# NCERT Exemplar Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

NCERT Exemplar Solutions Class 11 Maths Chapter 3 Trigonometric Functions are provided here. These NCERT Exemplar solutions are created by BYJUâ€™S expert faculty to help students’ exam preparation. These subject experts solve and provide the NCERT Exemplar Solution for Class 11, which will help students to solve problems effortlessly. They give a detailed and step-wise explanation of problems given in exercises in the NCERT Exemplar textbook for Class 11. In CBSE Class 11 Trigonometric Functions chapter, students are introduced to many important topics, which will be helpful for those who wish to pursue Mathematics in the future. These solutions help students prepare for their upcoming board exam by covering the whole syllabus in accordance with the NCERT guidelines. The PDF for Chapter 3, exemplar problems and solutions, can be downloaded and practised offline as well.

Chapter 3 of NCERT Exemplar Solutions for Class 11 Maths Trigonometric Functions explains the domain and range of trigonometric functions. Trigonometric functions can be defined as the functions of an angle. It is also called by other names, such as angle functions, circular functions and goniometric functions. The major functions that are most commonly known to students are sine, cosine, and tangent. Some functions, such as infinite series and differential equations, are used for whole numbers and complex numbers. In Chapter 3, students will learn and solve exemplar problems based on topics like

• The relation between degree and radian
• Trigonometric functions
• Domain and range of trigonometric functions
• Sine, cosine and tangent of some angles less than 90Â°
• Allied or related angles
• Functions of negative angles
• Some formulae regarding compound angles
• Trigonometric equations
• General solution of trigonometric equations

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Exercise Page No: 52

1. Prove thatÂ

Solution:

According to the question,

Using the identity,

sin2A + cos2A = 1, we get,

sin A + (1 â€“ cos A).

Hence, L.H.S = R.H.S

2. If [2sinÎ± / (1+cosÎ±+sinÎ±)] = y, then prove thatÂ [(1â€“ cosÎ±+sinÎ±) / (1+sinÎ±)]Â is also equal to y.

Solution:

According to the question,

y =2sinÎ± /(1+cosÎ±+sinÎ±)

Multiplying numerator and denominator by (1 â€“ cos Î± + sin Î±),

We get,

Hence Proved

3. If m sin Î¸ = n sin (Î¸ + 2Î±), then prove that

tan (Î¸ + Î±) cot Î± = (m + n)/(m â€“ n)

[Hints: Express sin(Î¸ + 2Î±) / sinÎ¸ = m/nÂ and apply componendo and dividend]

Solution:

According to the question,

m sin Î¸ = n sin (Î¸ + 2Î±)

To prove:

tan (Î¸ + Î±)cot Î± =(m + n)/(m â€“ n)

Proof:

m sin Î¸ = n sin (Î¸ + 2Î±)

â‡’Â sin(Î¸ + 2Î±) / sinÎ¸ = m/n

Applying componendo-dividendo rule, we have,

By transformation formula of T-ratios,

We know that,

sin A + sin B = 2 sin ((A+B)/2) cos ((A â€“ B)/2)

And,

sin A â€“ sin B = 2 cos ((A+B)/2) sin ((A â€“ B)/2)

On applying the formula, we get,

Therefore,Â tan (Î¸ + Î±) cot Î± =Â (m + n)/(m â€“ n)

Hence Proved

4. IfÂ Â where Î± lie between 0 and Ï€/4, find value of tan 2Î±
[Hint: Express tan 2Î± as tan (Î± + Î² + Î± â€“ Î²]

Solution:

According to the question,

cos(Î± + Î²) = 4/5 â€¦(i)

We know that,

sin x = âˆš(1 â€“ cos2x)

Therefore,

sin (Î± + Î²) = âˆš(1 â€“ cos2(Î± + Î²))

â‡’Â sin (Î± + Î²) = âˆš(1 â€“ (4/5)2) = 3/5Â â€¦(ii)

Also,

sin(Î± – Î²) = 5/13 {given} â€¦(iii)

we know that,

cos x = âˆš(1 â€“ sin2x)

Therefore,

cos (Î± – Î²) = âˆš(1 â€“ sin2(Î± – Î²))

â‡’Â cos (Î± – Î²) = âˆš(1 â€“ (5/13)2) = 12/13Â â€¦(iv)

Therefore,

tan 2Î± = tan (Î± + Î² + Î± â€“ Î²)

