Sponsor Area
Degrees to radians: Multiply by
Radians to degrees: Multiply by
Radians to degrees: Multiply by
Radian to degree: Multiply by
Radian to degree : Multiply by
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Number of revolutions in one minute or 60 seconds = 360 Number of revolutions in one second =
Degree measure of an angle turned in one second =
Radian measure of an angle turned in one second =
A wheel rotates through an angle of radians in one second. How many revolutions will it make in 5 minutes?
Radian measure of an angle turned in one second =
Degree measure of an angle turned in one second =
∴ Number of revolutions in one second =
Hence, the number of revolutions in 5 minutes =
The difference between two acute angles of a right angled triangle is Express the angles in: (a) degrees (b) radians.
Let the vertices of the triangle be A, B and C and
∴ ...(i)
Also, radians (given)
...(ii)
Adding (i) and (ii), we get
Now,
Hence,
(a)
(b)
Find the degree measure of the angle subtended at the centre of a circle of radius 100
cm by an arc of length 22 cm (Use ).
r = 100 cm, l = 22 cm
Now,
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Length of the circular wire of radius 7.5 cm
Which is cut and bent =
∴ For the circular hoop,
Let the angle be radians, Using,
we have
Hence, the angle subtended at the centre is
Radius of the circular track, r = 1500 m
Distance travelled in 1h or
Distance travelled by train in 1 sec =
Length of arc = Distance travelled by the train in 10 seconds =
Using we get
An athlete runs 4 times around a circular running path to describe 1760 metres.
What is the angle in (i) radians, (ii) degrees subtended at the centre of the circle, when
he had run a distance of 308 metres?
Distance travelled in one round =
Let r be the radius of the circular track.
∴
Let the angle described in covering a distance of 308 metres be radians.
Using we have
(i)
(ii)
Let be the three angles of a triangle in A.P. with d>0
Since the sum of the angles of a triangle is
∴
Now, number of degrees in the least angle = (60 - d)
Number of degrees in the greatest angle = (60 + d) Number of radians in the greatest angle =
According to the given condition,
4d = 120
d = 30
Hence, the angles of a triangle in degrees are: or
Let D km denote the diameter of the sun. Since the angle subtended by the sun is very small, therefore, its diameter is approximately equal to small arc of a circle with centre at the eye of the observer on the earth.
∴ l = D, r = 149,700,000 km
Since
Tips: -
Note:
1. The angle traced by hour hand in 1 hour is equal to .
2. The angle traced by minute hand in 1 minute is equal to
Here, the time is 7:20 or or
Angle traced by hour hand in 12 hours =
Angle traced by hour hand in 1 hour =
Angle traced by hour hand in =
Angle traced by minute hand in 60 minutes =
Angle traced by minute hand in 1 minute = =
Angle traced by minute hand in 20 minutes =
∴ The required angle between the hour hand and minute hand =
Here
sinx<0, cosx<0, secx<0, cosecx<0, tanx>0, cotx>0 (∵ x lies in quadrant III)
Now,
(∵ sinx is -ve)
,
Here, sinx =
sinx>0, cosx<0, tanx<0, cosecx>0, secx<0, cotx<0 (∵ x lies in quadrant II)
Now,
(∵ cosx is -ve)
Here,
sinx<0, cosx<0, tanx>0, cosecx<0, secx<0, cotx>0 (∵ x lies in quadrant III)
Now,
(∵ cosecx < 0)
sinx, cosx, secx, cosecx are periodic functions of period
∴ We may add or subtract an integral multiple of to convert the angle to an acute angle. Also, tanx and cotx are periodic functions of period
Therefore, we may add or subtract an integral multiple of to convert the angle to an acute angle.
(∴ The Period is
)
(From tables)
Hence,
sinx, cosx, secx, cosecx are periodic functions of period
∴ We may add or subtract an integral multiple of to convert the angle to an acute angle. Also, tanx and cotx are periodic functions of period
Therefore, we may add or subtract an integral multiple of to convert the angle to an acute angle.
