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Maths-

General

Easy

Question

Let $P equals left curly bracket theta colon sin space theta minus cos space theta equals square root of 2 cos space theta right curly bracket text and end text Q equals left curly bracket theta colon sin space theta plus cos space theta equals square root of 2 sin space theta right curly bracket$ be two sets. Then

$P \subset Q and P \neq \emptyset$
$Q \subset\subset P$
$P \subset Q$
P=Q

Hint:

In the two sets $theta$ is derived from the given trigonometric equations in the respective sets. So, we will first simplify the equation then will get the relation between them.

The correct answer is: P=Q

Given two sets $P equals left curly bracket theta colon sin space theta minus cos space theta equals square root of 2 cos space theta right curly bracket text and end text Q equals left curly bracket theta colon sin space theta plus cos space theta equals square root of 2 sin space theta right curly bracket$
$F r o m space P comma space sin theta space minus cos theta equals square root of 2 cos theta space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space F r o m space Q comma space sin theta plus cos theta equals square root of 2 sin theta rightwards double arrow sin theta equals square root of 2 cos theta plus cos theta space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow cos theta equals square root of 2 sin theta minus sin theta rightwards double arrow sin theta equals cos theta open parentheses square root of 2 plus 1 close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow cos theta equals sin theta open parentheses square root of 2 minus 1 close parentheses rightwards double arrow fraction numerator sin theta over denominator cos theta end fraction equals square root of 2 plus 1 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow fraction numerator sin theta over denominator cos theta end fraction equals fraction numerator 1 over denominator square root of 2 minus 1 end fraction cross times fraction numerator square root of 2 plus 1 over denominator square root of 2 plus 1 end fraction rightwards double arrow tan theta equals square root of 2 plus 1 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow tan theta equals fraction numerator square root of 2 plus 1 over denominator 2 minus 1 end fraction equals square root of 2 plus 1 H e n c e comma space P equals Q$

Related Questions to study

If $0 less than alpha less than pi over 6$ then $alpha left parenthesis cosec space alpha right parenthesis$ is

g i v e n comma space 0 less than alpha less than straight pi over 6..... space e q u a t i o n space 1 rightwards double arrow sin 0 less than sin alpha less than sin straight pi over 6 rightwards double arrow 0 less than sin alpha less than 1 half....... space e q u a t i o n space 2 d i v i d i n g space e q u a t i o n space 1 space a n d space 2 rightwards double arrow 0 less than fraction numerator alpha over denominator sin alpha end fraction less than straight pi over 3 rightwards double arrow 0 less than alpha. cos e c alpha less than straight pi over 3 S o comma space alpha space cos e c alpha space less than straight pi over 3 space

If $0 less than alpha less than pi over 6$ then $alpha left parenthesis cosec space alpha right parenthesis$ is

g i v e n comma space 0 less than alpha less than straight pi over 6..... space e q u a t i o n space 1 rightwards double arrow sin 0 less than sin alpha less than sin straight pi over 6 rightwards double arrow 0 less than sin alpha less than 1 half....... space e q u a t i o n space 2 d i v i d i n g space e q u a t i o n space 1 space a n d space 2 rightwards double arrow 0 less than fraction numerator alpha over denominator sin alpha end fraction less than straight pi over 3 rightwards double arrow 0 less than alpha. cos e c alpha less than straight pi over 3 S o comma space alpha space cos e c alpha space less than straight pi over 3 space

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

x subscript 1 end subscript

, path difference change by 2 xs.

therefore

the path difference between maxima and consecutive minima=

lambda divided by 2

2 x equals lambda divided by 2

Or
λ=4x=4×9 cm=36cm=0.36m

n equals fraction numerator v over denominator lambda end fraction equals fraction numerator 360 over denominator 0.36 end fraction equals 1000 blank H z

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

x subscript 1 end subscript

, path difference change by 2 xs.

