Hello, everyone.
Let us begin with our first lecture on trigonometry.
And before we start getting into trigonometry ratios and different identities, let us understand the
concept of an angle first.
Okay, so what is an angle?
Let's say if we have an article for Circle and these two are the.
Radius.
Let's say the center of the circle is all this is A and this is B, okay?
And this a B is the arc of the circle will center.
And this angle, A or B is defined as Peter.
Okay.
So if you have notice that this arc A, B has some length S and let's say the radius or A is of measure
R, then this angle theta is defined as S by R.
Okay, so angle is nothing but arc upon radius.
And we have to remember that if we divide the arc length by the radius, we get the angle in radian
measured.
Okay, so what is an angle?
Angle is the article for a circle divided by the radius of the circle.
And the angle that we get over here is an area.
Okay.
And you might be knowing the other measure of an angle like degree.
So what is one degree?
Or we can say that one two degree is equal to fight area, right?
So like, for example, if we have a complete circle and.
We can say that the circumference of the circle is two Bayer and the radius of the circle is R.
That means the angle in the complete circle is nothing but loop.
All right.
Are 362.
All right, so I hope you understand what is basically an angle.
And angle is nothing but the ratio of arc length divided by the radius.
And if we divide our land by the radius will get the angle in radium again.
We can convert degrees into radial and radial and to degree by using this formula that binary equals
to 180 degrees.
So I believe that you will be able to convert the angle from Radian to degree and degree to Radian.
So, for example, if you want to find out how much 120 degree is in Radian, so let's say we can see
that 180 degree is equal to by radium and we can say one degree equals two by by 180 radium.
And so we can say that 120 degree equals two by a by one and the into 120 radium which is nothing but
two by about three radium.
Okay.
So you can convert the angle from degree to radium and radium to degree very easily.
Okay.
So this is the first thing that we talked about.
What is an angle, the mathematical definition of an angle.
Now, one more thing I want to cover over here.
That is area of us atom.
So let's say we have some arc of a circle and some.
And substandard by this art and the center of the circle that is Peter or is the center in ebbs the
arc and let's say artist the radius first thing is what is this value of this arc land as that is are
and he does because we define what he does as a banana.
Now this area is what we are interested in the sector idea.
Okay and how much is the sector area?
So this area is half of our square table.
Or we can also say that this is half of orange BU and RTD is s.
So this is the area of this set.
Okay.
So these two are very basic things that you need to know regarding the sector, how an angle is defined
and how the area of a sector is defined.
And this arc, a B is an arc of a circle.
Okay?
You have to make sure of this thing.
It should not be an arc of any other curve.
Alright, now we're going to find one more thing that is perimeter of a sector and it is very easy to
see.
So let me just show it to you.
So let's say again, these two are the radius of a circle and this is a mark.
So this is again, 0eb, so this is of length R or B or is also a length R and ABS is of length s and
let's say this is theta.
So what does three metre off or a B, so perimeter of or a B is?
Or L.A. that is, ah, plus Abilene, that is S plus B all and that is again, ah, that is to ah plus
s and s can be written as ah, if he does so to ah plus arti diversity that has to be in really.
Okay.
You have to remember that, that whenever we are using the analyze our component areas we will get the
radium.
Okay.
So this can be done as ah and do two plus people.
This is the perimeter of OAB.
Okay.
I think you are clear with these three things.
First is what is an angle in Radian?
Second, how to convert Radian and degree.
Third is 80 alpha sector and the perimeter of the center.
Okay.
So if you are clear with all of these things, let's take a few questions based on it.
So questionable.
And I'm guessing number one, we are given that the perimeter.
So if the perimeter
of a certain sector.
Is then unit and radius off circle is three unit, then find 80 off center.
Okay, it's a very easy question.
So let us solve that.
So suppose this is the sector that we have.
Let's say again, call this as or maybe this is theta, this is R and this is offline s and is offline
R So what is the perimeter?
So that is to R plus s is equal to ten units and R this value we know is of three unit.
So six plus S is off then unit we can say S is of four units.