[∵ Period is
]
Hence,
As tanx and cotx are periodic functions of period
Therefore, we may add or subtract an integral multiple of to convert the angle to an acute angle.
(∵ Period is
)
(From tables)
Hence,
As sinx, cosx, secx, cosecx are periodic functions of
∴ We may add or subtract an integral multiple of to convert the angle to an acute angle.
(∵ Period is
)
Hence,
As tanx and cotx are periodic functions of period .
Therefore, we may add or subtract an integral multiple of to convert the angle to an acute angle.
(∵ Period is
)
(From tables)
Hence,
Since x lies in quadrant III.
∴ sinx <0, secx <0, tanx>0 ...(i)
Now, and
...(ii)
From (i) and (ii), we have
Also,
∴
If x, y are real, show that is valid only when
We know that
Now,
and
The equation is possible only if
or if (Provided
,
)
or if
or if
or if [∵
is not <0]
or if x - y = 0
or if x = y
Hence, the equation is possible when x = y
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Prove that:
L.H.S. =
Now, is in quadrant IV where
is -ve.
Now, is in quadrant III where
is -ve.
∴
Also,
∴ L.H.S. =
Prove that:
L.H.S. = is in quadrant II means
is +ve.
Also, is in quadrant II means
is -ve.
∴
Prove that:
In
∴ is in quadrant II, means
is -ve
∴ cos (A + B) = - cos C
Hence, cos (A + B) + cos C =
ABCD is a quadrilateral
x is quadrant IV means sinx is -ve.
∴ sin (A + B) = sin[2 - (C + D)] = - sin (C + D)
Hence, sin (A + B) + sin (C + D) = 0
A, B, C, D are the vertices of a cyclic quadrilateral. A + C =
and
∴
(∵ x is quadrant II means sinx is +ve)
Hence, sinA + sinB - sinC - SinD = 0
Prove the following:
Let (n+2)x = A and (n+1)x = B
L.H.S. = sin(n+1)x sin(n+2)x + cos(n+1)x cos(n+2)x
= sinB sinA + cosB cosA = cosA cosB + sinA sinB
= cos(A-B) = cos[(n+2)x - (n+1)x]
= cos (nx + 2x - nx - x) = cosx = R.H.S.
∴ L.H.S. = R.H.S.
Hence,
If and
where
and
find the
value of State the quadrant in which
terminates.
We know that
[∵
is -ve as
lies in IInd quadrant]
Now, and
We know that in Ist and IIIrd quadrant.
∴ lies in 1st quadrant.
Prove that:
R.H.S. = sin(A+B). sin(A-B)
= [sinA cosB + cosA sinB] [sinA cosB - cosA sinB]
=
=
=
=
Hence, sin (A + B). sin(A - B) =
Prove the following:
L.H.S. =
= sin (A + B). sin(A - B)
= sin(6x + 4x) . sin(6x - 4x)
= sin 10x. sin2x
= sin2x. sin10x = R.H.S.
Hence,
Prove the following:
L.H.S. =
=
= (Where A = 6x, B = 2x)
= sin(A + B) . sin(A - B)
= sin (6x + 2x) . sin(6x - 2x)
= sin 8x. sin4x
= sin 4x. sin 8x = R.H.S.
∴ L.H.S. = R.H.S.
Hence,
Show that;
x - y = (x - y + z - z) = -[(y - z) + (z - x)] tan(x - y) = tan[- {(y - z) + (z - x)}]
tan (x - y) = - tan[(y - z) + (z - x)] (∵
)
tan (x - y) = -
tan(x - y) - tan(x - y) tan(y - z) tan(z - x) = -tan(y - z) - tan(z - x)
Hence,
Sponsor Area
Find the maximum and minimum values of:
Let
Put
Squaring and adding, we get
∴
Now, (∵ cos x and sin x lie between -1 and 1)
Hence, max. value of {a cosx + b sinx} =
Min. value of {a cosx + b sinx} = -
Find the maximum and minimum values of:
Let
Put
Squaring and adding, we get
or
Now,
∴ Max. value of
Min. value of
Find the maximum and minimum values of:
Let
Put
Squaring and adding, we get
or r = 5
∴
Now,
Hence, max. value of
Min. value of
Show that:
Using sin 3A = 3 sinA - 4 sin3A
cos 3A = 4 cos3A - 3 cos A
We have, L.H.S. =
=
=
=
= 6 - 4(1) = 6 - 4 = 2 = R.H.S.
