Trigonometry Solution of TS & AP Board Class 10 Mathematics

Trigonometry Solution of  TS & AP Board Class 10 Mathematics

---

Exercise 11.1

Question 1.

In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA respectively. Then, find out sin A, cos A and tan A.

Answer:

We have

Given: ∠ABC = 90°, AB = 8 cm, BC = 15 cm and CA = 17 cm

We know sin A is given by,

⇒  [∵, perpendicular is the side opposite to the angle A & hypotenuse is the side opposite to the right angle of that triangle]

⇒  …(i)

Also, cos A is given by

⇒ 

⇒  …(ii)

Now, tan A can be found out by two ways:

Method 1: tan A is given by,

⇒ 

⇒ 

Method 2: tan A can also be written as,

⇒  [from equations (i) & (ii)]

⇒ 

Thus, ,  and .

Question 2.

The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25 cm and ∠Q = 90° respectively. Then find, tan P – tan R.

Answer:

We have

We don’t need to find hypotenuse (PR) in the ∆PQR as tan θ = perpendicular/base.

⇒ 

And 

⇒ 

tan P – tan R = 

⇒ tan P – tan R = 

⇒ tan P – tan R = 576/175

Thus, tan P – tan R = 576/175

Question 3.

In a right angle triangle ABC with right angle at B, in which a = 24 units, b = 25 units and ∠BAC = θ. Then, find cosθ and tanθ.

Answer:

We have

In ∆ABC, ∠ABC = 90° and ∠BAC = θ.

Using this information, we can say

AC = hypotenuse of the triangle

BC = perpendicular (side opposite to the angle θ or ∠BAC)

Using Pythagoras theorem,

(hypotenuse)² = (perpendicular)² + (base)²

⇒ (25)² = (24)² + (base)²

⇒ (AB)² = 625 – 576

⇒ (AB)² = 49

⇒ AB = √49 = 7 units

So, we have AB = 7 units, BC = 24 units and AC = 25 units.

Thus, 

⇒ 

And 

⇒ 

Thus,  and .

Question 4.

If  then find sin A and tan A

Answer:

Given that,

But 

⇒ 

⇒ base = 12 and hypotenuse = 13

So, using Pythagoras theorem, we can say

(hypotenuse)² = (perpendicular)² + (base)²

⇒ (perpendicular)² = (hypotenuse)² – (base)²

⇒ (perpendicular)² = (13)² – (12)²

⇒ (perpendicular)² = 169 – 144 = 25

⇒ perpendicular = √25 = 5

Using perpendicular = 5, base = 12 and hypotenuse = 13, we can find out sin A and tan A.

Sin A is given by

⇒ 

And, tan A is given by

⇒ 

Thus,  and .

Question 5.

If 3 tanA = 4, then find sin A and cos A.

Answer:

Given that, 3 tan A = 4

⇒ 

But 

⇒ 

⇒ perpendicular = 4 and base = 3

So, using Pythagoras theorem, we can say

(hypotenuse)² = (perpendicular)² + (base)²

⇒ (hypotenuse)² = (4)² + (3)²

⇒ (hypotenuse)² = 16 + 9 = 25

⇒ hypotenuse = √25 = 5

Using perpendicular = 4, base = 3 and hypotenuse = 5, we can find out sin A and cos A.

Sin A is given by

⇒ 

And, cos A is given by

⇒ 

Thus,  and .

Question 6.

If ∠A and ∠X are acute angles such that cos A = cos X then show that ∠A = ∠X.

Answer:

We have

In this ∆AOX,

cos A = cos X

and cos A is given by,

 [∵, ]

Similarly, 

⇒ 

⇒ AO = OX [clearly, since denominator from either sides cancel each other]

Now, if sides AO and OX of ∆AOX are equal.

Then, ∠A = ∠X [∵, In a triangle, angles opposite to the equal sides are also equal]

Hence, ∠A = ∠X

Question 7.

Given  then evaluate

A. 