Now area is we know that is half of square people and this little Red Raiders half are and to Arteta
that getting rid of the as S so one by two and two radius is three and as is four.
So this is 12 by two which is six square units.
Okay.
I hope this is very easy question and you are able to understand how to solve this question.
Okay.
Let's take one more question.
And in this question we are given that six equal circles
each of radius e.
Are pleased so that.
Each touch is.
Other two.
Okay, I'll draw a diagram and show that to you as well.
Each touches two others.
Their center.
Being all on the circumference.
Of another circle.
Okay.
Then we have to find the area.
That includes.
Okay, I'll draw the diagram as well.
So let's say we have a regular hexagon and on each side of the regular hexagon or on each vertex of
the regular hexagon, I'm going to draw a circle being.
And the center of the circle is the vertex of the hexagon.
Okay, so it is like this.
All right.
So let me reduce this a little bit.
Okay.
So we can draw the diagram in this way.
I hope you are clear.
And now we are interested in finding out the area of this region.
Okay.
And this land is of a unit because we are given that the radius of each circle is of a unit.
Okay.
So we want to find out the area of this region that is in between the circles.
Now, what we will do is first, let's say I will split this part and I will take that said, this is
the center of the regular hexagon and this is one of the triangle.
Right.
So let's take all this as a B and all.
So it will be as one triangle.
Similarly, we'll be having another triangle or b, c, we'll be having another triangle or C, B or
C, E or C, F and or FP.
So these six triangles will have equal area.
We can say by symmetry, and if I just find out this area and multiply that area by six, then we will
get the inequality.
Okay.
So what I can say is that this angle, angle A or B is nothing but 360 degree by six because all this
angle A or B and angle B or C, all of these angles will be equal.
We can say that symmetry.
So this angle, it will be is one sixth of the total angle that is 360 degree.
All right.
Or we can say that this is 120 degree or 62 degree or pi over three.
Really?
All right.
And since this angle and this angle and also same that symmetry, we can say that triangle A or B is
equal triangle.
All right.
And you can write down the area of the triangle.
It will be.
So what is the area of triangle?
It will be for an equilateral triangle.
What is the area ruled?
Three by four off side a square, and the side is of length to it.
So two is a square, which is root, three is square.
This is the area of this triangle.
Now, from this area, I will subtract this area and I will subtract this area.
And these two individual areas are same like this area and this area, they are same rate.
So I will calculate this area.
Let's call this as B and this Q so area of sector
B, B.
Q That is half order square.
That is a square and p de and this we know that this angle is also vibratory.
All right.
So similarly, area of sector
AP R
will be half a square and two by three.
Okay, so what is this area or Q B are so area of or QPR will be equal to area of the triangle or EAB
that is root three is square minus this area in this area.
So some of these two areas, a square in both by three.
Okay, so 80 of all QPR is equal to eight square times of root.
Three minus five.
Three.
All right.
So this is the blue area that we have figured out, this area.
All right, so what is the required area?
So required area is six times of area of all QPR.
And that is equal to six is a square and 2 to 3 minus by three.
Okay.
So this is all answer.
I hope you are able to understand this problem as well.
It is not a very tricky problem.
Okay.
So once you see the diagram and figure out how the curves are being placed, you can easily figure this
out.
Okay, so let's move on to the next part of trigonometry that is trigonometric ratios.
So first let us understand what are basically saying Peter Costello and then Peter and so on.
Okay, so suppose let's say we have an right angle triangle called this says ABC and it is right angled
at B.
Okay.
So if I assume let's set this angle as tighter than this side because the opposite side.
Okay, so I can call this as perpendicular or the opposite side.
This side AB is called base or I just inside and this side AC is called the hypotenuse.
Okay.
So if I'm writing sine theta, that is nothing but perpendicular up on hypotenuse.
So that is b c length upon ac length.
Okay.
So that will be the same value of this angle.
Similarly cost here will be based upon a hypotenuse that is ab on AC.
Don t don and bandleader can be written as perpendicular upon base, which is length of b c upon length
of AB.
Okay.
Similarly we can write because we can t though that is one by assigning video and we can say card theta
is one by 20.