∴ L.H.S. = R.H.S.
Hence,
Find in the following:
Find in the following:
is in quadrant III
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Prove the following:
L.H.S. =
= ...(i)
R.H.S. =
= ...(ii)
From (i) and (ii), we have
L.H.S. = R.HS.
If cos (A + B) sin (C - D) = cos (A - B) sin (C + D)
Show that: tan A tan B tan C + tan D = 0
cos (A + B) sin (C - D) = cos (A - B) sin (C + D)
Applying componendo and dividendo, we get
Hence,
x lies in quadrant II or III
or
or
Hence, the principal values of x are and
x lies in quadrant II or IV
or
or
Hence, the principal values of x are
is positive.
∴ x lies in quadrant I or quadrant IV.
∴ or
∴ The principal solutions are:
Now,
sin x + sin 3x + sin 5x = 0 (sin 5x + sin x) + sin 3x = 0
sin 3x (2 cos2x + 1) = 0
∴ either sin 3x = 0 or either
or
either
or
either
or
or
cos 3x + cos x - cos 2x = 0
either cos 2x = 0 or
either
or
either
or
x =
or
Find the general solution of:
Find the general solution of the equation:
Using ...(i)
Where s is the semi-perimeter of the triangle
Now,
∴ From (i),
In any triangle ABC, show that
In any
...(i)
Similarly, ...(ii)
and ...(iii)
Adding (i), (ii) and (iii), we get
L.H.S =
=
Hence,
Find the greatest angle of in which
and
Side b = AC = is the greatest
∴ is the greatest angle of the triangle.
Applying cosine formula
we have
∴
Hence, the greatest angle of the triangle is
Using formula,
, we have
In any if
show that the sides of the triangles are in A.P.
In (Given) ...(i)
Now,
Similarly,
Putting in (i), we get
By projection formula,
∴
Hence, the sides a, b, c of trianlge ABC are in A.P.
In any triangle ABC, show that
Using sine formula
∴ L.H.S.=
=
+
=
]
= k[0] = 0 = R.H.S.
Hence,
Angles A, B, C of a triangle ABC are in A.P. and Find the angles of the triangle.
A, B, C are in A.P. 2B = A + C ...(i)
Also, (∵ The sum of the angles of a
)
∴ ...(ii)
Now,
Also, (By sine formula)
(∴
)
Now,
or
Hence, the angles of the triangle are:
L.H.S. = a (Cos C - Cos B)
=
=
=
=
=
= ...(i)
Now, 2s = a + b + c
2s - 2a = a + b + c - 2a = b + c - a
By using these values in (i), we get
L.H.S. =
But,
∴ L.H.S. =
Hence,
Find when
a = 22, b = 28, c = 36
∴
s - a = 43 - 22 = 21,
s - b = 43 - 28 = 15,
s - c = 43 - 36 = 7
square units ...(i)
...(ii)
...(iii)
The minute hand of a watch is 10 cm long. How far does it tip move in 20 minutes?
Express the angular measurement of the angle of a regular decagon in (a) degrees (b) radians.
The difference between two acute angles of a right angled triangle is Express the angles in (a) degrees (b) radians.
A wheel makes 45 revolutions per minute. Through how many radians does it run in 1 second?
Find, in degree measure, the angle subtended at the centre of a circle of radius 3 cm by an arc of length 1 cm.
In a circle of 5 cm radius, what is the length of the arc which subtends an angle of at the centre?