We know that,

From equation i, ii, iii and iv we have,

Hence, tan 2Î± = 56/33

5. If tanx = b/aÂ then find the value ofÂ

Solution:

According to the question,

tan x = b/a

Let,

6. Prove that cos Î¸ cos Î¸/2 â€“ cos 3Î¸ cos 9Î¸/2 = sin7Î¸ sin4Î¸
[Hint: Express L.H.S. = Â½ [2cos Î¸cos Î¸/2 â€“ 2cos 3Î¸ cos 9Î¸ / 2]

Solution:

Using transformation formula, we get,

2 cos A cos B = cos(A + B) + cos (A â€“ B)

-2 sin A sin B = cos(A + B) – cos (A â€“ B)

Multiplying and dividing the expression by 2.

7. If a cos Î¸ + b sin Î¸ = m and a sin Î¸ â€“ b cos Î¸ = n, then show that a2Â + b2Â = m2Â + n2.

Solution:

According to the question,

a cos Î¸ + b sin Î¸ = m â€¦(i)

a sin Î¸ â€“ b cos Î¸ = n â€¦(ii)

Squaring and adding equation 1 and 2, we get,

(a cos Î¸ + b sin Î¸)2Â + (a sin Î¸ â€“ b cos Î¸)2Â = m2Â + n2

â‡’Â a2cos2Î¸ + b2sin2Î¸ + 2ab sin Î¸ cos Î¸ + a2sin2Î¸ + b2cos2Î¸ – 2ab sin Î¸ cos Î¸ = m2Â + n2

â‡’Â a2cos2Î¸ + b2sin2Î¸ + a2sin2Î¸ + b2cos2Î¸ = m2Â + n2

â‡’Â a2(sin2Î¸ + cos2Î¸) + b2(sin2Î¸ + cos2Î¸) = m2Â + n2

Using, sin2Î¸ + cos2Î¸ = 1,

We get,

â‡’Â a2Â + b2Â = m2Â + n2

8. Find the value of tan 22Â°30â€™.
[Hint: Let Î¸ = 45Â°, use

Solution:

Let, Î¸ = 45Â°

As we need to find: tan 22Â°30â€™ = tan (Î¸/2)

We know that,

sin Î¸ = cos Î¸ = 1/âˆš2 (for Î¸ = 45Â°)

Since,

Therefore,Â tan 22Â°30â€™ = âˆš2 â€“ 1

9. Prove that sin 4A = 4sinA cos3A â€“ 4 cosA sin3A.

Solution:

sin 4A = sin (2A + 2A)

We know that,

sin(A + B) = sin A cos B + cos A sin B

Therefore,Â sin 4A = sin 2A cos 2A + cos 2A sin 2A

â‡’Â sin 4A = 2 sin 2A cos 2A

From T-ratios of multiple angle,

We get,

sin 2A = 2 sin A cos A and cos 2A = cos2A â€“ sin2A

â‡’Â sin 4A = 2(2 sin A cos A)(cos2A â€“ sin2A)

â‡’Â sin 4A = 4 sin A cos3A â€“ 4 cos A sin3A

Hence, sin 4A = 4 sin A cos3A â€“ 4 cos A sin3A

10. If tan Î¸ + sin Î¸ = m and tan Î¸ â€“ sin Î¸ = n, then prove that m2Â â€“ n2Â = 4 sin Î¸ tan Î¸

[Hint: m + n = 2tanÎ¸, m â€“ n = 2 sin Î¸, then use m2Â â€“ n2Â = (m + n)(m â€“ n)]

Solution:

According to the question,

tan Î¸ + sin Î¸ = m â€¦(i)

tan Î¸ â€“ sin Î¸ = n â€¦(ii)

2 tan Î¸ = m + n â€¦(iii)

Subtracting equation ii from i,

We get,

2sin Î¸ = m â€“ n â€¦(iv)

Multiplying equations (iii) and (iv),

2sin Î¸ (2tan Î¸) = (m + n)(m â€“ n)

â‡’Â 4 sin Î¸ tan Î¸ = m2Â â€“ n2

Hence,

m2Â â€“ n2Â = 4 sin Î¸ tan Î¸

11. If tan (A + B) = p, tan (A â€“ B) = q, then show thatÂ tan 2A = (p + q) / (1 â€“ pq).
[Hint: Use 2A = (A + B) + (A â€“ B)]