And similarly, we can say that secant is equal to one by of course.
Okay.
I hope you already know these things.
So what will cause a candidate will be the candidate will be one by side and it also hypotenuse upon
perpendicular.
So we can write this value also that is AC upon BCE.
Okay.
So similarly we can find out what he does like p down, so on.
Now after this, let us see the nomadic identity you.
So in a right angle triangle, we know that Pythagoras theorem holds.
Okay, so let's say there's a right angle triangle, ABC and this is the angle pido.
This is perpendicular, this is base and this is hypotenuse.
Okay, so we can say that hypotenuse squared equals two perpendicular squared plus B squared.
Okay, this is the Pythagoras theorem that we have.
Now, if you divide both the sides by each Esquire B squared and B squared, we'll get different, different
identities.
Okay, so let's see if I divide both sides by edge, a square, then I get B by square less B by a square
as opposed to one, not B by just square assigned.
So this is nine squared plus cost square with the other that is equal to one.
Okay.
And similarly, we'll get the second identity that is one plus ten square with equal to second square.
All right.
This identity will get by dividing with base.
And third identity we'll get is one plus third squared is a postal because again square with this identity
we'll get by dividing this equation by B squared on both sides.
Okay.
So these three are trigonometric identities.
We must remember these identities and there are some more derived formula that can come from these identities
and write a few of them.
And it's good that a few remember those kinds of formulas and you should know how they are coming.
Also, because sometimes in problem using them directly gives us a very clear solution of the problem.
Okay, I'll be covering some problems.
If you use those formulas, it'll be very fast to solve those problems.
Okay, so first thing we can say that is sine squared equal to one minus called squared.
Now I can write it like this, that same square taken by the same time to sine theta and that is equal
to one minus cost.
We don't do one plus cost it.
Okay.
On the right hand side, I have applied a squared minus bs4 so I can say that I ended up on one minus
cost.
Peter is equal to one plus costly double sided.
Okay, so this is also trigonometric entity that we can derive from this original identity.
Okay?
Similarly, using this identity, we can prove that 60 plus Khatun Peter is equal to one upon second.
Dominus Dan.
Peter Okay, and this is precisely very useful for problems, okay?
And one more we can say that because you can Peter minus card Peter is equal to one upon because again.
Peter plus carpenter okay.
Similarly we can see that consequently the plus card equals two one of one.
Consequently, Dominus got it.
Okay, so these three derived identities can also be used.
Okay, now let us try and understand more about the behaviors of these functions sine theta called PDI
and so on, using a unit circle.
And let us say that if these are the coordinate axis and I am drawing a unit circle whose center is
origin, okay, somewhat like this.
Now if I take a point B such that let's say this point is e this center is all and this is, let's say
in ash, this one is B and this one is B that.
Okay.
And this angle angle B is Peter.
All right.
Now, if I drop A perpendicular from B on two sides, okay, let's say this is the perpendicular.
So I can say that this length or B's of one unit because it's a unit circle and let's call this point
does m so we can say and B up on or B is equal to sign.
Peter All right.
So and B is equal to or behind even or B is one so the same.
Peter And similarly we will be able to say or M That will be equal to cost.
Peter Now that means the coordinate of this point.
B will be Godspeed.
Saint Peter And if you have studied circle in corner geometry, then you might have known that position
of a or the point coordinate of a point on this are going to be squared plus less for equal to one is
considerable sine theta.
All right.
Now we will analyze the behavior of the x cardinal of this point being the Y coordinate of this point
B, and then we'll get to know the behavior.
Of course, Biden signed it.
Okay.
So let's try some.
EMS and let me know whether you are able to understand those warnings or not.
So first thing, we can tell you that if this plane B is on this circuit and it is moving on to the
circuit in anti-clockwise direction.
Okay.
This way.
So when I'm at point E, the coordinate of B will be one common zero, right?
Because this point X coordinate as well in order to zero.
And if we are moving from A to B and then a dash and to B dash, no where in between the X coordinate
is going to be more than one.
Okay.