A rail road curve is to be laid out on a circle. What radius should be used if the track is to change direction by angle at the centre in a distance of 40 metres ?
The angles of a triangle are in A.P. and the least angle is . Find the greatest angle in radian measure.
Solution not provided.
If cos x = lies in the third quadrant, find the values of other five trigonometric functions.
If lies in second quadrant, find the values of the other five trigonometric functions.
Find the values of other five trigonometric functions in the following: lies in fourth quadrant.
Find the values of other five trigonometric functions in the following: lies in second quadrant.
Find the values of the remaining trigonometric functions in the following: is in quadrant IV
Find the values of the remaining trigonometric functions in the following: is in quadrant IV
Find the values of the remaining trigonometric functions in the following: is in quadrant II
Find the values of the remaining trigonometric functions in the following: is in quadrant I
If is in quadrant II, then find the values of all trigonometric functions.
If find Cos A and also the quadrant in which A lies.
If find sin x and cos x. Also find the quadrant in which x lies.
Solution not provided.
Ans. x lies in quadrant I.
Find the general solutions of the following trigonometric equation:
Solution not provided.
Ans.
Find the general solutions of the following trigonometric equation:
Solution not provided.
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Find the general solutions of the following trigonometric equation:
Solution not provided.
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Find the general solutions of the following trigonometric equation:
Solution not provided.
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Find the general solutions of the following trigonometric equation:
Solution not provided.
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Find the general solutions of the following trigonometric equation:
Solution not provided.
Ans.
Find the general solutions of the following trigonometric equation:
tanx = -1
Solution not provided.
Ans. .
Find the general solutions of the following trigonometric equation:
Solution not provided.
Ans.
Find the general solutions of the following trigonometric equation:
Solution not provided.
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Find the general solutions of the following trigonometric equation:
Solution not provided.
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Find the general solution of the following trigonometric equation:
Solution not provided.
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Find the general solution of the following trigonometric equation:
Solution not provided.
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Find the general solution of the following trigonometric equation:
Solution not provided.
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Find the general solution of the following trigonometric equation:
Solution not provided.
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Find the general solution of the following trigonometric equation:
Solution not provided.
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Find the general solution of the following trigonometric equation:
Solution not provided.
Ans. No solution.
Find the general solution of the following trigonometric equation:
Solution not provided.
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Find the general solution of the following trigonometric equation:
Solution not provided.
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Find the general solution of the following trigonometric equation:
Solution not provided.
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Find the general solution of the following trigonometric equation:
Solution not provided.
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Find the general solution of the following trigonometric equation:
Solution not provided.
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Find the general solution of the following trigonometric equation:
Solution not provided.
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Find the general solution of the following trigonometric equation:
Solution not provided.
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Find the general solution of the following trigonometric equation:
Solution not provided.
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Find the general solution of the following trigonometric equation:
Solution not provided.
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Find the general solution of the following trigonometric equation:
Solution not provided.
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Find the general solution of the following trigonometric equation:
Solution not provided.
Ans.
In triangle ABC, a = 4, b = 6 and c= 8, show that 8 cosA + 16 cosB + 4 cos C = 17
Solution not provided.
A, B, C are angles of a triangle such that show that the trianlge is isosceles.
Solution not provided.
If triangle ABC, b cos B = c cos C, show that the triangle is either isosceles or right angled.
Solution not provided.
The area of triangle ABC is 54 cm2 and its perimeter is 9 cm, find the radius of its incircle.
Solution not provided.
Ans. 12 cm.
If D and C are respectively the number of degrees and radians in a given angle, show that:
Solution not provided.
Find the angle in degrees subtended at the centre of a circle by an arc whose length is 2.2 times the radius of the circle.
Solution not provided.
Ans.
Find the angle in: (a) radians (b) degrees between the minute hand and the hour hand at 20 minutes past 7.
Solution not provided.
Ans. (a) (b)
In a cyclic quadrilateral ABCD, show that (a) sin A = sin C (b) cos B + cos D = 0.