therefore

the path difference between maxima and consecutive minima=

lambda divided by 2

2 x equals lambda divided by 2

Or
λ=4x=4×9 cm=36cm=0.36m

n equals fraction numerator v over denominator lambda end fraction equals fraction numerator 360 over denominator 0.36 end fraction equals 1000 blank H z

The value of $cot space fraction numerator 7 pi over denominator 16 end fraction plus 2 cot space fraction numerator 3 pi over denominator 8 end fraction plus cot space fraction numerator 15 pi over denominator 16 end fraction$

cot space fraction numerator 7 pi over denominator 16 end fraction plus 2 cot space fraction numerator 3 pi over denominator 8 end fraction plus cot space fraction numerator 15 pi over denominator 16 end fraction equals tan open parentheses straight pi over 2 minus fraction numerator 7 straight pi over denominator 16 end fraction close parentheses plus 2 tan open parentheses straight pi over 2 minus fraction numerator 3 straight pi over denominator 2 end fraction close parentheses plus c o t open parentheses straight pi minus straight pi over 16 close parentheses space space space space open square brackets tan straight pi over 2 minus x equals c o t space x comma space c o t space straight pi minus straight x equals negative cotx close square brackets equals tan straight pi over 16 plus 2 tan straight pi over 8 minus c o t straight pi over 16 equals open parentheses tan straight pi over 16 minus c o t straight pi over 16 close parentheses plus 2 tan straight pi over 8 equals open parentheses fraction numerator sin begin display style straight pi over 16 end style over denominator cos begin display style straight pi over 16 end style end fraction minus fraction numerator cos begin display style straight pi over 16 end style over denominator sin begin display style straight pi over 16 end style end fraction close parentheses plus 2 fraction numerator sin begin display style straight pi over 8 end style over denominator cos begin display style straight pi over 8 end style end fraction equals open parentheses fraction numerator sin squared begin display style straight pi over 16 end style minus cos squared begin display style straight pi over 16 end style over denominator cos begin display style straight pi over 16 end style sin begin display style straight pi over 16 end style end fraction close parentheses plus 2 fraction numerator sin begin display style straight pi over 8 end style over denominator cos begin display style straight pi over 8 end style end fraction equals 2 fraction numerator negative cos begin display style straight pi over 8 end style over denominator sin begin display style straight pi over 8 end style end fraction plus 2 fraction numerator sin begin display style straight pi over 8 end style over denominator cos begin display style straight pi over 8 end style end fraction space space space space space space space space space space space space space space space space space space space space space space space space open square brackets cos squared x minus sin squared x equals c o x 2 x comma space 2 sin x space cos x space equals sin 2 x close square brackets equals negative 2 open parentheses fraction numerator cos squared begin display style straight pi over 8 end style minus sin squared begin display style straight pi over 8 end style over denominator sin begin display style straight pi over 8 end style cos begin display style straight pi over 8 end style end fraction close parentheses equals negative 2 open parentheses fraction numerator 2 cos begin display style straight pi over 4 end style over denominator sin begin display style straight pi over 4 end style end fraction close parentheses space space space space space space space space space space space space space space space space space space space open square brackets cos squared x minus sin squared x equals c o x 2 x comma space 2 sin x space cos x space equals sin 2 x close square brackets equals negative 4 c o t straight pi over 4 equals negative 4 cross times 1 space space space space space space space space space space space space space space open square brackets c o t straight pi over 4 equals 1 close square brackets equals negative 4

The value of $cot space fraction numerator 7 pi over denominator 16 end fraction plus 2 cot space fraction numerator 3 pi over denominator 8 end fraction plus cot space fraction numerator 15 pi over denominator 16 end fraction$