Solution:

We know that,

tan 2A = tan (A + B + A â€“ B)

And also,

12. If cosÎ± + cosÎ² = 0 = sinÎ± + sinÎ², then prove that cos 2Î± + cos 2Î² = â€“2cos (Î± + Î²).
[Hint: cosÎ± + cosÎ²)2Â â€“ (sinÎ± + sinÎ²)2Â = 0]

Solution:

According to the question,

cosÎ± + cosÎ² = 0 = sinÎ± + sinÎ² â€¦(i)

Since,Â LHS = cos 2Î± + cos 2Î²

We know that,

cos 2x = cos2x â€“ sin2x

Therefore,

LHS = cos2Î± â€“ sin2Î± + (cos2Î² â€“ sin2Î²)

â‡’Â LHS = cos2Î± + cos2Î² â€“ (sin2Î± + sin2Î²)

Also, since,

a2Â + b2Â = (a+b)2Â â€“ 2ab

â‡’Â LHS = (cosÎ± + cosÎ²)2Â â€“ 2cosÎ± cosÎ² â€“(sinÎ± + sinÎ²)2Â +2sinÎ± sinÎ²

From equation (i),

â‡’Â LHS = 0 – 2cosÎ± cosÎ² -0 + 2sinÎ± sinÎ²

â‡’Â LHS = -2(cosÎ± cosÎ² â€“ sinÎ± sinÎ²)

âˆµÂ cos (Î± + Î²) = cosÎ± cosÎ² â€“ sinÎ± sinÎ²

Therefore,Â LHS = -2 cos (Î± + Î²) = RHS

Hence, cos 2Î± + cos 2Î² = â€“2cos (Î± + Î²)

13.

IfÂ Â then show thatÂ
[Hint: Use componendo and Dividendo]

Solution:

According to the question,

14.

IfÂ Â then show that sinÎ± + cosÎ± = âˆš2 cos Î¸.
[Hint: Express tanÎ¸ = tan(Î± â€“ Ï€/2) Î¸ = Î± â€“ Ï€/4]

Solution:

We know that,

We know that,

tan(x-y) =Â (tan x â€“ tan y) / (1 + tan x . tan y)

Therefore, tan Î¸ = tan ( Î± â€“ Ï€/4)

â‡’Â Î¸ = Î± – Ï€/4

â‡’Â Î± = Î¸ + Ï€/4 â€¦(i)

To prove,

sinÎ± + cosÎ± = âˆš2 cos Î¸

âˆµÂ LHS = sinÎ± + cosÎ±

From equation (i)

â‡’Â LHS = sin(Î¸ + Ï€/4) + cos(Î¸ + Ï€/4)

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

And, cos(x + y) = cos x cos y â€“ sin x sin y

Therefore,Â LHS = sin Î¸ cos(Ï€/4) + sin(Ï€/4)cos Î¸ + cos Î¸ cos(Ï€/4) – sin(Ï€/4)sin Î¸

âˆµÂ sin(Ï€/4)=cos(Ï€/4) = 1/âˆš2

â‡’Â LHS =Â sin Î¸ (1/âˆš2) + (1/âˆš2) cos Î¸ + cos Î¸ (1/âˆš2) â€“ sin Î¸ (1/âˆš2)

â‡’Â LHS = 2 cos Î¸ (1/âˆš2)

â‡’Â LHS = âˆš2 cos Î¸ = RHS

Therefore, sinÎ± + cosÎ± = âˆš2 cos Î¸

15. If sin Î¸ + cos Î¸ = 1, then find the general value of Î¸.

Solution:

According to the question,

sin Î¸ + cos Î¸ = 1

As,Â sin Î¸ + cos Î¸ = 1

Since we know,

If sin Î¸ = sinÎ±Â â‡’Â Î¸ = nÏ€ + (-1)nÎ±

We get,

Î¸ + Ï€/4 = nÏ€ + (-1)n(Ï€/4)

â‡’Â Î¸ = nÏ€ + (Ï€/4)((-1)nÂ â€“ 1)

16. Find the most general value of Î¸ satisfying the equation tan Î¸ = â€“1 and cos Î¸ = 1/âˆš2

Solution:

According to the question,

We have,

tan Î¸ = -1

And cos Î¸ =1/âˆš2 .