Solution not provided.
Solve the following equation, giving the general solution:
sin mx + sin nx = 0
Solution not provided.
Ans.
The angles of a triangle are in A.P. and the greatest angle is 5 times the smallest angle. Find the angle in radians.
Solution not provided.
Ans.
If in any triangle, the angles are to one another as 1:2:3, prove that the corresponding sides are
Solution not provided.
If in a
prove that it is either a right angled or an isosceles triangle.
Solution not provided.
In any if
are in A.P., prove that cot A, cot B and cot C are also in A.P.
Solution not provided.
If in triangle ABC, prove that it is either a right angled or an isosceles triangle.
Solution not provided.
In any triangle ABC, then prove that the triangle is a right angled triangle.
Solution not provided.
In a triangle ABC, if a:b:c = 4:5:6, find the ratio of the radius of the circumcircle to that of incircle.
Solution not provided.
Ans. 16:7.
The angles of a triangle are in the ratio 1:2:7. Show that the ratio of the greatest side to the least side is
Solution not provided.
Proof: Draw a circle with centre at O and radius OA = r. Cut off an arc AB of length r. Join OA, OB and produce AO to meet the circle again in C.
Since the angles at the centre of a circle are proportional to the lengths of the arcs which subtend them.
∴
Draw a circle of radius OA = r with centre at O.
Cut off an arc AB of length r and join OB; then radian.
Cut off an another arc ABC of length l and join OC such that (radians). Since angles at the centre of the circle are proportional to the arcs on which they stand,
∴
Prove the following:
(a)
(b)
We know that:
...(i)
...(ii)
....(iii)
...(iv)
(a) Replacing x by x + y, in (iii), we get
...(v)
Also,
[By using (ii)]
= sinx cosy + cosx siny [By using (iii) and (iv)]
Hence, sin(x+y) = sinx cosy + cosx siny
(b) Replacing y by - y in (v), we get
sin(x-y) = sinx cos(-y) + cosx sin(-y)
= sinx cosy - cosx siny [∵ cos(-y) = cosy, sin(-y) = - siny]
Hence, sin(x-y) = sinx cosy - cosx siny
Prove that:
As we know that:
cos(x + y) = cosx cosy - sinx siny
Replacing x by and y by x, we get
Hence,
Prove that:,
We know that:
cos(x - y) = cosx cosy + sinx siny ...(i)
sin (x - y) = sinx cosy - cosx siny ...(ii) ...(iii)
...(iv)
Replacing x by and y by x in (i), we get
= - cosx [By using (iii) and (iv)]
Hence,
Replacing x by and y by x in (ii), we get
= sinx [By using (iii) and (iv)]
Hence,
Hence,
Prove the following:
We know that:
cos (x - y) = cosx cosy + sinx siny ...(i)
sin (x - y) = sinx cosy - cosx siny ...(ii)
...(iii)
...(iv)
Replacing x by and y by x in (i), we get
= - cosx [By using (iii) and (iv)]
Hence,
Replacing x by and y by x in (ii), we get
[By using (iii) and (iv)]
Hence,
Hence,
Prove the following:,
,
We know that, cos (x + y) = cosx cosy - sinx siny ...(i)
sin (x + y) = sinx cosy + cosx siny ...(ii)
...(iii)
...(iv)
Replacing x by and y by x in (i), we get
= - cosx [By using (iii) and (iv)]
Hence,
Replacing x by and y by x in (ii), we get
= - sinx [ By using (iii) and (iv)]
Hence,
Hence,
Prove the following:,
,
We know that: cos (x - y) = cosx cosy + sinx siny ...(i)
sin (x - y) = sinx cosy - cosx siny ...(ii)
...(iii)
....(iv)
Replacing x by and y by x in (i), we get
= - sinx [By using (iii) and (iv)]
Hence,
Replacing x by and y by x in (ii), we get
= - cosx [By using (iii) and (iv)]
Hence,
Hence,
Prove the following:
We know that: cos(x + y) = cosx cosy - sinx siny ...(i)
sin (x + y) = sinx cosy + cosx siny ...(ii)
...(iii)
...(iv)
Replacing x by and y by x in (i), we get
= sin x [By using (iii) and (iv)]
Hence,
Replacing x by and y by x in (ii), we get
= - cosx [By using (iii) and (iv)]
Hence,
Hence,
Prove the following:
We know that:
cos(x - y) = cosx cosy + sinx siny ...(i)
sin (x - y) = sinx cosy - cosx siny ...(ii)
...(iii)
...(iv)
Replacing x by and y by x in (i), we get
[By using (iii) and (iv)]
= cos x
Hence,
Replacing x by and y by x in (iii), we get
= - sinx [By using (iii) and (iv)]
Hence,
Hence,
Prove the following:
As sinx is periodic function of period Hence, an integral multiple of
can be added or subtracted from the angle.