cot space fraction numerator 7 pi over denominator 16 end fraction plus 2 cot space fraction numerator 3 pi over denominator 8 end fraction plus cot space fraction numerator 15 pi over denominator 16 end fraction equals tan open parentheses straight pi over 2 minus fraction numerator 7 straight pi over denominator 16 end fraction close parentheses plus 2 tan open parentheses straight pi over 2 minus fraction numerator 3 straight pi over denominator 2 end fraction close parentheses plus c o t open parentheses straight pi minus straight pi over 16 close parentheses space space space space open square brackets tan straight pi over 2 minus x equals c o t space x comma space c o t space straight pi minus straight x equals negative cotx close square brackets equals tan straight pi over 16 plus 2 tan straight pi over 8 minus c o t straight pi over 16 equals open parentheses tan straight pi over 16 minus c o t straight pi over 16 close parentheses plus 2 tan straight pi over 8 equals open parentheses fraction numerator sin begin display style straight pi over 16 end style over denominator cos begin display style straight pi over 16 end style end fraction minus fraction numerator cos begin display style straight pi over 16 end style over denominator sin begin display style straight pi over 16 end style end fraction close parentheses plus 2 fraction numerator sin begin display style straight pi over 8 end style over denominator cos begin display style straight pi over 8 end style end fraction equals open parentheses fraction numerator sin squared begin display style straight pi over 16 end style minus cos squared begin display style straight pi over 16 end style over denominator cos begin display style straight pi over 16 end style sin begin display style straight pi over 16 end style end fraction close parentheses plus 2 fraction numerator sin begin display style straight pi over 8 end style over denominator cos begin display style straight pi over 8 end style end fraction equals 2 fraction numerator negative cos begin display style straight pi over 8 end style over denominator sin begin display style straight pi over 8 end style end fraction plus 2 fraction numerator sin begin display style straight pi over 8 end style over denominator cos begin display style straight pi over 8 end style end fraction space space space space space space space space space space space space space space space space space space space space space space space space open square brackets cos squared x minus sin squared x equals c o x 2 x comma space 2 sin x space cos x space equals sin 2 x close square brackets equals negative 2 open parentheses fraction numerator cos squared begin display style straight pi over 8 end style minus sin squared begin display style straight pi over 8 end style over denominator sin begin display style straight pi over 8 end style cos begin display style straight pi over 8 end style end fraction close parentheses equals negative 2 open parentheses fraction numerator 2 cos begin display style straight pi over 4 end style over denominator sin begin display style straight pi over 4 end style end fraction close parentheses space space space space space space space space space space space space space space space space space space space open square brackets cos squared x minus sin squared x equals c o x 2 x comma space 2 sin x space cos x space equals sin 2 x close square brackets equals negative 4 c o t straight pi over 4 equals negative 4 cross times 1 space space space space space space space space space space space space space space open square brackets c o t straight pi over 4 equals 1 close square brackets equals negative 4

parallel

For any real $theta$ , the maximum value of $cos squared space left parenthesis cos space theta right parenthesis plus sin squared space left parenthesis sin space theta right parenthesis$ is

As cos is a decreasing function and sin is an increasing function so the function

cos squared space left parenthesis cos space theta right parenthesis plus sin squared space left parenthesis sin space theta right parenthesis

is maximum when

theta

90 degree

.
So, the maximum value of

cos squared space left parenthesis cos space theta right parenthesis plus sin squared space left parenthesis sin space theta right parenthesis

cos squared space left parenthesis cos space 90 degree right parenthesis plus sin squared space left parenthesis sin space 90 degree right parenthesis equals cos squared 0 plus sin squared 1 equals 1 plus sin squared 1

For any real $theta$ , the maximum value of $cos squared space left parenthesis cos space theta right parenthesis plus sin squared space left parenthesis sin space theta right parenthesis$ is

As cos is a decreasing function and sin is an increasing function so the function

cos squared space left parenthesis cos space theta right parenthesis plus sin squared space left parenthesis sin space theta right parenthesis

is maximum when

theta

90 degree

.
So, the maximum value of

cos squared space left parenthesis cos space theta right parenthesis plus sin squared space left parenthesis sin space theta right parenthesis

cos squared space left parenthesis cos space 90 degree right parenthesis plus sin squared space left parenthesis sin space 90 degree right parenthesis equals cos squared 0 plus sin squared 1 equals 1 plus sin squared 1