â‡’ Î¸ = â€“ Ï€/4

So, we know that,

Î¸ = 2Ï€ â€“ Ï€/4 = 7Ï€/4

So, general solution is Î¸ = 7Ï€/4 + 2 n Ï€, nâˆˆ Z

17. If cot Î¸ + tan Î¸ = 2 cosec Î¸, then find the general value of Î¸.

Solution:

According to the question,

â‡’Â 1 = 2 cosec Î¸ sin Î¸ cos Î¸

We know that,

sin Î¸ cosec Î¸ = 1

â‡’Â 1 = 2 cos Î¸

â‡’Â cos Î¸ = 1/2 = cos(Ï€/3)

Hence,

TheÂ solution of cos x = cos Î± can be given by,

x = 2mÏ€ Â± Î±Â âˆ€Â mÂ âˆˆÂ Z

â‡’Â Î¸ = 2nÏ€ Â± Ï€/3, nÂ âˆˆÂ Z

18. If 2sin2Î¸ = 3cos Î¸, where 0 â‰¤ Î¸ â‰¤ 2Ï€, then find the value of Î¸.

Solution:

According to the question,

2sin2Î¸ = 3cos Î¸

We know that,

sin2Î¸ = 1 â€“ cos2Î¸

Given that,

2 sin2 Î¸ = 3 cos Î¸

2 â€“ 2 cos2 Î¸ = 3 cos Î¸

2 cos2 Î¸ + 3 cos Î¸ â€“ 2= 0

(cos Î¸ + 2)(2 cos Î¸ â€“ 1) = 0

Therefore,

cos Î¸ = Â½ = cos Ï€/3

Î¸ = Ï€/3 or 2Ï€ â€“ Ï€/3

Î¸ = Ï€/3, 5Ï€/3

Therefore,Â 2(1 â€“ cos2Î¸) = 3cos Î¸

â‡’Â 2 â€“ 2cos2Î¸ = 3cos Î¸

â‡’Â 2cos2Î¸ + 3cos Î¸ – 2 = 0

â‡’Â 2cos2Î¸ + 4cos Î¸ – cos Î¸ – 2 = 0

â‡’Â 2cos Î¸ (cos Î¸+ 2) +1 (cos Î¸ + 2) = 0

â‡’Â (2cos Î¸ + 1)(cos Î¸ + 2) = 0

Since, cos Î¸Â âˆˆÂ [-1,1] , for any value Î¸.

So, cos Î¸ â‰  – 2

Therefore,

2 cos Î¸ – 1 = 0

â‡’Â cos Î¸ = Â½

= Ï€/3 or 2Ï€ â€“ Ï€/3

Î¸ = Ï€/3, 5Ï€/3

19. If sec x cos 5x + 1 = 0, where 0 < x â‰¤ Ï€/2, then find the value of x.

Solution:

According to the question,

sec x cos 5x = -1

â‡’Â cos 5x = -1/sec x

We know that,

sec x = 1/cos x

â‡’Â cos 5x + cos x = 0

By transformation formula of T-ratios,

We know that,

â‡’Â 2 cos 3x cos 2x = 0

â‡’Â cos 3x = 0 or cos 2x = 0

âˆµÂ 0 < x â‰¤ Ï€/2

Therefore,Â 0< 2x â‰¤ Ï€ or 0< 3x â‰¤ 3Ï€/2

Therefore,Â 2x = Ï€/2

â‡’Â x = Ï€/4

3x = Ï€/2

â‡’Â x = Ï€/6

Or 3x = 3Ï€/2

â‡’Â x = Ï€/2

Hence, x = Ï€/6, Ï€/4, Ï€/2.

20. If sin (Î¸ + Î±) = a and sin (Î¸ + Î²) = b, then prove that cos 2(Î± â€“ Î²) â€“ 4ab cos (Î± â€“ Î²) = 1 â€“ 2a2Â â€“ 2b2

Solution:

According to the question,

sin (Î¸ + Î±) = a and sin(Î¸ + Î²) = b

LHS = cos 2(Î± â€“ Î²) â€“ 4ab cos (Î± â€“ Î²)

Using cos 2x = 2cos2x â€“ 1,

Let us solve,

â‡’Â LHS = 2cos2(Î± â€“ Î²) – 1 â€“ 4ab cos(Î± â€“ Î²)