Hence,
Prove the following:
As cosx is a periodic function of period . Hence, an integral multiple of
can be added or subtracted from the angle.
Hence,
Prove the following:
As tanx is periodic function of period . Hence, an integral multiple of
can be added or subtracted from the angle.
Hence,
Prove that
We know that
...(i)
Replacing in (i), we get
∴ sin (2x) = sinx. cosx + cosx. sinx
∴ sin 2x = 2 sinx cosx ...(A)
Also,
Dividing numerator and denominator by we get
...(B)
From (A) and (B), we have
Prove that:
We know that
Replacing by x, we get
cos(x + x) = cosx. cosx - sinx. sinx
∴ cos 2x = ...(A)
Also, we know that
∴ ...(i)
...(ii)
Putting (i) in (A), we get
...(B)
Putting (ii) in (A), we get
...(C)
Now,
∴ ....(D)
From (A), (B), (C) and (D), we have
.
Prove that:
2 cosA sinB = sin (A + B) - sin (A - B)
R.H.S. = sin (A + B) - sin (A - B)
= [sinA cosB + cosA sinB] - [sinA cosB - cosA sinB]
= 2 cosA sinB = L.H.S.
Hence, 2 cosA sinB = [sin (A + B) - sin (A - B)]
Prove that:
2 cosA cosB = cos (A + B) + cos (A - B)
R.H.S. = cos(A + B) + cos(A - B)
= [cosA cosB - sinA sinB] + [cosA cosB + sinA sinB]
= 2 cosA cosB = L.H.S.
Hence, 2 cosA cosB = cos (A + B) + cos (A - B)
Prove that:
-2 sinA sinB = cos (A + B) - cos (A - B)
R.H.S. = cos(A + B) + cos(A - B)
= [cos A cosB - sinA sinB] - [cosA cosB + sinA sinB]
= - 2 sinA sinB = L.H.S.
Hence, -2 sinA sinB = cos (A + B) - cos (A - B)
Prove the following:
Let
R.H.S. =
= sin (x + y) - sin (x - y)
=
= sin C - sin D = L.H.S.
Hence,
Prove the following:
Let and
R.H.S. =
= cos (x + y) - cos (x - y)
=
= cos C - cos D = L.H.S.
Hence,
Find the general expression for all angles whose sine is zero.
Let be any angle
Draw PM X' OX
Then,
Now, sinx = 0 MP = 0
OP coincides with OX and OX'
which is true when
Hence, sinx = 0
Let be any angle, then
tanx = 0
PM = 0
OP coincides with OX and OX'
i.e.
So, general value of x is given by
Find the general solution or the general expression for all angles, whose cosine is zero.
Let be any angle.
Now, cos x = 0
OM = 0
OP coincides with OY or OY' which is true when
So, general value of x is given by
Find the general expression for all the angles whose cotangent is zero.
Let by any angle, then
cot x = 0
OM = 0
OP coincides OY or OY'
i.e.
