Introduction To Trigonometry Class 10th Mathematics CBSE Solution

Class 10^th Mathematics CBSE Solution

Exercise 8.1

1. In Fig. 8.13, find tan P - cot R. c p 12cm]
2. If sin a = 3/4 calculate Cos A and tan A
3. Given 15 cot A = 8, find sin A and sec A
4. Given sec theta = 13/12 calculate all other trigonometric ratios
5. If A and B are acute angles such that Cos A = Cos B, then show that A = B…
6. If cottheta = 7/8 evaluate: (i) (1+sintegrate heta) (1-sintegrate…
7. If 3 cot A = 4, check whether 1-tan^2a/1+tan^2a = cos^2a-sin^2a or not…
8. In triangle ABC, right-angled at B, if tana = 1/root 3 find the value of: (i)…
9. In PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values…
10. State whether the following are true or false. Justify your answer. (i) The…

Exercise 8.2

1. Evaluate the following: (i) 60^circle cos30^circle +sin30^circle cos60^circle…
2. Choose the correct option and justify your choice: 2tan30^circle
3. 1-tan^245^circle /1+tan^245^circle = Choose the correct option and justify your…
4. Sin 2A = 2 sin A is true when A = Choose the correct option and justify your…
5. 2tan30^circle /1-tan^230^circle = Choose the correct option and justify your…
6. If tan (a+b) = root 3 and tan (a-b) = 1/root 3 0^circle a+b less than equal to…
7. State whether the following are true or false. Justify your answer (i) Sin (A +…

Exercise 8.3

1. Evaluate: (i) sin18^circle /cos72^circle (ii) tan26^circle /cot64^circle (iii)…
2. Show that:(i) tan 48 tan 23 tan 42 tan 67 = 1(ii) cos 38 Cos 52 - sin 38 sin 52…
3. If tan 2A = cot (A - 18), where 2A is an acute angle, find the value of A…
4. If tan A = cot B, prove that A + B = 90
5. If sec 4A = cosec (A - 20), where 4A is an acute angle, find the value of A…
6. If A, B and C are interior angles of a triangle ABC, then show that sin (b+c/2)…
7. Express sin 67 + Cos 75 in terms of trigonometric ratios of angles between 0 and…

Exercise 8.4

1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A…
2. Write all the other trigonometric ratios of A in terms of sec A.
3. Evaluate: (i) sin^263^circle + sin^227^circle /cos^217^circle + cos^273^circle…
4. 9sec^2a-9tan^2a = Choose the correct option. Justify your choice.A. 1 B. 9 C. 8…
5. (1+tantheta +sectheta) (1+cottheta -costheta c theta) = Choose the correct…
6. Choose the correct option. Justify your choice.(Sec A + tan A) (1 - sin A) =A.…
7. 1+tan^2a/1+cot^2a = Choose the correct option. Justify your choice.A. sec^2a B.…
8. Prove the following identities, where the angles involved are acute angles for…

Testing

1. Draw the graph of 2x – 3y = 4. From the graph, find whether x = -1, y = -2 is a solution…

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Exercise 8.1

Question 1.

In Fig. 8.13, find tan P – cot R.

Answer:

Applying Pythagoras theorem for ΔPQR, we obtain

Pythagoras Theorem : the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

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

Question 2.

If sin  calculate Cos A and tan A

Answer:

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

According to Pythagoras theorem: (Hypotenuse)² = (Base)² + (Perpendicular)²

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

(4k)² = AB² + (3k)²

16k² - 9k² = AB²

7k² = AB²

Cos A = 

=  = 

And we know that 

Tan A = 

=  = 

So the ratios are cos A = √7/4 and tan A = 3/√7 
Question 3.

Given 15 cot A = 8, find sin A and sec A

Answer:

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.

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.

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

= (8k)² + (15k)²

= 64k² + 225k²

= 289k²

AC = 17k

Now we know that,

Sin A = 

= 

And also,

Sec A = 

= 

Question 4.

Given sec calculate all other trigonometric ratios

Answer:

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.

Now according to Pythagoras theorem,

The square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

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 = 

=  = 

Question 5.

If ∠ A and ∠ B are acute angles such that Cos A = Cos B, then show that ∠A = ∠B

Answer: 
To Prove: ∠ A = ∠ B
Given: cos A = cos B
Let there be a right angled triangle ABC, right angled at C.
Now we know that the side opposite to the angle of which we are taking trigonometric ratio is perpendicular, side opposite to right angle is hypotenuse and the side left is base.
So, now let cos of A and B
We know that,

Since for angle A, BC is perpendicular and AC is the base.
Now,

Since for angle B, BC is base and AC is perpendicular.
Given: cos A = cos B
Therefore,
AC = BC
Now, the angles opposite to the equal sides are also equal.
Therefore, ∠A = ∠B
Hence, Proved. 
Question 6.

If evaluate:

(i) 

(ii) 

Answer:

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

Given:

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

Now we have all the three sides of the triangle, 
Base = 7 k
Perpendicular = 8 k
Hypotenuse = √113 k
Now applying other trigonometric angle formulas

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

Question 7.

If 3 cot A = 4, check whether  or not

Answer:

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

Now,

Cos A = 

= 

= 

And also we know that,

Sin A = 

= 

= 

And ,

Tan A = 

= 

= 

Now putting the values so obtained we get,

= 7/25

Cos²A – Sin²A = 

= 

= 

Therefore,

 Cos²A – Sin²A
Hence, the expression is correct.

Question 8.