â‡’Â LHS = 2cos (Î± â€“ Î²) {cos (Î± â€“ Î²) â€“ 2ab} â€“ 1

Since,

cos (Î± â€“ Î²) = cos {(Î¸ + Î±) â€“ (Î¸ + Î²)}

cos (A â€“ B) = cos A cos B + sin A sin B

â‡’Â cos (Î± â€“ Î²) = cos(Î¸ + Î±)cos(Î¸ + Î²) + sin(Î¸ + Î±)sin(Î¸ + Î²)

Since,

sin(Î¸ + Î±) = a

â‡’Â cos(Î¸ + Î±) = âˆš(1 â€“ sin2(Î¸ + Î±) = âˆš(1 â€“ a2)

Similarly,

cos(Î¸ + Î²) = âˆš(1 â€“ b2)

Therefore,

cos(Î± â€“ Î²) = âˆš(1-a2)âˆš(1-b2) + ab

Therefore,

LHS = 2{ab + âˆš(1 â€“ a2)(1 â€“ b2)}{ab + âˆš(1 â€“ a2)(1 â€“ b2) -2ab} â€“ 1

â‡’Â LHS = 2{âˆš(1 â€“ a2)(1 â€“ b2) + ab}{âˆš(1 â€“ a2)(1 â€“ b2) – ab}-1

Using (x + y)(x â€“ y) = x2Â â€“ y2

â‡’Â LHS = 2{(1-a2)(1-b2) â€“ a2b2} â€“ 1

â‡’Â LHS = 2{1 â€“ a2Â â€“ b2Â + a2b2} â€“ 1

â‡’Â LHS = 2 â€“ 2a2Â â€“ 2b2Â â€“ 1

â‡’Â LHS = 1 â€“ 2a2Â â€“ 2b2Â = RHS

Therefore,

We get,

cos 2(Î± â€“ Î²) â€“ 4ab cos (Î± â€“ Î²) = 1 â€“ 2a2Â â€“ 2b2

21. If cos (Î¸ +Â Ï•) = m cos (Î¸ â€“Â Ï•), then prove that tanÂ Î¸ = ((1 â€“ m)/(1 + m)) cot Ï•
[Hint: Express cos (Î¸ + Ï•)/ cos (Î¸ â€“ Ï•) = m/lÂ and apply Componendo and Dividendo] Solution:

According to the question,

22. Find the value of the expression

Solution:

According to the question,

Let, y = 3[sin4 (3Ï€/2 â€“ Î±) + sin 4 (3Ï€ + Î±)] â€“ 2[sin6 (Ï€/2 + Î±) + sin6 (5Ï€ â€“ Î±)]

We know that,

sin(3Ï€/2 â€“ Î±) = -cos Î±

sin(3Ï€ + Î±) = -sin Î±

sin(Ï€/2 + Î±) = cos Î±

sin(5Ï€ â€“ Î±) = sin Î±

Therefore,

y =Â 3[(â€“ cos Î±)4 + (â€“ sin Î±)4] â€“ 2[cos6 Î± + sin6 Î±]

â‡’Â y = 3 [cos4Î± + sin4Î±] â€“ 2[sin6Î± + cos6Î±]

â‡’Â y = 3[(sin2Î± + cos2Î±)2Â â€“ 2sin2Î± cos2Î±] â€“ 2[(sin2Î±)3Â + (cos2Î±)3]

Since, we know that,

sin2Î± + cos2Î± = 1

Also, we know that,

a3+b3Â = (a+b)(a2Â â€“ ab + b2)

â‡’Â y = 3[1 â€“ 2sin2Î± cos2Î±] â€“ 2[(sin2Î± + cos2Î±)( cos4Î± + sin4Î±- sin2Î± cos2Î±)]

â‡’Â y = 3[1 â€“ 2sin2Î± cos2Î±] â€“ 2[cos4Î± + sin4Î±- sin2Î± cos2Î±]

â‡’Â y = 3[1 â€“ 2sin2Î± cos2Î±] â€“ 2[(sin2Î± + cos2Î±)2Â â€“ 2sin2Î± cos2Î± – sin2Î± cos2Î±]

â‡’Â y = 3[1 â€“ 2sin2Î± cos2Î±] â€“ 2[1 â€“ 3sin2Î± cos2Î±]

â‡’Â y = 3 â€“ 6sin2Î± cos2Î± â€“ 2 + 6 sin2Î± cos2Î±

â‡’Â y = 1

23. If a cos2Î¸ + b sin 2Î¸ = c has Î± and Î² as its roots, then prove thatÂ tan Î± + tan Î² = 2b/(a + c)

[Hint: Use the identitiesÂ cos 2Î¸ = (( 1 â€“ tan2 Î¸)/(1 + tan2 Î¸) and sin 2Î¸ = 2tan Î¸/(1 + tan2 Î¸)]

Solution:

According to the question,

a cos2Î¸ + b sin 2Î¸ = c

Î± and Î² are the roots of the equation.