Identify A and B

With H₂ / pd / CaCO₃ - is addition of ‘H’ know is oxidizing agent

Identify A and B

chemistry-General

With H₂ / pd / CaCO₃ - is addition of ‘H’ know is oxidizing agent

If $square root of 2 cos space A equals cos space B plus cos cubed space B text and end text square root of 2 sin text end text A equals sin space B minus sin cubed space B$ then sin (A-B) =

cos A = cos B + cos^{3} B..... equation 1

∴ cos A = 1/ \sqrt2 (cos B + cos^{3} B) ........ (Dividing both sides by \sqrt2)

√2 sin A = sin B - sin³ B..... equation 2

∴ sin A = 1/√2 (sin B - sin³ B) ........ (Dividing both sides by √2)
TO FIND-
Sin (A - B)
SOLUTION-
We know that-
Sin (A - B) = (Sin A * Cos B) - (Cos A * Sin B)
∴ Sin (A - B) = [1/√2 (sin B - sin³ B) * cos B] - [1/√2 (cos B + cos³ B) * sin B] ...... (From Equations i & ii)
∴ Sin (A - B) = [1/√2 * (sin B Cos B - sin³ B Cos B)] - [1/√2 * (sin B cos B + cos³ B * sin B)]
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin³ B Cos B - (1/√2 sin B cos B + 1/√2 cos³ B * sin B)
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin³ B Cos B - 1/√2 sin B cos B - 1/√2 cos³ B * sin B
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin³ B Cos B - 1/√2 sin B cos B - 1/√2 cos³ B * sin B
∴ Sin (A - B) = -1/√2 sin³ B Cos B - 1/√2 cos³ B * sin B

∴ Sin (A - B) = -1/√2 sin B Cos B (sin² B +cos² B)
= -1/2√2 sin2B
Now squaring and adding equation 1 and 2 we get

2 cos squared A plus 2 sin squared A equals cos squared B plus cos to the power of 6 B plus 2 cos to the power of 4 B plus sin squared B plus sin to the power of 6 B minus 2 sin to the power of 4 B rightwards double arrow 2 open parentheses cos squared A plus sin squared A close parentheses equals cos squared B plus sin squared B plus open parentheses cos squared B plus sin squared B close parentheses cubed minus 3 open parentheses cos squared B sin squared B close parentheses open parentheses cos squared B plus sin squared B close parentheses plus 2 open parentheses cos to the power of 4 B minus sin to the power of 4 B close parentheses rightwards double arrow 2 equals 1 plus 1 minus 3 over 4 sin squared 2 B plus 2 cos 2 B space space space space open square brackets 4 cos squared B sin squared B equals sin squared 2 B comma cos to the power of 4 B minus sin to the power of 4 B equals cos 2 B space close square brackets space space space open square brackets cos squared B plus sin squared B equals 1 close square brackets rightwards double arrow 0 equals negative 3 sin squared 2 B plus 8 cos 2 B rightwards double arrow 0 equals negative 3 open parentheses 1 minus cos squared 2 B close parentheses plus 8 cos 2 B space space space space open square brackets cos squared 2 B plus sin squared 2 B equals 1 close square brackets rightwards double arrow 0 equals 3 cos squared 2 B minus 3 plus 8 cos 2 B rightwards double arrow 3 cos squared 2 B plus 8 cos 2 B minus 3 equals 0 rightwards double arrow 3 cos squared 2 B plus 9 cos 2 B minus cos 2 B minus 3 equals 0 rightwards double arrow 3 cos 2 B open parentheses cos 2 B plus 3 close parentheses minus 1 open parentheses cos 2 B plus 3 close parentheses equals 0 rightwards double arrow open parentheses 3 cos 2 B minus 1 close parentheses open parentheses cos 2 B plus 3 close parentheses equals 0 rightwards double arrow cos 2 B equals 1 third space o r space cos 2 B equals negative 3 open parentheses n o t space p o s s i b l e close parentheses N o w comma space w e space k n o w space t h a t space open parentheses cos squared 2 B plus sin squared 2 B close parentheses equals 1 S o comma space 1 over 9 plus sin squared 2 B equals 1 rightwards double arrow sin squared 2 B equals 1 minus 1 over 9 equals 8 over 9 sin 2 B equals plus-or-minus fraction numerator 2 square root of 2 over denominator 3 end fraction t h e r e f o r e comma space sin open parentheses A minus B close parentheses equals fraction numerator negative sin 2 B over denominator 2 square root of 2 end fraction equals plus-or-minus fraction numerator 2 square root of 2 over denominator 3 end fraction cross times fraction numerator 1 over denominator 2 square root of 2 end fraction equals plus-or-minus 1 third