Answer:

We have been given that,

And we have to solve for .

We know the formula: (x + y)(x – y) = x² – y²

Using this,

Also, we know the relationship between cos θ and sin θ which is given by

cos² θ + sin² θ = 1

⇒ cos² θ = 1 – sin² θ

So,

⇒ 

As we know,

So, B = 7 and P = 8

By Pythagoras theorem,

H² = P² + B²
H² = 8² + 7²
= 64 + 49
=113

H = √113

As we know,

So,

Now,

Question 8.

Given  then evaluate

B. 

Answer:

Given,

We know that, cosec²θ = 1 + cot²θ

Now, To Find:

Dividing both numerator and denominator by sin θ, we have

Question 9.

In a right angle triangle ABC, right angle is at B, if  then find the value of

A. sin A cosC + cos A sin C

Answer:

We have

Given that, tan A = √3/1

And tan A is given by,

⇒ 

⇒ perpendicular = √3x and base = x

Then, we can use Pythagoras theorem in ∆ABC,

(hypotenuse)² = (perpendicular)² + (base)²

⇒ (hypotenuse)² = (√3x)² + (x)²

⇒ (hypotenuse)² = 3x² + x² = 4x²

⇒ hypotenuse = √(4x²) = 2x

We have, AB = √3x, BC = x and AC = 2x.

Using these values,

⇒ 

⇒ 

⇒  …(i)

Similarly, 

⇒ 

⇒  …(ii)

Also,

⇒ 

⇒ 

⇒  …(iii)

Similarly, 

⇒ 

⇒  …(iv)

We have to solve: sin A cos C + cos A sin C.

Substituting equations (i), (ii), (iii) & (iv) in above,

sin A cos C + cos A sin C = 

⇒ sin A cos C + cos A sin C = 1/4 + 3/4

⇒ sin A cos C + cos A sin C = 4/4 = 1

Thus, sin A cos C + cos A sin C = 1.

Question 10.

In a right angle triangle ABC, right angle is at B, if  then find the value of

B. cos A cos C – sin A sin C

Answer:

To find: cos A cos C – sin A sin C.

From previous part of the question, we have

Using these values, we get

cos A cos C – sin A sin C = 

⇒ cos A cos C – sin A sin C = √3/4 - √3/4 = 0

Thus, cos A cos C – sin A sin C = 0.

---

Exercise 11.2

Question 1.

Evaluate the following.

A. sin 45° + cos 45°

Answer:

By trigonometric identities, we can say

sin 45° = 1/√2

and cos 45° = 1/√2

Adding them, we get

sin 45° + cos 45° = 1/√2 + 1/√2

⇒ sin 45° + cos 45° = 2/√2 = √2

Thus, sin 45° + cos 45° = √2.

Question 2.

Evaluate the following.

B. 

Answer:

Trigonometric identities:

cos 45° = 1/√2

sec 30° = 1/cos 30° = 2/√3 [∵, cos 30° = √3/2]

cosec 60° = 1/sin 60° = 2/√3 [∵, sin 60° = √3/2]

Putting the values we get,

Question 3.

Evaluate the following.

C. 

Answer:

By trigonometric identities,

cot 45° = 1/tan 45° = 1/1 = 1

sec 30° = 1/cos 30° = 2/√3 [∵, cos 30° = √3/2]

cosec 60° = 1/sin 60° = 2/√3 [∵, sin 60° = √3/2]

sin 30° = 1/2

tan 45° = 1

cos 60° = 1/2

Putting all these values in , we get

Since, numerator is equal to denominator in the above calculation, we can say

Question 4.

Evaluate the following.

D. 2tan² 45° + cos² 30° - sin² 60°

Answer:

By trigonometric identities,

sin 60° = √3/2

tan 45° = 1

cos 30° = √3/2

Putting these values in 2 tan² 45° + cos² 30° - sin² 60°, we get

2 tan² 45° + cos² 30° - sin² 60° = 2(1)² + (√3/2)² – (√3/2)²

⇒ 2 tan² 45° + cos² 30° - sin² 60° = 2

Thus, 2 tan² 45° + cos² 30° - sin² 60° = 2.

