**Chapter 3 Trigonometric Functions Miscellaneous Exercise 3**

**Chapter 3 Trigonometric Functions Miscellaneous Exercise 3**

**I) Select the correct option from the given alternatives.Question 1.The principal of solutions equation sinθ = −1/2 are ________.**

**Solution:**

(b) 7π6,11π6

(b) 7π6,11π6

**Question 2.The principal solution of equation cot θ = √3 ___________.**

**Solution:**

**Question 3.The general solution of sec x = √2 is __________.**

**Question 4.If cos pθ = cosqθ, p ≠ q rhen ________.**

**Question 5.If polar co-ordinates of a point are (2,π/4) then its cartesian co-ordinates are ______.**

**Question 6.If √3 cosx – sin x = 1, then general value of x is _________.**

**Question 7.In ∆ABC if ∠A = 45°, ∠B = 60° then the ratio of its sides are _________.**

**Question 8.In ∆ABC, if c ^{2} + a^{2} – b^{2} = ac, then ∠B = __________.**

**Question 9.In ABC, ac cos B – bc cos A = ____________.**

(a) a

^{2}– b

^{2}

(b) b

^{2}– c

^{2}

(c) c

^{2}– a

^{2}

(d) a

^{2}– b

^{2}– c

^{2}

**Solution:**

(a) a

^{2}– b

^{2}

**Question 10.If in a triangle, the are in A.P. and b : c = √3 : √2 then A is equal to __________.**

(a) 30°

(b) 60°

(c) 75°

(d) 45°

**Solution:**

(c) 75°

**Question 11.**

**Question 12.The value of cot (tan ^{-1} 2x + cot^{-1} 2x) is __________.**
(a) 0

(b) 2x

(c) π + 2x

(d) π – 2x

**Solution:**

(a) 0

**Question 13.Solution:**

**Question 14.**

**Question 16.**

**Question 17.**

**Question 18.The principal value branch of sec ^{-1} x is __________.Solution:**

**Question 20.If tan θ + tan 2θ + tan 3θ = tan θ∙tan 2θ∙tan 3θ, then the general value of the θ is _______.**

**Question 21.If any ∆ABC, if a cos B = b cos A, then the triangle is ________.**

(a) Equilateral triangle

(b) Isosceles triangle

(c) Scalene

(d) Right angled

**Solution:**

(b) Isosceles triangle

**II: Solve the followingQuestion 1.Find the principal solutions of the following equations :**
(i) sin2θ = −1/2

Solution:

sin2θ = −1/2

Since, θ ∈ (0, 2π), 2∈ ∈ (0, 4π)

**(ii) tan3θ = -1Solution:**

Since, θ ∈ (0, 2π), 3∈ ∈ (0, 6π)

… [∵ tan(π – θ) = tan(2π – θ) = tan(3π – θ)

= tan (4π – θ) = tan (5π – θ) = tan (6π – θ) = -tan θ]

**(iii) cotθ = 0Solution:**

**Question 2.Find the principal solutions of the following equations :Solution:**

**(ii) tan5θ = -1Solution:**

**(iii) cot2θ = 0Solution:**

**Question 3.Which of the following equations have no solutions ?(i) cos 2θ = 1/3Solution:**

cos 2θ = 1/3

Since 1/3 ≤ cosθ ≤ 1 for any θ

cos2θ = 1/3 has solution

**(ii) cos ^{2} θ = -1Solution:**
cos2θ = -1

This is not possible because cos2θ ≥ 0 for any θ.

∴ cos2θ = -1 does not have any solution.

**(iii) 2 sinθ = 3Solution:**

2 sin θ = 3 ∴ sin θ = 3/2

This is not possible because -1 ≤ sin θ ≤ 1 for any θ.

∴ 2 sin θ = 3 does not have any solution.

**(iv) 3 sin θ = 5Solution:**
3 sin θ = 5

∴ sin θ = 5/3

This is not possible because -1 ≤ sin θ ≤ 1 for any θ.

∴ 3 sin θ = 5 does not have any solution.

Question 4.