Using the formula of multiple angles,

We know that,

â‡’Â a(1 â€“ tan2Î¸) + 2b tan Î¸ – c(1 + tan2Î¸) = 0

â‡’Â (-c â€“ a)tan2Î¸ + 2b tan Î¸ – c + a = 0 â€¦(i)

We know that,

The sum of roots of a quadratic equation, ax2Â + bx + c = 0 is given by (-b/a)

Therefore,

tan Î± + tan Î² =Â â€“2b/â€“(c + a) = 2b/(c + a)

Hence, tan Î± + tan Î² = 2b/(c + a)

24. If x = secÂ Ï•Â â€“ tanÂ Ï•Â and y = cosecÂ Ï•Â + cotÂ Ï•, then show that xy + x â€“ y + 1 = 0.

[Hint: Find xy + 1 and then show tan x â€“ y = â€“(xy + 1)]

Solution:

According to the question,

x = secÂ Ï•Â â€“ tanÂ Ï•Â and y =Â cosecÂ Ï•Â + cotÂ Ï•

Given that, LHS = xy + x â€“ y + 1

Thus, LHS = xy + x â€“ y + 1 = 0

25. If Î¸ lies in the first quadrant and cos Î¸ = 8/17,Â then find the value of cos(30Â° + Î¸) + cos (45Â° â€“ Î¸) + cos (120Â° â€“ Î¸)

Solution:

According to the question,

cos Î¸ = 8/17

sin Î¸ = Â±âˆš(1 â€“ cos2Î¸)

Since, Î¸ lies in first quadrant, only positive sign can be considered.

â‡’Â sin Î¸ = âˆš(1 â€“ 64/289) = 15/17

Let, y = cos(30Â° + Î¸) + cos (45Â° â€“ Î¸) + cos (120Â° â€“ Î¸)

We know that,

cos(x + y) = cos x cos y â€“ sin x sin y

Therefore,

y = cos30Â° cos Î¸ â€“ sin30Â° sin Î¸ + cos45Â° cos Î¸ + sin45Â°sin Î¸ +cos120Â° cos Î¸ + sin120Â° sin Î¸

Substituting values of cos30Â°, sin30Â°, cos 120Â°, sin120Â° and cos 45Â°

26. Find the value of the expression cos4(Ï€/8) + cos4(3Ï€/8)Â + cos4(5Ï€/8)Â + cos4(7Ï€/8).
[Hint: Simplify the expression to

Solution:

According to the question,

Let y = cos4(Ï€/8) + cos4(3Ï€/8)Â + cos4(5Ï€/8)Â + cos4(7Ï€/8).

â‡’ y = cos4(Ï€/8) + cos4(3Ï€/8)Â + cos4(Ï€ â€“ 3Ï€/8)Â + cos4(Ï€ â€“ Ï€/8).

Since we know that, cos (Ï€ â€“ x) = â€“ cos x, we get,

= 2 â€“ (1/âˆš2)2

= 2 â€“ Â½

= 3/2

27. Find the general solution of the equation

5cos2Î¸ + 7sin2Î¸ â€“ 6 = 0

Solution:

According to the question,

5cos2Î¸ + 7sin2Î¸ â€“ 6 = 0

We know that,

sin2Î¸ = 1 â€“ cos2Î¸

Therefore,Â 5cos2Î¸ + 7(1 â€“ cos2Î¸) â€“ 6 = 0

â‡’Â 5cos2Î¸ + 7 â€“ 7cos2Î¸ â€“ 6 = 0

â‡’Â -2cos2Î¸ + 1 = 0

â‡’Â cos2Î¸ = Â½

Therefore,Â cos Î¸ = Â±1/âˆš2

Therefore,Â cos Î¸ = cos Ï€/4 or cos Î¸ = cos 3Ï€/4

Since,Â solution of cos x = cos Î± is given by

x = 2mÏ€ Â± Î±Â âˆ€Â mÂ âˆˆÂ Z

Î¸ = nÏ€ Â± Ï€/4, nÂ âˆˆÂ Z

28. Find the general solution of the equation sin x â€“ 3sin2x + sin3x = cos x â€“ 3cos2x + cos3x

Solution:

According to the question,

sin x â€“ 3sin2x + sin3x = cos x â€“ 3cos2x + cos3x

Grouping sin x and sin 3x in LHS and, cos x and cos 3x in RHS,

We get,

sin x + sin3x â€“ 3sin2x = cos x + cos3x â€“ 3cos2x

Applying transformation formula,

cos A + cos B = 2cos ((A + B)/2) cos((A â€“ B)/2)

sin A + sin B = 2sin ((A + B)/2) cos((A â€“ B)/2)

â‡’

â‡’Â 2sin 2x cos x â€“ 3sin 2x = 2cos 2x cos x â€“ 3cos 2x

â‡’Â 2sin 2x cos x â€“ 3sin 2x â€“ 2cos 2x cos x + 3cos 2x = 0

â‡’Â 2cos x (sin 2x â€“ cos 2x) â€“ 3(sin 2x â€“ cos 2x) = 0

â‡’Â (sin 2x â€“ cos 2x)(2cos x â€“ 3) = 0

â‡’Â cos x = 3/2 or sin 2x = cos 2x

As cos xÂ âˆˆÂ [-1,1]

Hence,Â no value of x exists for which cos x = 3/2

Therefore,Â sin 2x = cos 2x

â‡’Â tan 2x = 1 = tan Ï€/4

We know solution of tan x = tan Î± is given by,

x= nÏ€ + Î± , nÂ âˆˆÂ Z

Therefore,Â 2x = nÏ€ + (Ï€/4)

â‡’ x = nÏ€/2 + (Ï€/8), nÂ âˆˆÂ Z

29. Find the general solution of the equation (âˆš3 â€“ 1) cos Î¸ + (âˆš3 + 1) sin Î¸ = 2

[Hint: Put âˆš3 â€“ 1 = rÂ sin Î±, âˆš3 + 1 = rÂ cos Î± which gives tan Î± = tan((Ï€/4) â€“ (Ï€/6)) Î± = Ï€/12]

Solution:

Let, r sinÎ± = âˆš3 â€“ 1 and r cosÎ± = âˆš3 + 1

Therefore, r = âˆš{(âˆš3 â€“ 1)2 + (âˆš3 + 1)2} = âˆš8 = 2âˆš2

And,Â tan Î± = (âˆš3 â€“ 1) / (âˆš3 + 1)

Therefore,Â r(sinÎ± cos Î¸ + cosÎ± sin Î¸) = 2

â‡’ r sin (Î¸+Î±) = 2

â‡’ sin (Î¸+Î±) = 1/âˆš2

â‡’ sin (Î¸+Î±) = sin (Ï€/4)

â‡’ Î¸+Î± = nÏ€ + (â€“ 1)n (Ï€/4), nÂ âˆˆÂ Z

â‡’ Î¸ = nÏ€ + (â€“ 1)n (Ï€/4) â€“ (Ï€/12), nÂ âˆˆÂ Z

Objective Type Questions

30. If sin Î¸ + cosec Î¸ = 2, then sin2Î¸ + cosec2Î¸ is equal to
A. 1

B.4
C. 2
D. None of these

Solution:

C. 2

Explanation:

According to the question,

sin Î¸ + cosec Î¸ = 2

Squaring LHS and RHS,

We get,

â‡’Â (sin Î¸ + cosec Î¸)2Â = 22

â‡’Â sin2Î¸ + cosec2Î¸ + 2 sin Î¸ cosec Î¸ = 4

â‡’Â sin2Î¸ + cosec2Î¸ + 2 sin Î¸ (1/sin)Î¸ = 4

â‡’Â sin2Î¸ + cosec2Î¸ + 2 = 4

â‡’Â sin2Î¸ + cosec2Î¸ = 2

Thus, option (C) 2 is the correct answer.

31. If f(x) = cos2x + sec2x, then
A. f(x) < 1
B. f(x) = 1
C. 2 < f(x) < 1
D. f(x) â‰¥ 2
[Hint: A.M â‰¥ G.M.]

Solution:

D. f(x) â‰¥ 2

Explanation:

According to the question,

We have, f(x) = cos2x + sec2x

We know that, A.M â‰¥ G.M.

