Chapter 3 Trigonometric Functions Ex 3.1 Solutions

Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 3 Trigonometric Functions Ex 3.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Trigonometric Functions Ex 3.1

(ii) sec θ =
Solution:

(iii) cot θ =
Solution:
The given equation is cot θ = which is same as tan θ = .
We know that,

Hence, the required principal solution are
θ = and θ = .

(iv) cot θ = 0.
Solution:

(ii) tanθ = -1
Solution:
We know that,
tan = 1 and tan(π – θ) = -tanθ,
tan(2π – θ) = -tanθ

Hence, the required principal solutions are
θ = and θ = .

(iii) cosecθ + 2 = 0.
Solution:

(ii) cosθ =
Solution:
The general solution of cos θ = cos ∝ is
θ = 2nπ ± ∝, n ∈ Z
Now, cosθ = = cos …[∵ cos = ]
∴ the required general solution is
θ = 2nπ ± , n ∈ Z.

(iii) tanθ =
Solution:
The general solution of tan θ = tan ∝ is
θ = nπ + ∝, n ∈ Z
Now, tan θ = = tan …[tan = ]
∴ the required general solution is
θ = nπ + , n ∈ Z.

(iv) cotθ = 0.
Solution:
The general solution of tan θ = tan ∝ is
θ = nπ + ∝, n ∈ Z
Now, cot θ = 0 ∴ tan θ does not exist
∴ tanθ = tan [∵ tan does not exist]
∴ the required general solution is
θ = nπ + , n ∈ Z.

(ii) cosecθ = –
Solution:
The general solution of sinθ = sin∝ is

(iii) tanθ = -1
Solution:
The general solution of tanθ = tan∝ is

(ii) tan =
Solution:
The general solution of tan θ = tan ∝ is

(iii) cot 4θ = -1
Solution:
The general solution of tan θ = tan ∝ is

(ii) 4 sin²θ = 1
Solution:
The general solution of sin²θ = sin²∝ is
θ = nπ ± ∝, n ∈ Z
Now, 4 sin²θ = 3

(iii) cos 4θ = cos 2θ
Solution:
The general solution of cos θ = cos ∝ is
θ = 2nπ ± ∝, n ∈ Z
∴ the general solution of cos 4θ = cos 2θ is given by
4θ = 2nπ ± 2θ, n ∈ Z
Taking positive sign, we get
4θ = 2nπ + 2θ, n ∈ Z
∴ 2θ = 2nπ, n ∈ Z
∴ θ = nπ, n ∈ Z
Taking negative sign, we get
4θ = 2nπ – 2θ, n ∈ Z
∴ 6θ = 2nπ, n ∈ Z
∴ θ = , n ∈ Z
Hence, the required general solution is
θ = , n ∈ Z or ∴ θ = nπ, n ∈ Z.
Alternative Method:
cos 4θ = cos 2θ
∴ cos4θ – cos 20 = 0
∴ -2sin∙sin = 0
∴ sin3θ∙sinθ = 0
∴ either sin3θ = 0 or sin θ = 0
The general solution of sin θ = 0 is
θ = nπ, n ∈ Z.
∴ the required general solution is given by
3θ = nπ, n ∈ Z or θ = nπ, n ∈ Z
i.e. θ = , n ∈ Z or θ = nπ, n ∈ Z.

(ii) tan³θ = 3tanθ
Solution:
tan³θ = 3tanθ
∴ tan³θ – 3tanθ = 0
∴ tan θ (tan²θ – 3) = 0
∴ either tan θ = 0 or tan²θ – 3 = 0
∴ either tanθ = 0 or tan²θ = 3
∴ either tan θ = 0 or tan²θ = ( )³
∴ either tan θ = 0 or tan²θ = (tan)³…[tan = ]
∴ either tanθ = 0 or tan²θ = tan²
The general solution of
tanθ = 0 is θ = nπ, n ∈ Z and
tan²θ = tan²∝ is θ = nπ ± ∝, n ∈ Z.
∴ the required general solution is given by
θ = nπ, n ∈ Z or θ = nπ ± , n ∈ Z.

(iii) cosθ + sinθ = 1.
Solution:
cosθ + sinθ = 1

(ii) cos²θ = -1
Solution:
cos²θ = -1
This is not possible because cos²θ ≥ 0 for any θ.
∴ cos²θ = -1 does not have any solution.

(iii) 2 sinθ = 3
Solution:
2 sin θ = 3 ∴ sin θ =
This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 2 sin θ = 3 does not have any solution.

(iv) 3 tanθ = 5
Solution:
3tanθ = 5 ∴ tanθ =
This is possible because tan θ is any real number.
∴ 3tanθ = 5 has solution.