In triangle ABC, right-angled at B, if find the value of:

(i) Sin A Cos C + Cos A Sin C

(ii) Cos A Cos C – Sin A Sin C

Answer:

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,

Now, for angle A, Perpendicular will be side opposite to angle A that is BC and Hypotenuse will be AC

Sin A = 

=  = 

And also by cosine formula, 

For angle A, base will be AB and hypotenuse will be AC

Cos A = 

=  = 

For angle C, Perpendicular will be AB and hypotenuse will be AC

Sin C = 

=  = 

For angle C, base will be equal to BC and hypotenuse equal to AC

Cos C = 

=  = 

(i) sin A cos C + cos A sin C

= 

= 

= 

(ii) cos A cos C - sin A sin C

= () () – () ()

= 

= 0

Question 9.

In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of Sin P, Cos P and tan P.

Answer:

Given that, PR + QR = 25

PQ = 5

Let PR be x

Therefore,

QR = 25 - x

Pythagoras Theorem: the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

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

and we know, 

sin P = 

cos P = 

tan P = 

Question 10.

State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1

(ii) Secfor some value of angle A

(iii) Cos A is the abbreviation used for the cosecant of angle A

(iv) Cot A is the product of cot and A.

(v) for some angle θ

Answer:

(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

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Exercise 8.2

Question 1.

Evaluate the following:
(i) 

(ii) 

(iii) 

(iv) 

(v) 

Answer: 

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)

As the denominator has an irrational number, we need to rationalize the fraction

(iv)

(v)

Question 2.

Choose the correct option and justify your choice:

A. sin 60°

B. cos 60°

C. tan 60°

D. sin 30°

Answer:

(i) We know that,


Putting values in given equation, 

[Since, √3 × √3 = 3]

in options we have,

sin 60° = √3/2

cos 60° = ½

tan 60° = √3

sin 30° = ½

Only (A) is correct.

Question 3.

Choose the correct option and justify your choice:


A. tan 90°

B. 1

C. sin 45°

D. 0

Answer:

= 

= 

= 0

Hence, (D) is correct

Question 4.

Choose the correct option and justify your choice:

Sin 2A = 2 sin A is true when A =
A. 0°

B. 30°

C. 45°

D. 60°

Answer:

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

Question 5.

Choose the correct option and justify your choice:


A. Cos 60°

B. sin 60°

C. tan 60°

D. sin 30°

Answer:

Now according to options,

Out of the given alternatives, only tan 60° = 

Hence, (C) is correct

Question 6.

If  and  find A and B

Answer:

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)

And

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°

Question 7.

State whether the following are true or false. Justify your answer

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

(ii) The value of sin θ increases as θ increases

(iii) The value of cos θ increases as θ increases.

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

(v) Cot A is not defined for A = 0°

Answer:

(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

Clearly the value increases.

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

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Exercise 8.3

Question 1.

Evaluate:

(i) 

(ii) 

(iii) Cos 48° – sin 42°

(iv) Cosec 31° – sec 59°

Answer:

(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

Question 2.

Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° Cos 52° – sin 38° sin 52° = 0

Answer:

(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

Question 3.

If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Answer:

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°

Question 4.

If tan A = cot B, prove that A + B = 90°

Answer:

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

Question 5.

If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A

Answer:

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

Therefore,

90° – 4A = A - 20°

110°– 4A = A

5A = 110°

A = 22°

Question 6.

If A, B and C are interior angles of a triangle ABC, then show that

Answer:

To Prove: 

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

∠A + ∠B + ∠C = 180°

⇒ B + C = 180°- A

Taking sin on both sides of equation

Therefore,

As, sin(90 - θ) = cos θ
Therefore, 

Hence proved

Question 7.

Express sin 67° + Cos 75° in terms of trigonometric ratios of angles between 0° and 45°

Answer:

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)

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Exercise 8.4

Question 1.

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A

Answer:

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 =

Question 2.

Write all the other trigonometric ratios of ∠A in terms of sec A.

Answer:

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
=

Question 3.

Evaluate:
(i) 
(ii) Sin 25° Cos 65° + Cos 25° Sin 65°

Answer:

(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

Question 4.

Choose the correct option. Justify your choice.


A. 1

B. 9

C. 8

D. 0

Answer:

Following is the explanation:

9 sec² A – 9 tan² A

= 9 (sec² A – tan² A)

= 9 x 1

= 9

Question 5.

Choose the correct option. Justify your choice.


A. 0

B. 1

C. 2

D. –1

Answer:

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

=

= 

Multiplying both terms, we get

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

Question 6.

Choose the correct option. Justify your choice.
(Sec A + tan A) (1 – sin A) =
A. sec A
B. sin A
C. cosec A
D. Cos A

Answer:

= 

= 

= (1 – sin²A)/cosA

= cos²A/cosA

= cos A

Question 7.

Choose the correct option. Justify your choice.


A. 

B. -1

C. 

D. 

Answer:

We know,

1 + tan²θ = sec²θ

and 
1 + cot²θ = cosec²θ

Therefore, 



⇒

Therefore, option (D) is correct

Question 8.

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) 
(ii) 
(iii) 
(iv) 
(v) using the identity 
(vi) 
(vii) 
(viii)
(ix)  [Hint: Simplify LHS and RHS separately]
(x) 

Answer:

(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.

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Testing

Question 1.

Draw the graph of 2x – 3y = 4. From the graph, find whether x = -1, y = -2 is a solution or not. (Final)

Answer:

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,

X	-4	-1	2	5
Y	-4	-2	0	2

On plotting these points we have the following graph,

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