â‡’Â cos02x + sec2x â‰¥ 2

â‡’Â f(x) â‰¥ 2

Thus, option (D) f(x) â‰¥ 2 is the correct answer.

32. If tan Î¸ = 1/2 and tanÂ Ï•Â = 1/3, then the value of Î¸ +Â Ï•Â is
A. Ï€/6
B. Ï€
C. 0
D. Ï€/4

Solution:

D. Ï€/4

Explanation:

According to the question,

Thus, option (D) Ï€/4 is the correct answer.

33. Which of the following is not correct?
A. sin Î¸ = â€“ 1/5
B. cos Î¸ = 1
C.Â sec Î¸ = Â½
D. tan Î¸ = 20

Solution:

C.Â sec Î¸ = Â½

Explanation:

According to the question,

We know that,

a)Â sin Î¸ = â€“ 1/5 is correct since Sin Î¸Â âˆˆÂ [-1,1]

b) cos Î¸ = 1 is correct since Cos Î¸Â âˆˆÂ [-1,1]

c)Â sec Î¸ = Â½

â‡’ (1/cos Î¸) = Â½

â‡’cosÂ Î¸=2 is incorrect since CosÂ Î¸Â âˆˆÂ [-1,1]

d) tan Î¸ = 20 is correct since tan Î¸Â âˆˆÂ R.

Thus, option (C) sec Î¸ = Â½ is the correct answer.

34. The value of tan 1Â° tan 2Â° tan 3Â°â€¦ tan 89Â° is
A. 0
B. 1
C. Â½
D. Not defined

Solution:

B. 1

Explanation:

According to the question,

tan 1Â° tan 2Â° tan 3Â°â€¦ tan 89Â°

= tan 1Â° tan 2Â° â€¦ tan 45Â° tan (90-44Â°) tan(90-43Â°)â€¦tan (90-1Â°)

= tan 1Â°tan 2Â° â€¦ tan 45Â°cot 44Â°cot 43Â°â€¦cot 1Â° [âˆµtan (90-Î¸)=cot Î¸]

= tan 1Â° cot 1Â° tan 2Â° cot 2Â°â€¦tan45Â°â€¦ tan 89Â° cot 89Â°

=1.1â€¦.1 = 1

Thus, option (B) 1 is the correct answer.

35. The value of (1 â€“ tan215o)/(1 + tan215o) is
A. 1

B.Â âˆš3

C. âˆš3/2
D. 2

Solution:

C. âˆš3/2

Explanation:

According to the question,

Let Î¸ = 15Â°Â â‡’Â 2Î¸ = 30Â°

Now, since we know that,

Thus, option (C) âˆš3/2 is the correct answer.

36. The value of cos 1Â° cos 2Â° cos 3Â°â€¦ cos 179Â° is
A.Â 1/âˆš2
B. 0
C. 1
D. â€“1

Solution:

B. 0

Explanation:

According to the question,

Since cos90Â° =0

We get,

â‡’Â cos 1Â° cos 2Â° cos 3Â°â€¦ cos90Â°â€¦ cos 179Â° = 0

Thus, option (B) 0 is the correct answer.

37. If tan Î¸ = 3 and Î¸ lies in third quadrant, then the value of sin Î¸ is
A. 1/âˆš10
B. â€“ 1/âˆš10
C.Â â€“ 3/âˆš10
D.Â 3/âˆš10

Solution:

C.Â â€“ 3/âˆš10

Explanation:

According to the question,

Given that, tan Î¸ = 3 and Î¸ lies in third quadrant

â‡’ cot Î¸ = 1/3

We know that,

Cosec2Î¸ = 1+cot2Î¸

Thus, option (C)Â â€“ 3/âˆš10 is the correct answer.

38. The value of tan 75Â°â€“ cot 75Â° is equal to
A. 2âˆš3

B.Â 2 + âˆš3

C. 2 â€“ âˆš3
D. 1

Solution:

A. 2âˆš3

Explanation:

According to the question,

We have,

tan 75Â°â€“ cot 75Â°

= -2cot150Â°

= -2 cot (180Â°-30Â°)

= 2cot30Â°

=2âˆš3

Thus, option (A)Â 2âˆš3 is the correct answer.