Introduction To Trigonometry

Applying Pythagoras theorem for ΔPQR, we obtain

PR² = PQ² + QR²

(13)² = (12)² + QR²

169 cm² = 144 cm² + QR²

25 cm² = QR²

QR = 5 cm

tan P =

=

cot R =

=

tan P - cot R =

= 0
tan P - cot R = 0

Let ΔABC be a right-angled triangle, right-angled at point B

Given that:

As we know trigonometrical ratios give ratio between sides of right angled triangle instead of their actual values. So from sin A we get the ratio of Perpendicular and Hypotenuse of the triangle. Now as we don't know the exact value let Perpendicular = 3 k and Hypotenuse = 4 k

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

(4k)² = AB² + (3k)²

16k² - 9k² = AB²

7k² = AB²

Cos A =

= =

Tan A =

= =

Consider a right-angled triangle, right-angled at B

Given: 15 cot A = 8
To find: sin A and sec A



Cot A =

It is given that,

Cot A =

Let AB be 8k. Therefore, BC will be 15k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

= (8k)² + (15k)²

= 64k² + 225k²

= 289k²

AC = 17k

Sin A =

=

Sec A =

=

Consider a right-angle triangle ΔABC, right-angled at point B

sec =

=

If AC is 13k, AB will be 12k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

(AC)² = (AB)² + (BC)²

(13 k)² = (12 k)² + (BC)²

169 k² = 144 k² + BC²

25 k² = BC²

BC = 5 k

Sin =

= =

Cos θ =

= =

Tan θ =

= =

Cot θ =

= =

Cosec θ =

= =

Let us consider a right triangle ABC, right-angled at point B

As, a trigonometric Ratio shows the ratio between different sides, cot also shows the ratio of Base and Perpendicular. As we don't know the absolute value of Base and Perpendicular, let the base and perpendicular be multiplied by a common term, let it be k
Therefore,
Base = 7 k
Perpendicular = 8k
According to pythagoras theorem,
(Hypotenuse)² = (Base)² + (Perpendicular)²

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

= (8k)² + (7k)²

= 64k² + 49k²

= 113k²

AC = √113 k

Sin =

=

=

Cos θ =

=

=

(i)
Putting the obtained trigonometric ratios into the expression we get,
= (1 – sin² θ)/(1 – cos² θ)

= 49/64

(ii) Cot² θ = (cot θ)²

It is given that 3 cot A = 4

Consider a right triangle ABC, right-angled at point B

To Find : Cos²A – Sin²A or not

Cot A =

=

This fraction shows that if the length of base will be 4 then the altitude will be 3. And base and altitude will both increase with this propotion only. Let AB be 4k then BC will be 3k

In ΔABC,
Pythagoras theorem : It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
By pythagoras theorem

(AC)² = (AB)² + (BC)²

= (4k)² + (3k)²

= 16k² + 9k²

= 25k²

AC = 5k

Cos A =

=

=

Sin A =

=

=

Tan A =

=

=

= 7/25

Cos²A – Sin²A =

=

=

Therefore,

Cos²A – Sin²A
Hence, the expression is correct.

For finding value of sin A cos C + cos A sin C , we need to find values of sin A, cos C, cos A, sin C

tan A = =

If BC is k, then AB will be, where k is a positive integer

In ΔABC,

AC² = AB² + BC²

= ()² + (k)²

= 3k² + k² = 4k²

AC = 2k
For finding the perpendicular and base for an angle of a right angled triangle, always take the side opposite to the angle as perpendicular and the other side as base

We know that,

Sin A =

= =

Cos A =

= =

Sin C =

= =

Cos C =

= =

(i) sin A cos C + cos A sin C

=

=

=

(ii) cos A cos C - sin A sin C

= () () – () ()

=

= 0

Given that, PR + QR = 25

PQ = 5

Let PR be x

Therefore,

QR = 25 - x

Applying Pythagoras theorem in ΔPQR, we obtain

PR² = PQ² + QR²

x² = (5)² + (25 - x)²

x² = 25 + 625 + x² - 50x [as, (a + b)² = a² + b² + 2ab]