Question 5.

Evaluate the following.

E. 

Answer:

We have to solve:

Recall the trigonometric identities,

sin² θ + cos² θ = 1

& sec² α – tan² α = 1

Put θ = 30° and α = 60°, we get

sin² 30° + cos² 30° = 1

& sec² 60° - tan² 60° = 1

So,

Thus,

Question 6.

Choose the right option and justify your choice -

A) 

A. sin 60°

B. cos 60°

C. tan 30°

D. sin 30°

Answer:

we know,

tan 30° = 1/√3

tan 45° = 1

Then, putting these values in the question, we get

⇒ 

And we know, tan 30° = 1/√3

But, sin 60° = √3/2

cos 60° = 1/2

& sin 30° = 1/2

Thus, option (C) is correct.

Question 7.

Choose the right option and justify your choice -

B) 

A. tan 90°

B. 1

C. sin 45°

D. 0

Answer:

We know that,

tan 45° = 1

So, using this value of tangent, we can write

The answer has come out to be 0.

Thus, option (D) is correct.

Question 8.

Choose the right option and justify your choice -

C) 

A. cos 60°

B. sin 60°

C. tan 60°

D. sin 30°

Answer:

We know that,

tan 30° = 1/√3

So, using this value of tangent, we can write

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

And we know, tan 60° = √3

But, cos 60° = 1/2

sin 60° = √3/2

& sin 30° = 1/2

Thus, option (C) is correct.

Question 9.

Evaluate sin 60° cos 30° + sin 30°cos 60°. What is the value of sin(60° + 30°). What can you conclude?

Answer:

Let us first solve sin 60° cos 30° + sin 30° cos 60°.

We know,

sin 60° = √3/2

cos 30° = √3/2

sin 30° = 1/2

& cos 60° = 1/2

So, sin 60° cos 30° + sin 30° cos 60° = 

⇒ sin 60° cos 30° + sin 30° cos 60° = 

⇒ sin 60° cos 30° + sin 30° cos 60° = 1 …(i)

Now, for sin (60° + 30°):

sin (60° + 30°) = sin 90°

⇒ sin (60° + 30°) = 1 [∵, sin 90° = 1] …(ii)

By equations (i) & (ii), we can conclude that

sin (60° + 30°) = sin 60° cos 30° + sin 30° cos 60°

And infact in general, let 60° = x and 30° = y. Then,

sin (x + y) = sin x cos y + sin y cos x

Question 10.

Is it right to say cos(60° + 30°) = cos 60° cos 30° - sin 60° sin 30°.

Answer:

Let us solve Left-Hand-Side:

cos (60° + 30°) = cos 90°

⇒ cos (60° + 30°) = 0 [∵, cos 90° = 0]

Now, solve for Right-Hand-Side:

⇒ 

⇒ cos 60° cos 30° - sin 60° sin 30° = 0

Question 11.

In right angle triangle ΔPQR, right angle is at Q and PQ = 6cms ∠RPQ = 60°. Determine the lengths of QR and PR.

Answer:

To find QR:

Since, tan θ = perpendicular/base

We know that,

⇒  [∵, PQ = 6 cm & tan 60° = √3]

⇒ QR = 6√3

Now, PR can be found by two ways -

1^st method: In ∆PQR, using Pythagoras theorem,

PR² = PQ² + QR² [∵, (hypotenuse)² = (perpendicular)² + (base)²]

⇒ PR² = 6² + (6√3)²

⇒ PR² = 36 + 108 = 144

⇒ PR = √144 = 12

2^nd Method:

Since, cos θ = base/hypotenuse

We know that,

⇒ 

⇒ 

⇒ PR = 2 × PQ

⇒ PR = 2 × 6 = 12

Thus, QR = 6√3 cm and PR = 12 cm.

Question 12.