In fact, I might add it as the X coordinators minus one and zero.
So we can say that the x coordinate of point B is always going to be in between minus one and one.
So we can say that cost de is always less than or equal to one and then equals two minus one.
Okay.
And similarly, we can say that the coordinate of this point B will be 001.
So the maximum value of playing it is going to be one of the minimum by the upside and it is going to
be minus one.
And so we can say that S.E. De is also less than or equal to or greater than equals to minus one.
All right.
And one more thing we can say at this point, a course of zero is one and we can say that sine of zero
zero.
Similarly, at this point, B, the angle is by virtue, of course, of by way to zero and signoff by
two is one.
All right.
And similarly for this point, also, we can say and for this point also, we consider.
All right, the third thing we can say is if we are moving from A to B, okay, in P belonging from
zero to PI, if we are moving from A to B, the Y coordinate of B is increasing.
Okay.
So we can say that sine theta is increasing.
All right.
That means like if I take two, I notice warranted are two in this interval and if I say so, both.
He doesn't know that all belongs from 0 to 5 and P is less than t double.
Then what can I say?
Considered sine theta one is also less than.
Okay.
These two angles must belong in this interval.
All right.
I hope you are clear what I am trying to convey on here.
And similarly in P belonging from 0 to 5 to consider decreases and y decreases.
Because if I go from A to B, the X coordinate is decreasing.
All right.
So positive is decreasing.
So we can conclude a lot of things about 99 Costa de using this unit circle.
Now I suppose if I am at some point over here on this point, does.
Q So the X coordinator's negative in the Y coordinate is positive.
All right.
So Sino de is positive and second quadrant and cause he does so sine theta is positive and second quadrant
because it is going to be positive upon positive and positive and negative in the second quarter.
Okay.
So we can write a general rule that in first quadrant, all econometric functions are positive because
sine didn't cause they are positive.
So all functions will be positive because then they will be positive, then it will be positive and
so on in second quadrant only sine and because it can be a positive in third quadrant that in court
are positive and in fourth quadrant cause and second second are positive.
And how we can prove that thing, we can say that let's say if I take a point are in third quadrant.
So X coordinator's negative and y coordinate is also negative.
Okay, so anti and B sign upon cost will be positive and completely lock to be positive.
So that is how we can conclude.
So there are many things we can talk in this unit circle.
So like if I draw a line of 45 degree in this case, this point will be one by root two and one by two.
So we can say that signed by about four or 45 degree is one by road two and caused by therefore is of
course to improve these things we can see and one more thing we can say that in this region in 025 3
to 0 two, pi therefore is an okay x coordinate as more than wireguard in it.
Right?
So we can say if he lands from zero to pi before then, cause he does more than say that.
Right.
But because the x coordinate is more than the bike, so we can then do a lot of things.
Okay, all.
Only depends on what all you want to control.
All right.
Now, let us move on to the next part, and I will try to understand these functions.
I ended up on state and landed more.
Okay.
So we'll study graphs of trigonometry functions and we will only study sine sedan.
And as of now and in the later part of this course, we will study the rest of the route of normal functions.
Okay.
So first we will draw a graph of y is equal to sign x.
Okay.
Now let's say these are the coordinate axis and I am drawing the graph of visible to say an X, Y and
z of the values zero.
And it looks like this.
Okay.
So it's a periodic graph.
And on the left hand side also.
Okay.
All of these points are of the same height.
Now at five by two, the values one at zero, the values zero at by the values zero at three by two.
The value is minus one.
Okay, this one is minus one and so on.
And we can say that this graph is periodic with period two.
By that means.
The entire graph is nothing but the repetition of this red graph.
Okay.
So like we can say that sine of x plus two is nothing word signings.
Okay.
At any angle x at an angle to buy plus x have the same output.
All right.
So that's what we mean by this equation.
Alright, now let's look at the next graph and the second graph that we will cover is Y is equal to
cause X.
Okay.
So if I in any question, if I want you to recall the graph of Y as opposed to say an X, I hope you
will be able to recall this thing.
Okay, and remember these basic values.
So make it minus by y it is minus one and so on.