Find the general solutions of the following equations :

**(ii) tan ^{2}θ = 3Solution:**

**(iii) sin θ – cosθ = 1Solution:**
∴ cosθ – sin θ = -1

**(iv) sin ^{2}θ – cos^{2}θ = 1**

Solution:

sin

^{2}θ – cos

^{2}θ = 1

∴ cos

^{2}θ – sin

^{2}θ = -1

∴ cos2θ = cosπ …(1)

The general solution of cos θ = cos ∝ is

θ = 2nπ ± ∝, n ∈ Z

∴ the general solution of (1) is given by

2θ = 2nπ ± π, n ∈ Z

∴ θ = nπ ± π2, n ∈ Z

**Question 5.Solution:**

By the sine rule,

**Question 6.**

**Question 7.**

**Question 8.In ∆ABC if cosA = sin B – cos C then show that it is a right angled triangle.Solution:**

cos A= sin B – cos C

∴ cos A + cos C = sin B

∴ A – C = B

∴ A = B + C

∴ A + B + C = 180° gives

A + A = 180°

∴ 2A = 180 ∴ A = 90°

∴ ∆ ABC is a rightangled triangle.

**Question 9.**

∴ sin [π – (B + C)] ∙ sin (B – C)

= sin [π – (A + B)]∙sin (A – B) … [∵ A + B + C = π]

∴ sin(B + C) ∙ sin(B – C) = sin (A + B) ∙ sin (A – B)

∴ sin

^{2}B – sin

^{2}C = sin

^{2}A – sin

^{2}B

∴ 2 sin

^{2}B = sin

^{2}A + sin

^{2}C

∴ 2k

^{2}b

^{2}= k

^{2}a

^{2}+ k

^{2}c

^{2}

∴ 2b

^{2}= a

^{2}+ c

^{2}

Hence, a

^{2}, b

^{2}, c

^{2}are in A.P.

**Question 10.**

∴ sin A = sin 60° cos 45° + cos 60° sin 45°

and sin B = sin 60° cos 45° – cos 60° sin 45°

∴ sin A = sin (60° + 45°) – sin 105°

and sin B = sin (60° – 45°) = sin 15°

∴ A = 105° and B = 15°

Hence, A = 105°, B 15° and C = √6 units.

**Question 11.In ∆ABC prove the following :(i) a sin A – b sin B = c sin (A – B)Solution:**

By sine rule,

∴ a = ksinA, b = ksinB, c = ksinC,

LHS = a sin A – b sinB

= ksinA∙sinA – ksinB∙sinB

= k (sin

^{2}A – sin

^{2}B)

= k (sin A + sin B)(sin A – sin B)

= k × sin (A + B) × sin (A – B)

= ksin(π – C)∙sin(A – B) … [∵ A + B + C = π]

= k sinC∙sin (A – B)

= c sin (A – B) = RHS.

**Solution:**

**(iii) a ^{2} sin (B – C) = (b^{2} – c^{2}) sinA**

Solution:

By sine rule,

∴ a = ksinA, b = ksinB, c = ksinC

RHS = (b

^{2}– c

^{2}) sin A

= (k

^{2}sin

^{2}B – k

^{2}sin

^{2}C)sin A

= k

^{2}(sin

^{2}B – sin

^{2}C) sin A

= k

^{2}(sin B + sin C)(sin B – sin C) sin A

= k

^{2}× sin (B + C) × sin (B – C) × sin A

= k

^{2}∙sin(π – A)∙sin(B – C)∙sinA … [∵ A + B + C = π]

= k

^{2}sin A∙sin (B – C)∙sin A

= (k sin A)

^{2}∙sin (B – C)

= a

^{2}sin (B – C) = LHS.

**(iv) ac cos B – bc cos A = (a ^{2} – b^{2}).**

Solution:

Solution:

**Solution:**

By sine rule,

**Question 12.Solution:**

a, b, c, are in A.P.

∴ 2b = a + c …(1)

**Question 13.Solution:**

In ∆ABC, if ∠C = 90º

∴ c

^{2}= a

^{2}+ b

^{2}…(1)

By sine rule,

**Question 14.**

By sine rule,

∴ sin A cos B = cos A sinB

∴ sinA cosB – cosA sinB = 0

∴ sin (A – B) = 0 = sin0

∴ A – B = 0 ∴ A = B

∴ the triangle is an isosceles triangle.

**Question 15.In ∆ABC if sin ^{2}A + sin^{2}B = sin^{2}C then prove that the triangle is a right angled triangle.Question is modifiedIn ∆ABC if sin^{2}A + sin^{2}B = sin^{2}C then show that the triangle is a right angled triangle.Solution:**
By sine rule,

∴ sin A = ka, sinB = kb, sin C = kc

∴ sin

^{2}A + sin

^{2}B = sin

^{2}C

∴ k

^{2}a

^{2}+ k

^{2}b

^{2}= k

^{2}c

^{2}

∴ a

^{2}+ b

^{2}= c

^{2}

∴ ∆ABC is a rightangled triangle, rightangled at C.