If $square root of 2 cos space A equals cos space B plus cos cubed space B text and end text square root of 2 sin text end text A equals sin space B minus sin cubed space B$ then sin (A-B) =

cos A = cos B + cos^{3} B..... equation 1

∴ cos A = 1/ \sqrt2 (cos B + cos^{3} B) ........ (Dividing both sides by \sqrt2)

√2 sin A = sin B - sin³ B..... equation 2

∴ sin A = 1/√2 (sin B - sin³ B) ........ (Dividing both sides by √2)
TO FIND-
Sin (A - B)
SOLUTION-
We know that-
Sin (A - B) = (Sin A * Cos B) - (Cos A * Sin B)
∴ Sin (A - B) = [1/√2 (sin B - sin³ B) * cos B] - [1/√2 (cos B + cos³ B) * sin B] ...... (From Equations i & ii)
∴ Sin (A - B) = [1/√2 * (sin B Cos B - sin³ B Cos B)] - [1/√2 * (sin B cos B + cos³ B * sin B)]
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin³ B Cos B - (1/√2 sin B cos B + 1/√2 cos³ B * sin B)
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin³ B Cos B - 1/√2 sin B cos B - 1/√2 cos³ B * sin B
∴ Sin (A - B) = 1/√2 sin B Cos B - 1/√2 sin³ B Cos B - 1/√2 sin B cos B - 1/√2 cos³ B * sin B
∴ Sin (A - B) = -1/√2 sin³ B Cos B - 1/√2 cos³ B * sin B

∴ Sin (A - B) = -1/√2 sin B Cos B (sin² B +cos² B)
= -1/2√2 sin2B
Now squaring and adding equation 1 and 2 we get

2 cos squared A plus 2 sin squared A equals cos squared B plus cos to the power of 6 B plus 2 cos to the power of 4 B plus sin squared B plus sin to the power of 6 B minus 2 sin to the power of 4 B rightwards double arrow 2 open parentheses cos squared A plus sin squared A close parentheses equals cos squared B plus sin squared B plus open parentheses cos squared B plus sin squared B close parentheses cubed minus 3 open parentheses cos squared B sin squared B close parentheses open parentheses cos squared B plus sin squared B close parentheses plus 2 open parentheses cos to the power of 4 B minus sin to the power of 4 B close parentheses rightwards double arrow 2 equals 1 plus 1 minus 3 over 4 sin squared 2 B plus 2 cos 2 B space space space space open square brackets 4 cos squared B sin squared B equals sin squared 2 B comma cos to the power of 4 B minus sin to the power of 4 B equals cos 2 B space close square brackets space space space open square brackets cos squared B plus sin squared B equals 1 close square brackets rightwards double arrow 0 equals negative 3 sin squared 2 B plus 8 cos 2 B rightwards double arrow 0 equals negative 3 open parentheses 1 minus cos squared 2 B close parentheses plus 8 cos 2 B space space space space open square brackets cos squared 2 B plus sin squared 2 B equals 1 close square brackets rightwards double arrow 0 equals 3 cos squared 2 B minus 3 plus 8 cos 2 B rightwards double arrow 3 cos squared 2 B plus 8 cos 2 B minus 3 equals 0 rightwards double arrow 3 cos squared 2 B plus 9 cos 2 B minus cos 2 B minus 3 equals 0 rightwards double arrow 3 cos 2 B open parentheses cos 2 B plus 3 close parentheses minus 1 open parentheses cos 2 B plus 3 close parentheses equals 0 rightwards double arrow open parentheses 3 cos 2 B minus 1 close parentheses open parentheses cos 2 B plus 3 close parentheses equals 0 rightwards double arrow cos 2 B equals 1 third space o r space cos 2 B equals negative 3 open parentheses n o t space p o s s i b l e close parentheses N o w comma space w e space k n o w space t h a t space open parentheses cos squared 2 B plus sin squared 2 B close parentheses equals 1 S o comma space 1 over 9 plus sin squared 2 B equals 1 rightwards double arrow sin squared 2 B equals 1 minus 1 over 9 equals 8 over 9 sin 2 B equals plus-or-minus fraction numerator 2 square root of 2 over denominator 3 end fraction t h e r e f o r e comma space sin open parentheses A minus B close parentheses equals fraction numerator negative sin 2 B over denominator 2 square root of 2 end fraction equals plus-or-minus fraction numerator 2 square root of 2 over denominator 3 end fraction cross times fraction numerator 1 over denominator 2 square root of 2 end fraction equals plus-or-minus 1 third