Okay, now let's draw y is equal to cause x.
So let's say these are the coordinate axis and y is equal to cause x looks like this.
Okay.
And this is the value at zero.
The value is one at five by two.
The value is zero and by the value is minus one.
Okay.
At three by value the value is zero.
And at two point the value is one.
So again, we can say that with cost also period is too bad.
Okay, so if I kind of repeat this graph, I'm being able to draw the entire course x graph.
So cost of x plus two is of course x.
Okay.
And similarly, if we can look at these two graph, we can see that by shifting this graph by some angle,
I can make this graph.
Okay.
So like you can say that goes off or sign of x plus by two is equal to six.
Okay.
We can realize this thing on the unit circle also that it is very easy to see like suppose let's say
this is the angle x, right?
So this is the entire angle pi medical ethics.
All right.
Now this angle will be five to minus X.
Okay.
And we can write the coordinate of this point and then we can just compare.
Okay?
So like this distance is minus nine x because we can say that this length, okay, let me use a different
color and show this tree and this land is going to be signing.
Right.
And it is going to be negative also.
So we can say this is a minus sign.
It's and this land is going to be cause x.
Okay.
So what will be the coordinate of this point?
The coordinate of this point will be minus nine x come across x.
And we can also see that the coordinate of this point with this rotation is going to be this point.
Coordinate goes off by value plus X and sign off by plus six.
Okay.
So if we compare the Y coordinate, we can see signed by our two plus axis Cossacks.
Right?
So it is very easy to prove this thing.
Now, let us look at the next graph.
That is why the calls to Benning's.
Okay.
So again, suppose we have some coordinate axis and the graph of panics looks like this.
Okay.
And I'll explain this to you.
What this dotted line is called.
Okay.
And what does it mean?
So this is, again, the graph of that.
It's in the second part.
And let's say this is another dotted line and it's just the repetition of this graph.
Okay.
So I hope you are clear.
How does this look like?
Now, let me give you some initial values.
So I'd zero the values zero at 5329 of five is not defined are infinity.
Okay.
It can be like if we are writing ten of just less than two, it is very close to infinity when we are
writing ten of just more than by value.
Then it is minus infinity.
Okay, over here.
So I will say panics is not defined at three by two.
Also panics is not defined surveillance the next point.
So ten x is defined for all real values except are the multiple of by random because there are multiple
of 5 to 10 x is shooting to infinity or minus infinity.
Okay, so at those point it is not defined at minus five where two it is again minus not defined.
Okay.
And one more value we know over here at before the value of annexes one because sign hidden passport
one by two.
Okay so this is the graph of panics and we can say that at every pie like the graph is repeating.
So that means from minus by whether to pivot to part.
I just repeat that part and I can draw the entire planet graph.
So begins at that tone of x plus five.
I do x plus buyers equals two panics.
Okay, because I have every by the graph this repeating.
Okay.
So let us gather one more thing before we complete the theory part of this lecture, and that is called
Allied Angles.
Okay.
And it is kind of very easy to understand and very important also.
Okay.
So we have written this result, dark sign of pi over two blessings.
Okay, so aligning those will help us write more results like this and we will write a lot and that
will work on all the cases.
Okay.
So first of all, what is a light and also to when meals are allowed, if they are some are different
than multiple of pilot.
So like for example he does and he does minus five were two other light because their difference is
a multiple of pi by two or we can see that he doesn't he plus by is a allowed handle because again their
difference is a multiple of by two or three done by a minus.
He does also like Daniel or Pi, where two minus three does alternate allowed.
And so those angles are related angle which says that there some are differences a multiple of by by
two.
Okay.
Now we will write a rule to find out the value of same and by a very two plus three down.
Okay so using this rule, we will be able to find out.
Okay, so what do we do in this is first we have to know a few things.
Like first thing we need to know is what are complementary functions.
Okay, so if I say complementary of same side axis causes complementary of an axis cortex and so on
and complementary oblique indexes, positive index.
Okay, this is the first thing you need to know.
These are called complementary function.
The second thing you need to know is which trigonometric functions are positive and which quadrant.