So, general value of x is given by
We have sinx = siny sinx - siny = 0
either
either
or
either
or
either
or
The general solution is:
COROLLARY: If cosecx = cosec y, then six = siny
So,
Find general value of x, when cos x = cos y
We have cosx = cosy or cosx - cosy = 0
either
or
either
or
where
either
or
where
either
or
, where
where
Hence, the general solution is
COROLLARY: If secx = secy cosx = cosy
So, when secx = secy, the general solution of where
To find general value of x when tanx = tany
We have tanx = tany
sinx cosy - cosx siny = 0
sin (x - y) = 0
where
where
Hence, when tanx = tany,
then general value of x is given by
where
COROLLARY: If cotx = coty tanx = tany
So, when cotx = coty, then x = where
Prove that:
where
The given equation is: ....(i)
Dividing (i) throughout by we get
if
if
if
if
In any triangle ABC, prove that
(i)
(ii)
(iii)
(i) R.H.S. =
[∵ ]
Similarly, we can prove
(ii)
and (iii)
We have to show that
R.H.S. = b cosC + c cosB = k sinB cos C + k sin C cos B
= k [sin (B + C)] = k sin(-A) [By sine formula]
= k sinA = a = L.H.S.
In any triangle, we have to show that
a = b cosC + c cosB
b = c cosA + a cosC
c = a cosB + b cosA
R.H.S. =
=
=
Similarly, we can show that
b = c cosA + a cosC
c = a cosB + b cosA
In any triangle ABC, we have to show that
(i) (ii)
(iii)
(i) R.H.S. =
[∵ b sinC = c sinB]
[By sine formula]
= - cos(B + C) [∵ k sin B = b]
= - cos ( - A)
= cos A = L.H.S.
Similarly, we can show (ii) and (iii)
We have to deduce cosine formula from sine formula i.e. we have to show
R.H.S. =
[By sine formula]
[∵ sin (A + B) + sin (A - B) = 2 sinA cosB]
= L.H.S.
Prove in any triangle ABC
(i) (ii)
(iii)
(i) We have
or
Similarly, we can prove that
(ii)
(iii)
We have,
[∵ sin A is positive as
]
Similarly, we can show
Hence,
Note: Area of the traingle ABC is denoted by A.
Proof:
Let angle C be acute in figure (i), obtuse in figure (ii) and the right angle in figure (iii)
In each triangle, draw AD perpendicular on BC i.e. (on produced if necessary)
In each figure,
or AD = c sinB ...(i)
[∵ AB = C]
In figure (i),
or AD = b sinC [∵ AC = b]
In figure (ii),
or AD = b sinC
In figure (iii),
or AD = b sinC
In all figures, AD = b sin C ...(ii)
By (i) and (ii), we get
Similarly,
Hence,
Proof:
Let angle C be acute in figure (i), obtuse in figure (ii) and right angle in figure
(iii) Draw AD BC in all the triangles (on produced if necessary)
....(i)
In figure (i),
In figure (ii),
In figure (iii),
∴ In each figure, AD = b sin C ....(ii)
In figure (ii),
In figure (i),
In figure (iii),
∴ In each figure, ....(iii)
Using (ii) and (iii) in (i), we get
Similarly,
Hence proved.
In any triangle ABC, prove that:
(i) a = b cos C + c cos B (ii) b = c cos A + a cos C (iii) c = a cos B + b cos A
Let angle C be acute in figure (i), obtuse in figure (ii) and right angle in figure
(on produced if necessary)
(iii) Draw AD ⊥ BC in all the triangles.
In figure (i),
= c cos B + b cos C ...(i)
In figure (ii),
= c cos B + b cos C ...(ii)
In figure (iii),
= c cos B + b cos C ...(iii)
From (i), (ii) and (iii), we have
a = b cos C + c cos B
Simillarly, we can prove
b = c cos A + a cos C
c = a cos B + b cos A
Prove that in any triangle ABC.