In right angle is at Y, YZ = x and XZ = 2x then determine ∠YXZ and ∠YZX.

Answer:

We have

With given values, YZ = x and XZ = 2x, we can find out both angles.

For ∠YXZ:

Let ∠YXZ = θ, then

⇒ 

⇒ 

⇒ θ = sin^-1(1/2)

⇒ θ = 30° [∵, sin 30° = 1/2]

For ∠YZX = α, then

⇒ 

⇒ 

⇒ α = cos^-1(1/2)

⇒ α = 60° [∵, cos 60° = 1/2]

Question 13.

Is it right to say that sin(A + B) = Sin A + Sin B? Justify your answer.

Answer:

No, it is not correct to say that sin (A + B) = sin A + sin B.

Justification: Let’s justify it by showing contradiction.

Let it be true that, sin (A + B) = sin A + sin B.

Now, let A = 30° and B = 60°

Then,

sin (30° + 60°) = sin 30° + sin 60°

⇒ sin 90° = sin 30° + sin 60°

⇒ 1 = 1/2 + √3/2

⇒ 1 = (1 + √3)/2

But, it’s not true.

1 ≠ (1 + √3)/2

Hence, we have a contradiction.

And therefore, it’s not right to say that sin (A + B) = sin A + sin B.

---

Exercise 11.3

Question 1.

Evaluate

A) 

Answer:

By trigonometric identity, we have

tan (90° - θ) = cot θ

Replace θ = 54°

⇒ tan (90° - 54°) = cot 54°

⇒ tan 36° = cot 54° [∵, 90° - 54° = 36°]

Now, 

⇒ 

Question 2.

Evaluate

B) cos 12° - sin 78°

Answer:

By trigonometric identity, we can say

cos (90° - θ) = sin θ

Now, just replace θ by 78°.

We get

cos (90° - 78°) = sin 78°

⇒ cos 12° = sin 78° [∵, 90° - 78° = 12°]

Now, cos 12° - sin 78° = sin 78° - sin 78°

⇒ cos 12° - sin 78° = 0

Question 3.

Evaluate

C) cosec 31° - sec 59°

Answer:

By trigonometric identity, we can say

cosec (90° - θ) = sec θ

Replace θ by 59°.

We get

cosec (90° - 59°) = sec 59°

⇒ cosec 31° = sec 59°

Now, cosec 31° - sec 59° = sec 59° - sec 59°

⇒ cosec 31° - sec 59° = 0

Question 4.

Evaluate

D) sin 15° sec 75°

Answer:

By trigonometric identities, we can say

sin (90° - θ) = cos θ

And sec θ = 1/cos θ

Replace θ by 75°.

We get

sin (90° - 75°) = cos 75°

⇒ sin 15° = cos 75° [∵, 90° - 75° = 15°]

And sec 75° = 1/cos 75°

Using these values, we can solve the given expression.

sin 15° sec 75° = sin 15°/cos 75°

⇒ sin 15° sec 75° = cos 75°/cos 75°

⇒ sin 15° sec 75° = 1

Question 5.

Evaluate

E. tan 26° tan 64°

Answer:

By trigonometric identities, we can say

tan (90° - θ) = cot θ

And tan θ = 1/cot θ

Replace θ by 64°.

We get

tan (90° - 64°) = cot 64°

⇒ tan 26° = cot 64° [∵, 90° - 64° = 26°]

And tan 64° = 1/cot 64°

Using these values, we can solve for the given expression.

tan 26° tan 64° = cot 64° tan 64°

⇒ tan 26° tan 64° = cot 64°/cot 64°

⇒ tan 26° tan 64° = 1

Question 6.