50x = 650

x = 13

Therefore,

PR = 13 cm

QR = (25 - 13) cm = 12 cm

sin P =

cos P =

tan P =

(i) Consider a ΔABC, right-angled at B

Tan A =

But > 1

Tan A > 1

So, Tan A < 1 is not always true

Hence, the given statement is false

(ii) Sec A =

Let AC be 12k, AB will be 5k, where k is a positive integer

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

(12k)² = (5k)² + BC²

144k² = 25k² + BC²

BC² = 119k²

BC = 10.9k

It can be observed that for given two sides AC = 12k and AB = 5k,

BC should be such that,

AC - AB < BC < AC + AB

12k - 5k < BC < 12k + 5k

7k < BC < 17 k

However, BC = 10.9k. Clearly, such a triangle is possible and hence, such value of sec A is possible

Hence, the given statement is true

(iii) Abbreviation used for cosecant of angle A is cosec A. And Cos A is the abbreviation used for cosine of angle A

Hence, the given statement is false

(iv) Cot A is not the product of cot and A. It is the cotangent of ∠A

Hence, the given statement is false

(v) sin θ =

In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin θ is not possible

Hence, the given statement is false

Given is the table of Trigonometrical ratios for different values. Use the values from table for solving the questions.

(i) sin60° cos30° + sin30° cos 60°

= 1

(ii) 2 tan²45° + cos²30° - sin²60°

= 2 (1)² + (√3/2)² - (√3/2)²

= 2 + 3/4 - 3/4

= 2

(iii)

(iv)

(v)

(i) We know that,


Putting values in given equation,

in options we have,

sin 60° = √3/2

cos 60° = ½

tan 60° = √3

sin 30° = ½

Only (A) is correct.

=

=

= 0

Hence, (D) is correct

Out of the given alternatives, only A = 0° is correct

As

sin 2A = sin 0° = 0

2 sin A = 2sin 0° = 2(0) = 0

Hence, (A) is correct

Now according to options,

Out of the given alternatives, only tan 60° =

Hence, (C) is correct

Now, we know that



So, from the question we can tell,

tan (A + B) =

tan (A + B) = tan 60^o

A + B = 60° .......eq (i)

tan (A – B) =

tan (A - B) = tan 30°

A - B = 30 ...........eq(ii)

On adding both equations, we obtain

2 A = 90°

⇒ A = 45

From equation (i), we obtain

45 + B = 60

B = 15^o

Therefore, ∠A = 45° and ∠B = 15°

(i) sin (A + B) = sin A + sin B

Let A = 30° and B = 60°

sin (A + B) = sin (30° + 60°) = sin 90°

= 1

sin A + sin B = sin 30° + sin 60°

=

=

Clearly, sin (A + B) ≠sin A + sin B

Hence, the given statement is false

(ii) The value of sin θ increases as θ increases in the interval of 0° < θ < 90° as sin 0° = 0

sin 30^o =

sin 45^o = = 0.707

sin 60^o = = 0.886

sin 90° = 1

Hence, the given statement is true

(iii) cos 0° = 1

cos 30^o =

cos 45^o =

cos 60^o =

cos 90° = 0

It can be observed that the value of cos θ does not increase in the interval of 0° < θ < 90°

Hence, the given statement is false

(iv) sin θ = cos θ for all values of θ.