Draw AD BC, then
Similarly, we can show that
Further, since
In any triangle ABC, prove that
Proof:
We have,
[By sine formula]
Similarly, we can show that
Prove that for any triangle ABC
where R is the radius of the circumcircle.
Proof:
Let be acute in figure (i), obtuse in figure (ii) and right angle in figure (iii)
Join BO and produce it to meet the circle in A' and join CA',
then triangle BCA' is a right angle triangle in (i) and (ii), but C and A' coincide in (iii).
In figure (i),
Since angles in the same segment are equal
∴
[By sine formula]
[∵ BCA' =
]
In figure (ii),
(∵ Sum of opposite angles of a cyclic quadrilateral is )
i.e.
In figure (iii),
[∵
i.e. angle is semi-circle]
In all cases, we have
But,
Hence,
Let I be incentre of the triangle ABC
Join ID, IE and IF which are perpendicular to BC, CA and AB respectively
ID = IE = IF = r (In radius)
Now, area of the triangle ABC =
= Area of triangle IBC + Area of triangle ICA + Area of triangle IAB
or
Proof: (i) Let I be the incentre and D, E, F be the points of contact of the incircle with the sides.
Join ID, IE, IF such that
ID BC, IE
CA and IF
AB
Now, 2s - 2a = AB + CA - BC
= (AF + FB) + (CE + EA) - (BD + DC)
= (AF + BD) + (CD + AF) - (BD + DC)
[∵ AE = AF, BF = BD and CD = CE]
= 2AF AF = s - a ...(i)
Now, from right angle triangle AFI,
or [By using (i)]
Similarly, we can show other two results.
(ii) Now, [From figure]
or
Similarly, we can show other two results.
Length of arc AB, l = 21 cm
Radius of pendulum, r = 75 cm
Using we have
∴ (a)
(b)
The 4th term in the expansion of is
Find the values of a and n.
The expression can be written as
sin A cos A+1
sec A cosec A+1
tan A + cot A
Sec A + cosec A
B.
sec A cosec A+1
Given expression is
Statement − 1: For every natural number n ≥ 2
Statement −2: For every natural number n ≥ 2,
Statement −1 is false, Statement −2 is true
Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1
Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.
Statement − 1 is true, Statement − 2 is false.
C.
Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.
If A = and I =
, then which one of the following holds for all n ≥ 1, by the principle of mathematical induction
An = nA – (n – 1)I
An = 2n-1A – (n – 1)I
An = nA + (n – 1)I
An = 2n-1A + (n – 1)I
A.
An = nA – (n – 1)I
By the principle of mathematical induction (1) is true.
Let S(K) = 1 +3+5+..... (2K-1) = 3+K2. Then which of the following is true?
S(1) is correct
Principle of mathematical induction can be used to prove the formula
S(K) ≠S(K+1)
S(K)⇒ S(K+1)
D.
S(K)⇒ S(K+1)
S(K) = 1 + 3 + 5 + ...... + (2K - 1) = 3 + K2
Put K = 1 in both sides
∴ L.H.S = 1 and R.H.S. = 3 + 1 = 4 ⇒ L.H.S. ≠ R.H.S.
Put (K + 1) on both sides in the place of K L.H.S. = 1 + 3 + 5 + .... + (2K - 1) + (2K + 1)
R.H.S. = 3 + (K + 1)2 = 3 + K2 + 2K + 1
Let L.H.S. = R.H.S.
1 + 3 + 5 + ....... + (2K - 1) + (2K + 1) = 3 + K2 + 2K + 1
⇒ 1 + 3 + 5 + ........ + (2K - 1) = 3 + K2 If S(K) is true, then S(K + 1) is also true. Hence, S(K) ⇒ S(K + 1)
Maximum sum of coefficient in the expansion of (1 – x sinθ + x2 )n is
1
2n
3n
0
C.
3n
Sum of coefficients in (1 – x sinθ + x2 )n is (1 – sinθ + 1)n
(putting x = 1)
This sum is greatest when sinθ = –1, then maximum sum is 3n .
Sponsor Area
Sponsor Area