So in first quadrant or trigonometry, functions are positive.
Second quadrant sign in because I got a positive and then third quarantine and called are positive and
in fourth quadrant cause and second are positive.
Okay so these two things you need to know now let's write the rule for the same end.
Pi over two plus theta.
Okay, rule number one if end is hard then function converge and do complementary.
Okay.
Like sine will convert into cause and then convert it to cause and so on.
If and even function remains simple.
Okay, so this is the first rule.
No second rule is check quadrant of end by a very do plus Peter okay assuming
he does acute okay and use same sign.
As part of the original function in that quote, we'll get explaining the meaning of this in that quadrant.
Okay.
So what do we mean by this?
Let's say, for example, I want to write sign off by way two plus freedom.
Okay.
This is a good example to understand.
So, first of all, what do we have to do?
We have to find out the value of N and another multiple of by value.
So what here?
One is a multiple of pirate also.
And as odd then and as our function will convert into, it's complementary that we inside will convert.
And of course.
So you just might consider.
Now, second point say is the check the quadrant of empire plus Tito.
So whereas spider two plus p done this is the zero degree over here or zero radian so this is by way
to Radian plus.
Peter that means we are in the second quadrant and in the second quadrant use the same sign as that
of original function.
So what is the original function saying?
This is the new function.
Okay, remember that we have to talk only about the original function.
So what is the sign of sine in the second quarter?
It is positive.
So this value is plus positive.
So signed by two plus three days plus cost, you know.
Okay, let me give you a few more examples
than three by word, two minus three down.
Okay.
So if we look at the value of N the value of energies, three of them will convert into card.
Now we have to look for the angle three by two, minus three.
Then which quadrant does this angle exist?
Okay, so we are zero over here.
This is by value.
This is by.
And this is three by two minus.
Peter that means we are somewhere in the third quarter and then third quadrant ten is positive.
So divide plus start.
Peter Okay, similarly let's say we are writing secant of seven by two plus.
Peter Now let us look at the value of N for the value of n is seven seven as hard.
So we'll write it as Peter.
Now let's look at where the angle seven by two plus key delays.
Okay, so by very two, two by two, three, five, four, five, six, seven, five by two plus.
Peter Okay then we move an anticlockwise direction that he does positive.
Okay.
I hope you know this convention.
So in the fourth quadrant, rehab.
Okay.
So in fourth quadrant, second is positive, right?
Because we know that in the fourth quadrant, secant and cosine positive.
So this is going to be plus cosecant.
Peter Okay.
I hope you are able to understand how to find out the values of the allied angles.
Okay.
And other than these things, you need to know one more thing.
I hope you are already introduced to that, and that is that regular trig values of trigonometric functions
sine the basketball bandido court three down second time because it can be done for the angle p equals
to zero t equals to 30 degrees or five as extended equals to 40 degrees t equals to 60 degrees then
t equals to 90 degrees.
I hope you know these values.
Okay.
And just put of these values.
If you don't know this, just remember these values.
So this then I am writing and it means this value is not defined.
Okay.
So I am writing that by name.
All right.
So after this, let us take a few questions based on this part, and then we will finish our lecture
them in trigonometry.
So first question we are taking is find the values of.
And a part as and one degree in due time to degree and so on up to 1089 degrees.
And in the B part they're asking the value of sine squared, a five degree less lane squared and degree
and so on as one of those lines flooded five degrees.
Okay.
So let us solve these questions.
So in the first part, we can see that band of 89 degrees is that of 90 degrees, minus one degrees,
which is of this is one times of pilot.
Right.
Okay.
So I hope you understand this 90 degree thing.
So it is one times of 90 degree minus one degree.
So the value of N is one.
So it will become caught of one degree.
And let's look at the quadrant.
So it is in the first quadrant.
So 90 degrees, minus one degrees, it is in the first one.
So it is plus quadrant building on it.
So similarly, we can say time off, it'd be eight degrees plus two degrees and so on and so on.
Okay.
So we can say, Don, of one degree in the town of 88 degree, 89 degrees, in the town of two degree
in the town of 88 degrees and so on and so on.