This is true when θ = 45°

As,

sin 45^o =

cos 45^o =

It is not true for all other values of θ

As sin 30^o = and cos 30^o =

Hence, the given statement is false

(v) cot A is not defined for A = 0°

cot A =

cot 0^o =

= =undefined

Hence, the given statement is true

(i)
(18^o = 90^o- 72^o)

= 1

(ii)

(iii) cos 48^o - sin 42^o

= cos (90^o - 42^o) – sin 42^o

= sin 42^o - sin 42^o

= 0

(iv) cosec 31^o - sec 59^o

= cosec(90^o - 59^o) – sec 59^o

= sec 59^o - sec 59^o

= 0

(i) tan 48^o tan 23^o tan 42^o tan 67^o = 1

LHS: tan 48^o tan 23^o tan 42^o tan 67^o

= tan (90^o - 42^o) tan 23^o tan 42^o tan (90^o - 23^o)
As tan(90 - θ) = cot θ

= cot 42^o tan 23^o tan 42^o cot 23^o

= cot 42^o tan 42^o tan 23^o cot 23^o
As, tanθ cotθ = 1, because

= 1 x 1

= 1 = RHS

Hence tan 48^o tan 23^o tan 42^o tan 67^o = 1, Proved

(ii) cos 38^o cos 52^o - sin 38^o sin 52^o = 0

LHS= cos(90^o - 52^o) cos(90^o - 38^o) - sin 38^o sin 52^o

As we know,
cos(90 - θ) = sin θ
= sin 38^o sin 52^o - sin 38^o sin 52^o

= 0
=RHS

Hence cos 38^o cos 52^o - sin 38^o sin 52^o = 0 , proved

Given: tan 2A = cot(A - 18)
By complementary angle formula: cot(90 - A) = tan A
From the question,
tan 2 A = cot (90 - 2A)
Therefore,
cot(90 - 2A) = cot(A - 18)
Now both the angles will be equal, so

90° – 2A = A – 18°

108° – 2A = A

3A = 108°

A = 108/3
= 36°
A = 36°

To Prove: A + B = 90°
Formula to use = [tan (90° - A) = cot A] and [cot(90° - A) = tan A]
Proof:
tan A = cot (90°- A) (by complementary angles formula)
tan A = cot B

cot (90°- A) = cot B [ As, cot(90° - A) = tan A]
Therefore,

90° - A = B

A + B = 90°

Hence proved

sec 4A = cosec (90° - 4A) = cosec (A - 20°)

Therefore,

90° – 4A = A - 20°

110°– 4A = A

5A = 110°

A = 22°

To Prove:

Proof:
We know, By angle sum property of a triangle

∠A + ∠B + ∠C = 180°

⇒ B + C = 180°- A

Therefore,

As, sin(90 - θ) = cos θ
Therefore,

Hence proved

sin 67° + cos 75°

For expressing this into between 0° and 45° we need to know complementary angle formulas:

sin(90° - A) = cos A

cos(90° - A) = sin A

Hence,

Now we can break 67° as 90° – 23 °.

And 75° as 90° – 15°.

So,

sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°)

sin 67° + cos 75° = cos 23° + sin 15°

(Since, sin(90°- A) = cos A)

sin A can be expressed in terms of cot A as:

And we know that: cosec² A - cot² A = 1, cosec²A=1+cot²A , so cosec A=√(1+cot²A)
Therefore,

Now,

Sec A can be expressed in terms of cot A as:
We know that: sec² A - tan² A = 1, sec A = √(1 + tan² A)
And also, tan A = 1/ cot A
Therefore,

tanA can be expressed in terms of cotA as:

tanA =

SinA can be expressed in terms of sec A as:

sinA = √sin²A

sin A = √(1 -cos²A)

Now,

cos A can be expressed in terms of secA as:

cosA =

tanA can be expressed in the form of sec A as:
As, 1 + tan²A = sec²A

⇒ tan A =± √(sec²A -1)
As A is acute angle,
And tan A is positive when A is acute,
So,
tan A = √(sec²A -1)
cosec A can be expressed in the form of sec A as:

cosec A =

cot A can be expressed in terms of sec A as:

cot A =
as, 1 + tan²A = sec²A
=

(i)

(ii) sin 25°cos 65°+ cos 25°sin 65°

= sin 25° cos (90°-25°) + cos 25°sin 65°
= sin 25°sin 25° + cos 25°sin 65°
= sin² 25° + cos 25° sin (90°-25°)

= sin² 25° + cos² 25°

=1

Following is the explanation:

9 sec² A – 9 tan² A

= 9 (sec² A – tan² A)

= 9 x 1

= 9

Consider ( 1 + tanθ + secθ)(1 + cotθ - cosecθ) ,

=

=

Multiplying both terms, we get

= 2
Therefore, ( 1 + tanθ + secθ)(1 + cotθ - cosecθ) = 2

=

=

= (1 – sin²A)/cosA

= cos²A/cosA

= cos A

We know,

1 + tan²θ = sec²θ

and
1 + cot²θ = cosec²θ

Therefore,



⇒

Therefore, option (D) is correct

(i)
To Prove:

Proof:

=

=

Since,

=

= RHS
Hence, proved

(ii)
To Prove:

Proof:
LHS = +

Use the identity sin²θ + cos²θ = 1

=

=

=

=

= 2 sec A

= RHS

(iii)

[Hint: Write the expression in terms of sin θ and Cos θ]

=

Use the formula a³ - b³ = (a² + b² + ab)(a - b)

Cancelling (sin θ - cos θ) from numerator and denominator


As cosθ = 1/secθ and sinθ = 1/cosecθ

= 1 + sec θ cosec θ

= RHS

(iv)

[Hint: Simplify LHS and RHS separately]

LHS

use the formula secA = 1/cosA

= cos A + 1

RHS

Use the identity sin²θ + cos²θ = 1

Use the formula a² - b² = (a - b) ( a + b )

= cos A + 1

LHS = RHS

(v)using the identity

LHS

Dividing Numerator and Denominator by sin A


Use the formula cotθ = cosθ / sinθ

Using the identity


Use the formula a² - b² = (a - b) ( a + b )



= cot A + cosec A

= RHS

(vi)

Dividing numerator and denominator of LHS by cos A

As cosθ = 1/secθ and tanθ = sinθ /cosθ

Rationalize the square root to get,

Use the formula a² - b² = (a - b) ( a + b ) to get,


Use the identity sec²θ = 1 + tan²θ to get,

= sec A + tan A

= RHS

(vii)
To Prove:

Proof:
LHS


Since



As tanθ = sinθ/cosθ
= tanθ
= R.H.S
Hence, Proved.
(viii)
To Prove:

Proof:
LHS = (sin A + cosec A)² + (cos A + sec A)²

Use the formula (a+b)² = a² + b² + 2ab to get,

= (sin²A + cosec²A + 2sin A cosec A) + (cos²A + sec²A + 2 cos A sec A)
Since sinθ=1/cosecθ and cosθ = 1/secθ

= sin²A + cosec²A + 2 + cos²A + sec²A + 2
=(sin²A + cos²A)+cosec²A+sec²A+2+2
Use the identities sin²A + cos²A = 1, sec²A = 1 + tan²A and cosec²A = 1 + cot²A to get

= 1+ 1 + tan²A + 1 + cot²A + 2 + 2

= 1 + 2 + 2 + 2 + tan²A + cot²A

= 7 + tan²A + cot²A

= RHS

(ix)
To Prove:

[Hint: Simplify LHS and RHS separately]

Proof:
LHS = (cosec A – sin A) (sec A – cos A)
Use the formula sinθ = 1/cosecθ and cosθ = 1/secθ

= cos A sin A

RHS

use the formula tanθ=sinθ/cosθ and cotθ=cosθ/sinθ



Use the identity sin²A + cos²A = 1

= cos A sin A

LHS = RHS

(x)
To Prove:

Proof:
Taking left most term
Since, cot A is the reciprocal of tan A, we have

= right most part
Taking middle part:

= right most part

Hence, Proved.

Given equation, 2x – 3y = 4

⇒ 3y = 2x-4

When x=-4, then,

⇒ y = -4

When x=-1, then,

⇒ y = -2

When x=2, then,

⇒ y = 0

When x=5, then,

⇒ y = 2

Thus we have the following table,

On plotting these points we have the following graph,

Clearly, from the graph (-1, -2) is the solution of the line 2x – 3y = 4