Cos B)] -Â [1/âˆš2 * (sin BÂ cosÂ BÂ +Â cos^{3}Â B * sin B)]âˆ´Â Sin (A - B) =Â 1/âˆš2 sin B Cos B - 1/âˆš2 sin

^{3}Â B Cos B - (1/âˆš2 sin BÂ cosÂ BÂ +Â 1/âˆš2Â cos^{3}Â B * sin B)âˆ´Â Sin (A - B) =Â 1/âˆš2 sin B Cos B - 1/âˆš2Â sin

^{3}Â B Cos B - 1/âˆš2 sin BÂ cosÂ BÂ -Â 1/âˆš2Â cos^{3}Â B * sin Bâˆ´Â Sin (A - B) =Â 1/âˆš2 sin B Cos BÂ - 1/âˆš2Â sin

^{3}Â B Cos B - 1/âˆš2 sin BÂ cosÂ BÂ -Â 1/âˆš2Â cos^{3}Â B * sin Bâˆ´Â Sin (A - B) = -1/âˆš2Â sin

^{3}Â B Cos B -Â 1/âˆš2Â cos^{3}Â B * sin Bâˆ´Â Sin (A - B) = -1/âˆš2Â sinÂ B Cos B (sin

^{2}Â B +cos^{2}B)Â Â Â Â Â Â Â Â Â Â =Â -1/2âˆš2Â sin2B

Now squaring and adding equation 1 and 2 we get

chemistry-

### Observer the following reactions and predict the nature of A and B

Conceptual

### Observer the following reactions and predict the nature of A and B

chemistry-General

Conceptual

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### on ozonolysis gives

### on ozonolysis gives

chemistry-General

chemistry-

### Which of the following reactions will yield 2, 2-dibromopropane?

### Which of the following reactions will yield 2, 2-dibromopropane?

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Maths-

### IfÂ Â thenÂ Â is equal to

### IfÂ Â thenÂ Â is equal to

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### IfÂ Â whereÂ Â being an acute angle then Â Â is equal to

### IfÂ Â whereÂ Â being an acute angle then Â Â is equal to

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maths-

### Let A,B,CÂ be three angles such that Â andÂ . Then, all possible values of P such that A,B,C are the angles of a triangle is

### Let A,B,CÂ be three angles such that Â andÂ . Then, all possible values of P such that A,B,C are the angles of a triangle is

maths-General

chemistry-

### is an example for

### is an example for

chemistry-General