parallel

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

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

chemistry-General

on ozonolysis gives

on ozonolysis gives

chemistry-General

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

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

chemistry-General

parallel

If $fraction numerator cos space x over denominator a end fraction equals fraction numerator cos begin display style space end style left parenthesis x plus theta right parenthesis over denominator b end fraction equals fraction numerator cos begin display style space end style left parenthesis x plus 2 theta right parenthesis over denominator c end fraction equals fraction numerator cos begin display style space end style left parenthesis x plus 3 theta right parenthesis over denominator d end fraction$ then $fraction numerator a plus c over denominator b plus d end fraction$ is equal to

If $fraction numerator cos space x over denominator a end fraction equals fraction numerator cos begin display style space end style left parenthesis x plus theta right parenthesis over denominator b end fraction equals fraction numerator cos begin display style space end style left parenthesis x plus 2 theta right parenthesis over denominator c end fraction equals fraction numerator cos begin display style space end style left parenthesis x plus 3 theta right parenthesis over denominator d end fraction$ then $fraction numerator a plus c over denominator b plus d end fraction$ is equal to

If $tan space alpha equals P over q$ where $alpha equals 2 beta comma alpha$ being an acute angle then $1 half$ $left square bracket p cosec space 2 beta plus q sec space 2 beta right square bracket$ is equal to

If $tan space alpha equals P over q$ where $alpha equals 2 beta comma alpha$ being an acute angle then $1 half$ $left square bracket p cosec space 2 beta plus q sec space 2 beta right square bracket$ is equal to

Let A,B,C be three angles such that $A equals pi over 4$ and $tan space B tan space C equals p$ . Then, all possible values of P such that A,B,C are the angles of a triangle is

Let A,B,C be three angles such that $A equals pi over 4$ and $tan space B tan space C equals p$ . Then, all possible values of P such that A,B,C are the angles of a triangle is

parallel

is an example for

is an example for

chemistry-General

If $y equals left parenthesis 1 plus tan space A right parenthesis left parenthesis 1 minus tan space B right parenthesis$ where $A minus B equals pi over 4$ then $left parenthesis y plus 1 right parenthesis to the power of y plus 1 end exponent$ =

If $y equals left parenthesis 1 plus tan space A right parenthesis left parenthesis 1 minus tan space B right parenthesis$ where $A minus B equals pi over 4$ then $left parenthesis y plus 1 right parenthesis to the power of y plus 1 end exponent$ =

What is the hybridization state of benzyl carbonium ion

All carbons are sp²

What is the hybridization state of benzyl carbonium ion

chemistry-General

All carbons are sp²

parallel

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