Question 16.

In ∆ABC prove that a^{2}(cos^{2}B – cos^{2}C) + b^{2}(cos^{2}C – cos^{2}A) + c^{2}(cos^{2}A – cos^{2}B) = 0.

Solution:

By sine rule,

LHS = a^{2}(cos^{2}B – cos^{2}C) + b^{2}( cos^{2}C – cos^{2}A) + c^{2}(cos^{2}A – cos^{2}B)

= k^{2}sin^{2}A [(1 – sin^{2}B) – (1 – sin^{2}C)] + k^{2}sin^{2}B [(1 – sin^{2}C) – (1 – sin^{2}A)] + k^{2}sin^{2}C[(1 – sin^{2}A) – (1 – sin^{2}B)]

= k^{2}sin^{2}A (sin^{2}C – sin^{2}B) + k^{2}sin^{2}B(sin^{2}A – sin^{2}C) + k^{2}sin^{2}C (sin^{2}B – sin^{2}A)

= k^{2}(sin^{2}A sin^{2}C – sin^{2}Asin^{2}B + sin^{2}A sin^{2}B – sin^{2}B sin^{2}C + sin^{2}B sin^{2}C – sin^{2}A sin^{2}C)

= k^{2}(0) = 0 = RHS.

**Question 17.With usual notations show that (c ^{2} – a^{2} + b^{2}) tan A = (a^{2} – b^{2} + c^{2}) tan B = (b^{2} – c^{2} + a^{2}) tan C.Solution:**

By sine rule,

∴ a = fksinA, b = ksinB, c = ksinC

From (1), (2) and (3), we get

(c

^{2}– a

^{2}+ b

^{2}) tan A = (a

^{2}– b

^{2}+ c

^{2}) tan B

= (b

^{2}– c

^{2}+ a

^{2}) tan C.

**Question 18.**

∴ a + c + b = 3b …[∵ a cos C + c cos A = b]

∴ a + c = 2b

Hence, a, b, c are in A.P.

**Question 19.**

**Solution:**

**Question 21.**

**Question 22.Solution:**

We have to show that

**Question 23.Show that**

**Solution:**

**Question 24.**

**Question 25.If tan ^{-1}(x−1x−2) + tan^{-1}(x+1x+2) = π4 then find the value of x.Solution:tan^{-1}(x−1x−2) + tan^{-1}(x+1x+2) = π4**

**Question 26.If 2 tan ^{-1}(cos x ) = tan^{-1}(cosec x) then find the value of x.Solution:**

2 tan

^{-1}(cos x ) = tan

^{-1}(cosec x)

**Question 27.**

**Question 28.If sin ^{-1}(1 – x) – 2sin^{-1}x = π/2, then find the value of x.Solution:**

sin

^{-1}(1 – x) – 2sin

^{-1}x = π/2

**Question 29.**

**Question 30.**

**Question 31.**

**Question 32.Solution:**

We have to show that

**Question 33.Solution:**

**Question 34.Solution:**

**Question 35.Prove the following :Solution:**

Solution:

**Question 36.Solution:**

Let tan

^{-1}x = y

Then, x = tany

**Question 37.Solution:**

Question 38.

If tan^{-1} x + tan^{-1} y + tan^{-1} z = π/2 then, show that xy + yz + zx = 1

Solution:

tan^{-1} x + tan^{-1} y + tan^{-1} z = π/2

∴ 1 – xy – yz – zx = 0

∴ xy + yz + zx = 1.

**Question 39.If cos ^{-1} x + cos^{-1} y + cos^{-1} z = π then show that x^{2} + y^{2} + z^{2} + 2xyz = 1.Solution:**

0 ≤ cos

^{-1}x ≤ π and

cos

^{-1}x + cos

^{-1}y+ cos

^{-1}z = 3π

∴ cos

^{-1}x = π, cos

^{-1}y = π and cos

^{-1}z = π

∴ x = y = z = cosπ = -1

∴ x

^{2}+ y

^{2}+ z

^{2}+ 2xyz

= (-1)

^{2}+ (-1)

^{2}+ (-1)

^{2}+ 2(-1)(-1)(-1)

= 1 + 1 + 1 – 2

= 3 – 2 = 1.