Show that

A. tan 48° tan16° tan 42° tan 74° = 1

Answer:

We have

LHS = tan 48° tan 16° tan 42° tan 74° = (tan 48° tan 42°)(tan 16° tan 74°)

We know the trigonometric identities, we can say

tan (90° - θ) = cot θ

And tan θ = 1/cot θ

First, replace θ by 42°.

tan (90° - 42°) = cot 42°

⇒ tan 48° = cot 42° …(1)

And tan 42° = 1/cot 42° …(2)

Now, replace θ by 74°.

tan (90° - 74°) = cot 74°

⇒ tan 16° = cot 74° …(3)

And tan 74° = 1/cot 74° …(4)

Using equations (1), (2), (3) & (4), we get

LHS = tan 48° tan 16° tan 42° tan 74° = (tan 48° tan 42°)(tan 16° tan 74°)

= (cot 42°/cot 42°)(cot 74°/cot 74°)

= 1 = RHS

Thus, tan 48° tan 16° tan 42° tan 74° = 1.

Question 7.

Show that

B. cos36°cos54° - sin36°sin54°

Answer:

We have

LHS = cos 36° cos 54° - sin 36° sin 54°

We know the trigonometric identity,

cos (90° - θ) = sin θ

First, replace θ by 54°.

We get, cos (90° - 54°) = sin 54°

⇒ cos 36° = sin 54° …(1)

Now, replace θ by 36°.

We get, cos (90° - 36°) = sin 36°

⇒ cos 54° = sin 36° …(2)

Using equations (1) & (2), we get

LHS = cos 36° cos 54° - sin 36° sin 54° = sin 54° sin 36° - sin 54° sin 36°

= 0 = RHS

Thus, cos 36° cos 54° - sin 36° sin 54° = 0.

Question 8.

If tan2A = cot(A – 18°) where 2A is an acute angle. Find the value of A.

Answer:

Given that, 2A is an acute angle.

⇒ 2A < 90°

So, using trigonometric identity, we can say that

cot (90° - 2A) = tan 2A [∵, cot (90° - θ) = tan θ]

Now, replace tan 2A by cot (90° - 2A) in the given question.

tan 2A = cot (A – 18°)

⇒ cot (90° - 2A) = cot (A – 18°)

Now, we can compare the degrees from above, we get

90° - 2A = A – 18°

⇒ 2A + A = 90° + 18°

⇒ 3A = 108°

⇒ A = 108°/3

⇒ A = 36°

Thus, the value of A is 36°.

Question 9.

If tan A = cot B where A and B are acute angles, prove that A + B = 90°

Answer:

Given that, A and B are acute angles.

⇒ A < 90° & B < 90°

So, using trigonometric identity, we can say

tan (90° - B) = cot B [∵, tan (90° - θ) = cot θ]

Replace cot B of RHS by tan (90° - B) in the given question.

tan A = cot B

⇒ tan A = tan (90° - B)

Now, comparing the degrees from the above, we get

A = 90° - B

⇒ A + B = 90°

Hence, proved that A + B = 90°.

Question 10.

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

Answer:

If A, B and C are interior angles of ∆ABC, then we can say that

A + B + C = 180° (by angle sum property of a triangle)

⇒ A + B = 180° - C

Take LHS:

⇒ 

⇒ 

[∵, tan (90° - θ) = cot θ, where θ = C/2 here]

Hence, .

Question 11.

Express sin 75° + cos 65° in terms of trigonometric ratios of angles between 0^o and 45^o.

Answer:

Given: sin 75° + cos 65°.

We can write,

75° = 90° - 15°

& 65° = 90° - 25°

Then, sin 75° + cos 65° = sin (90° - 15°) + cos (90° - 25°)

⇒ sin 75° + cos 65° = cos 15° + sin 25°

[∵, sin (90° - θ ) = cos θ & cos (90° - θ) = sin θ]

In cos 15° + sin 25°, 15° & 25° both are angles between 0° and 45°.

Thus, answer is sin 75° + cos 65° = cos 15° + sin 25°.

---

Exercise 11.4

Question 1.

