Test: Trigonometric Identities - 1 - SAT MCQ

Concept:
sec² x - tan² x = 1
Calculation:
Given sec x + tan x = 2     ....(i)
∵ sec² x - tan² x = 1
(sec x + tan x)(sec x - tan x) = 1
2(sec x - tan x) = 1
sec x - tan x = 1/2    ....(ii)
Adding the equation (i) and (ii)
2 sec x = 2 + 1/2
2/cos⁡x = 5/2
cos x = 4/5

Given:
Tan⁴θ + Tan²θ = 1
Formula:
Sec²θ - Tan²θ = 1
Sin²θ + Cos²θ = 1
Calculation:
Tan⁴θ + Tan²θ = 1
⇒ Tan²θ (Tan²θ + 1) = 1
⇒ Tan²θ. Sec²θ = 1      {∵ sec²θ = 1 + tan²θ}
⇒ (sin² θ/cos⁴ θ) = 1
⇒ 1 – cos²θ = cos⁴ θ      {∵ sin² θ = 1 – cos²θ}
⇒ cos⁴ θ + cos² θ = 1
∴ Then the value of cos⁴ θ + cos² θ is 1.

Formula used:
tan(90° - θ) = cot θ
tan θ × cot θ = 1

Calculation:
Given that,
tan 48° tan 23° tan 42° tan 67° = tan(A + 30∘)
⇒ tan (90° - 42°)tan(90 - 67°)tan 42°tan 67° = tan(A +  30°)
⇒ cot 42° cot 67° tan 42°tan 67° = tan(A + 30°)
∵ tan θ × cot θ = 1
⇒ 1 × 1 = tan(A + 30°)
⇒ tan 45° = tan(A + 30°)    (∵ tan 45° = 1)
⇒ 45° = A + 30°
⇒ A = 15°

Concept:
a² - b² = (a - b) (a + b)
sec² x - tan² x = 1

Calculation:
sec⁴ x - tan⁴ x
=(sec² x - tan² x) (sec² x + tan² x)          (∵ a² - b² = (a - b) (a + b))
= 1 × (1 + tan² x + tan² x) (∵ sec² x - tan² x = 1)
= 1 + 2tan² x

Concept:
sin² x + cos² x = 1
Calculation:
Given: sin θ + cos θ = 7/5 
By, squaring both sides of the above equation we get,
⇒ (sin θ + cos θ)² = 49/25
⇒ sin² θ + cos² θ + 2sin θ.cos θ = 49/25
As we know that, sin² x + cos² x = 1
⇒ 1 + 2sin θcos θ = 49/25
⇒ 2sin θcos θ = 24/25
∴ sin θcos θ = 12/25

Calculation:
m = sin θ + cos θ 
⇒ m² = (sin θ + cos θ)²
⇒ m² = sin²θ + cos²θ + 2sinθ.cosθ 
⇒ m² = 1 + 2sinθcosθ 

⇒ n(m² - 1) = 2m
⇒ n(m + 1)(m - 1) = 2m

Concept:
The question employs the concept of complementary trigonometric identities.
tanθ = cot (90 - θ)
sinθ = cos (90 - θ)
secθ = cosec (90 - θ)

Calculation:
Since tanθ and cot (90 - θ), forms a complementary pair-
⇒ 2θ + 3θ + 10°  = 90° 
⇒ 5θ + 10° = 90°
⇒ 5θ = 80°
∴ θ = 16° 
The value of θ is 16° 

Formula:
sin2θ = 2sinθcosθ 
sin(A - B) = sinAcosB - cosAsinB

Calculation:
Multiplying the left hand side with cosθ, we get-

sin2θ can also be written as sin(3θ - θ)
 
Applying sin(A-B) formula in the numerator part, we get-

⇒ tan3θ - tanθ 
Equating LHS with RHS, we get-
⇒ tan3θ - tanθ = tan 270° - tan θ
⇒ 3θ = 270° 
⇒ θ = 90°
∴ The value of θ is 90°

Formula Used:

a² - b² = (a - b) (a + b)
Calculation:
We have to find the value of

tan θ & cot θ can be written as,

Hence,
(sin θ + cos θ)

Formula used:
sin(360° - θ) = - sin θ

Calculation:
sin 0° + sin 10° + sin 20° + sin 30° + ⋯ + sin 360°
⇒ sin 0° + sin 10° + ....+ sin 180° + sin (360 - 170°) +  sin (360 - 160°) + .....+ sin (360 - 10°) + sin (360 - 0°)  