A town of 44 degrees in the town of 46 degree and town of 45 degree.
Okay.
So I'm combining it like that sort of one degree in the town of 89 degrees, nothing out of one degree.
So this product is one.
This product is one.
So it is one into one into one.
So this is equal to one.
Okay.
I hope you are able to understand this.
Now, in b part, we have to find out the value of science squared five degrees plus science squared
and degrees and so on up science, let it be five degree.
So we will say that science squared defined degree is because it can cause a square five degree.
Okay, so sine squared five degrees plus called squared five degree will become one.
So again, last two terms the summers one and how many times do we have over here?
So we have a total of 17 times from five and two, 1 to 5 and to 17.
Okay.
So and the sum of two terms is one.
So this and this like we have eight pairs, so I'll write those pairs and I hope you'll be able to get
it.
Plus science squared be five degree.
The second pair will be sine squared and degree plus sign squared degree and the last pair that will
be formed will be sine squared four two degree plus line squared 50 all and cosine squared 45 will be
separately.
There will be no one to pair it up.
So this some is one, this some is one and we have eight ones like this.
So eight plus one by two and that is equal to 17 by two is the answer for this.
All right.
I hope you are able to understand these two questions.
Let us take one more question.
So in this question, we are given that 71 -20 de is equal to X, then we have to find sine theta in
terms of X.
Okay.
So we are going to use that relation that secant plus 20 though is equal to one up, consequently the
minus standard.
Okay.
And we know that's the candidate minus and this thing is X, so we can say that 60 plus bandito is equal
to one X and the original relation is the candidate -20 days.
All right.
Now, if we are these two things, then we get a second hitter that is equal to half of X plus one max.
So that can be extra squared plus one abont two x.
Okay.
And if we subtract these two things we get 20 to is equal to half of one by x minus x, right?
So this is equal to one minus X squared upon fluids.
And we want to find out the value of signed it, right.
So we have to find out the value of sign the line of X assigned to this.
You can write signed it are signed up on cost He died to save Peter again sign the cost it is then Peter
supplanted by Peter and the value of 20 that is this and the value of selected as this sword is going
to be one minus x squared upon one plus x squared.
All right.
So it's a pretty good problem if you understand this thing.
Let us take one more question.
And in this question we are given that if sine x plus sign squared x is equal to one.
So if this is given to us, then find the value of y is equal to cause raise to 12 x plus three cost
raise to ten of x plus three.
Cause that raise to eight of x plus quantities to six of x minus one.
Okay.
So we are given an equation in Synnex and this equation will get us some value of x.
And for that value of x, we ought to find out the value of this function.
All right.
So if we observe the relation carefully, then we can say that sine x is equal to one minus sign squared.
So this gives us that sine x is equal to cause squares.
Okay.
And in this entire Y, wherever we have causes correctly line that does sign is.
So this y will become signed to six of X plus three signed to five of x plus three sign these two four
of x plus sign cubed x minus one.
Okay.
And this y is equal to if you notice this, this is nothing.
What a perfect cube.
So this is science squared x plus sign x whole cube minus one and the value of science square x plus
nine x is nothing but one.
So the value of y is equal to one cubed minus one.
So that is equal to zero.
So the value of y is nothing.
What?
Zero.
Okay.
I hope you are able to understand this thing.
Now, before I end this lecture, I will would like to add two important formulas that generally we
use in problems.
Okay, I am putting up all of these formulas separately in a Formula three does well.
So first one is sign this two four of x, let me separate this thing out plus cost that is two four
of x.
This is nothing work one minus two sign squad x and causes flags.
And second one we can say that sign this to six of x plus cost for two sets of x.
This is nothing work.
One minus three signs, clerics, cosmetics.
Okay.
And you can easily prove this formulas you can take this is e squared plus B's blood basically plus
B squared minus two B and this one you can use the formula of a Q plus Q.
Okay, so remember these two formulas.
They are generally useful in problem solving.
Okay, so that's it for data for the first lecture.
We'll see you in the second lecture.
Thank you.