Evaluate the following :

A. (1 + tanθ + secθ)(1 + cotθ – cosecθ)

Answer:

We have

⇒ 

⇒ 

⇒  [∵, (a + b)(a – b) = a² – b²]

⇒  [∵, (a + b)² = a² + b² + 2ab]

⇒  [∵, sin² θ + cos² θ = 1]

⇒ 

Thus, (1 + tan θ + sec θ)(1 + cot θ – cosec θ) = 2.

Question 2.

Evaluate the following :

B. (sinθ + cosθ)² + (sinθ - cosθ)²

Answer:

We have

(sin θ + cos θ)² + (sin θ – cos θ)² = ((sin θ + cos θ) + (sin θ – cos θ))² – 2(sin θ + cos θ)(sin θ – cos θ) [∵, a² + b² = (a + b)² – 2ab]

⇒ (sin θ + cos θ)² + (sin θ – cos θ)² = (sin θ + cos θ + sin θ – cos θ)² – 2(sin² θ – cos² θ)

[∵, (a + b)(a – b) = a² – b²]

⇒ (sin θ + cos θ)² + (sin θ – cos θ)² = (2 sin θ)² – 2 sin² θ + 2 cos² θ

⇒ (sin θ + cos θ)² + (sin θ – cos θ)² = 4 sin² θ – 2 sin² θ + 2 cos² θ

⇒ (sin θ + cos θ)² + (sin θ – cos θ)² = 2 sin² θ + 2 cos²θ

⇒ (sin θ + cos θ)² + (sin θ – cos θ)² = 2 (sin² θ + cos² θ)

⇒ (sin θ + cos θ)² + (sin θ – cos θ)² = 2 [∵, sin² θ + cos² θ = 1]

Thus, (sin θ + cos θ)² + (sin θ – cos θ)² = 2.

Question 3.

Evaluate the following :

C. (sec²θ – 1) (cosec²θ – 1)

Answer:

We have

⇒ 

⇒  [∵, (1 – cos² θ) = sin² θ & (1 – sin² θ = cos² θ]

⇒ (sec² θ – 1)(cosec² θ – 1) = 1

Thus, (sec² θ – 1)(cosec² θ – 1) = 1.

Question 4.

Show that 

Answer:

Using trigonometric identities,

cosec θ = 1/sin θ & cot θ = cos θ/sin θ

LHS = (cosec θ – cot θ)²

As we know, sin² θ = 1 – cos² θ

And (1 – cos² θ) = (1 + cos θ)(1 – cos θ) [by (a² – b²) = (a + b)(a – b)]

⇒ sin² θ = (1 + cos θ)(1 – cos θ) …(i)

So, using equation (i),

LHS = (cosec θ – cot θ)² = 

 = RHS

Hence, we have got

(cosec θ – cot θ)² = .

Question 5.

Show that 

Answer:

Using trigonometric identity,

sin² A + cos² A = 1

⇒ cos² A = 1 – sin² A

Take Left hand side:

LHS = 

= sec A + tan A = RHS

Thus, .

Question 6.

Show that 

Answer:

Using trigonometric identity, we have

cot A = 1/tan A

Take left hand side,

LHS = 

= tan² A = RHS

Thus,  = tan² A.

Question 7.

Show that 

Answer:

Take left hand side of the given question:

LHS = 1/cos θ – cos θ

= (1 – cos² θ)/cos θ

= sin² θ/cos θ [∵, sin² θ + cos² θ = 1 ⇒ sin² θ = 1 – cos² θ]

= sin θ × sin θ/cos θ

= sin θ × tan θ [∵, tan θ = sin θ/cos θ]

= tan θ sin θ = RHS

Thus, .

Question 8.

Simplify secA(1 – sinA)(secA + tanA)

Answer:

By trigonometric identities, sec A = 1/cos A & tan A = sin A/cos A

Using these identities, we have

sec A (1 – sin A)(sec A + tan A) = 

⇒ sec A (1 – sin A)(sec A + tan A) = 

⇒ sec A (1 – sin A)(sec A + tan A) = 

⇒ sec A (1 – sin A)(sec A + tan A) = 

⇒ sec A (1 – sin A)(sec A + tan A) =  [∵, sin² A + cos² A = 1 ⇒ cos² A = 1 – sin² A]

⇒ sec A (1 – sin A)(sec A + tan A) = 1

Thus, sec A (1 – sin A)(sec A + tan A) = 1.