By using the  above formula
⇒ sin 0° + sin 10° + ......- sin 10° - sin 0° = 0
∴ sin 0° + sin 10° + sin 20° + sin 30° + ⋯ + sin 360° = 0

Concept:
I. sin (90° - θ) = cos θ
II. sin² θ + cos² θ = 1

Calculation:
⇒ sin² 5° + sin² 10° + sin² 15° + …… + sin² 85° + sin² 90°
= (sin² 5° + sin² 85°) + (sin² 10° + sin² 80°) + ….. +(sin² 40° + sin² 50°)+ sin² 45° + sin² 90°
= (sin² 5° + sin² (90° - 5°)) + (sin² 10° + sin² (90° - 10°)) + ….. +(sin² 40° + sin² (90° - 40°))+ sin² 45° + sin² 90°
As we know that, sin (90° - θ) = cos θ
= (sin² 5° + cos² 5°) + (sin² 10° + cos² 10°) + ….. +(sin² 40° + cos² 40°)+ sin² 45° + sin² 90°
As we know that, sin² θ + cos² θ = 1
= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + (1/√2)² + 1
= 9 + (1/2) = 19/2

Concept:
I. sec² θ – tan² θ = 1
II. a² – b² = (a - b) (a + b)

Calculation:
Given:
tan θ + sec θ = 4     ...(1)
As we know that, sec² θ – tan² θ = 1
⇒ sec² θ – tan² θ = 1
⇒ (sec θ – tan θ) (sec θ + tan θ) = 1
By substituting the value of tan θ + sec θ = 4, in the above equation, we get
⇒ sec θ – tan θ = 1/4     ... (2)
Adding equation (1) and (2), we get
⇒ 2 sec θ = 17/4
⇒ sec θ = 17/8
cos θ = 8/17

Concept:
cosec² x – cot² x = 1

Calculation:
Given: p = cosec θ – cot θ and q = (cosec θ + cot θ)^-1
⇒ cosec θ + cot θ = 1/q
As we know that, cosec² x – cot² x = 1
⇒ (cosec θ + cot θ) × (cosec θ – cot θ) = 1
1/q × p = 1
⇒ p = q

Given:
tanA + tan(45^∘) + tanC = tan(45^∘) × tanA × tanC,

Concept used:
(1.) If A + B + C = 180∘, tanA + tan(B) + tanC = tanA × tanB × tanC,
(2.) Sum of all internal angles of a triangle is 180∘.

Calculation:
According to the question,
⇒ tanA + tan(45^∘) + tanC = tan(45^∘) × tanA × tanC
As per the above concept.
⇒ A + B + C = 180^∘
⇒ A + 45∘ + C = 180^∘
⇒ A + C = 180^∘ - 45^∘
⇒ A + C = 135^∘
Therefore, 
⇒ tan(A + C) = tan(135^∘)
⇒ tan(135∘) = tan(90^∘ + 45^∘) = -1
Therefore, '-1' is the required answer.

Formula:
1 + cot²θ = cosec²θ

Calculation:
Squaring both equations, we get-
⇒ (3 - 4cot)² = (cosecθ)²
⇒ 9 + 16cot²θ - 2 × 3 × 4cotθ = cosec²θ
⇒ 9 + 16cot²θ - 24cotθ = cosec²θ (Eq 1)
Similarly, 
⇒ (4 + 3cotθ)² = (kcosecθ)²
⇒ 16 + 9cot²θ + 2 × 4 × 3cotθ = k²cosec²θ 
⇒ 16 + 9cot²θ + 24cotθ = k²cosec²θ (Eq 2)
Adding Eq 1 and Eq 2, we get-
⇒  9 + 16cot²θ - 24cotθ + 16 + 9cot²θ + 24cotθ = cosec²θ + k²cosec²θ
⇒ 25 + 25cot2θ = cosec²θ (1 + k²)
⇒ 25 (1 + cot²θ) = cosec²θ (1 + k²)
Substituting the value of 1 + cot²θ as cosec²θ, we get-
⇒ 25 × cosec²θ = cosec2θ (1 + k²)
⇒ 25 = 1 + k²
∴ k² = 24 or k = 2√6
The value of k is 2√6