Question 9.

Prove that (sin A + cosec A)² + (cos A + sec A)2 = 7 + tan² A + cot² A

Answer:

Take left hand side of the given equation:

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

Expanding the squares by formula: (a + b)² = a² + b² + 2ab

= sin² A + cosec² A + 2 sin A cosec A + cos² A + sec² A + 2 cos A sec A

Rearranging the terms, we get,

= (sin² A + cos² A) + 2 sin A cosec A + 2 cos A sec A + cosec² A + sec² A

we know that,

= 

= 

= 5 + 1/(sin² A cos² A) …(i)

Now, take right hand side of the equation:

RHS = 7 + tan² A + cot² A

= 

= 

= 

= 

= 5 + 1/(sin² A cos² A) …(ii)

From equation (i) & (ii),

LHS = RHS

Hence, proved.

Question 10.

Simplify (1 – cosθ) (1 + cosθ) (1 + cot²θ)

Answer:

We have

(1 – cos θ)(1 + cos θ)(1 + cot² θ) = [(1 – cos θ)(1 + cos θ)](1 + cot² θ)

⇒ (1 – cos θ)(1 + cos θ)(1 + cot² θ) = (1 – cos² θ)(1 + cot² θ) [∵, (a + b)(a – b) = a² – b²]

⇒ (1 – cos θ)(1 + cos θ)(1 + cot² θ) = sin² θ × (1 + cos² θ/sin² θ) [∵, (1 – cos² θ) = sin² θ & cot² θ = cos² θ/sin² θ]

⇒ (1 – cos θ)(1 + cos θ)(1 + cot² θ) = sin² θ × (sin² θ + cos² θ)/sin² θ

⇒ (1 – cos θ)(1 + cos θ)(1 + cot² θ) = sin² θ + cos² θ

⇒ (1 – cos θ)(1 + cos θ)(1 + cot² θ) = 1 [∵, sin² θ + cos² θ = 1]

Thus, (1 – cos θ)(1 + cos θ)(1 + cot² θ) = 1.

Question 11.

If secθ + tanθ = p, then what is the value of secθ – tanθ ?

Answer:

Given that, sec θ + tan θ = p.

By trigonometric identity, we have

sec² θ – tan² θ = 1

So, sec² θ – tan² θ = 1

⇒ (sec θ – tan θ) (sec θ + tan θ) = 1

⇒ sec θ – tan θ = 1/(sec θ + tan θ)

⇒ sec θ – tan θ = 1/p [given]

Hence, sec θ – tan θ = 1/p.

Question 12.

If cosesθ + cotθ = k then prove that 

Answer:

Given that, cosec θ + cot θ = k

⇒  [∵, cosec θ = 1/sin θ & cot θ = cos θ/sin θ]

⇒ (1 + cos θ)/sin θ = k

⇒ 1 + cos θ = k sin θ

Squaring both sides, we get

(1 + cos θ)² = (k sin θ)²

⇒ (1 + cos θ)² = k² sin² θ

⇒ (1 + cos θ)² = k² (1 – cos² θ) [∵, sin² θ + cos² θ = 1 ⇒ sin² θ = 1 – cos² θ]

⇒ (1 + cos θ)² = k² (1 – cos θ) (1 + cos θ) [∵, a² – b² = (a + b) (a – b)]

⇒ 1 + cos θ = k² (1 – cos θ)

⇒ 1 + cos θ = k² – k² cos θ

⇒ k² cos θ + cos θ = k² – 1

⇒ cos θ (k² + 1) = k² – 1

⇒ cos θ = (k² – 1)/(k² + 1)

Thus, cos θ = (k² – 1)/(k² + 1).

Hence, proved.