Test: Trigonometric Identities - 3 - SAT MCQ

It is given that, tan θ = 4/3
⇒ sin θ / cos θ = 4/3
So sin θ = 4, and cos θ = 3
Now, on putting the values of sin θ and cos θ in (3 sin θ + 2 cos θ) / (3 sin θ - 2 cos θ), we will get -
= 3x4 + 2x3/ 3x4 - 2x3
= 18/6
= 3

Given A = tan 11⁰ . tan 29⁰, and B = 2 cot 61⁰ . cot 79⁰
A / B = tan 11⁰ . tan 29⁰ / 2 cot 61⁰ . cot 79⁰
= [tan 11⁰ . tan 29⁰] / [2 cot (90⁰ - 29⁰) . cot (90⁰ - 11⁰)]
= tan 11⁰ . tan 29⁰ / 2 tan 11⁰ . tan 29⁰
= 1/2
So, A/B = 1/2
Or, 2A = B

Given sin (θ + 30⁰) = 3/√12
It can be written as sin (θ + 30⁰) = 3/2√3
Or, sin (θ + 30⁰) = √3/2
⇒ sin (θ + 30⁰) = sin 60⁰
⇒ θ + 30⁰ = 60⁰
⇒ θ = 30⁰
On putting θ = 30⁰, in cos² θ, we will get
cos² 30⁰ = (√3/2)²
= 3/4

cos² 55⁰ + cos² 35⁰ + sin² 65⁰ + sin² 25⁰
⇒ cos² (90⁰ - 35⁰) + cos² 35⁰ + sin² 65⁰ + sin² (90⁰ - 65⁰)
⇒ (sin² 35⁰ + cos² 35⁰) + (sin² 65⁰ + cos² 65⁰)
⇒ 1 + 1
= 2

Given cot 45⁰.sec 60⁰ = A tan 30⁰.sin 60⁰
So, 1 x 2 = A 1/√3 x √3/2
⇒ 2 = A/2
So, A = 4

(3 / 1+tan² θ) + 2 sin² θ + (1 / 1+cot² θ) = ?
According to the trigonometric identities, the given equation can be written as -
= 3/sec² θ + 2 sin² θ + 1/cosec² θ
= 3cos² θ + 2 sin² θ + sin² θ
= 3cos² θ + 3sin² θ
= 3(cos² θ + sin² θ)
= 3

Given: (4 sec² 30⁰ + cos² 60⁰ - tan² 45⁰) / (sin² 30⁰ + cos² 30⁰)
We have to put the numerical values,
= [4 (2/√3)² + (½)² - (1)²] / 1
⇒ sec² A - 1 = 1 + 2 (sec² B - 1)
⇒ sec² A - 1 = 1 + 2 sec² B - 2
⇒ sec² A - 1 = 2 sec² B - 1
⇒ 1/cos² A = 2/cos² B
⇒ cos² B = 2cos² A
⇒ or, cos B = √2 cos A
⇒ So, √2 cos A - cos B = 0

(5 / sec² θ) + 3 sin² θ + (2 / 1+cot² θ) = ?
According to the trigonometric identities, the given equation can be written as -
= 5cos² θ + 3 sin² θ + 2/cosec² θ
= 5cos² θ + 3 sin² θ + 2sin² θ
= 5cos² θ + 5sin² θ
= 5(cos² θ + sin² θ)
= 5

 tan 60⁰ = √3, cot 30⁰ = √3
So, cot 30⁰/tan 60⁰ = √3 / √3
= 1

It is given that, cos θ + sin θ = √2 cos θ …..(i)
On squaring both sides, we will get,
(cos θ + sin θ)² = (√2 cos θ)²
⇒ cos² θ + sin² θ + 2 sin θ cos θ = 2 cos² θ
Or, 2cos² θ - cos² θ - sin² θ = 2 sinθ cosθ
⇒ cos² θ - sin² θ = 2 sin θ cos θ
⇒ (cos θ + sin θ) (cos θ - sin θ) = 2 sin θ cos θ
⇒ (√2 cos θ) (cos θ - sin θ) = 2 sin θ cos θ [from equation (i)]
⇒ (cos θ - sin θ) = 2 sinθ cosθ / √2 cos θ
= √2 sin θ

Here, we can apply the formula -
cot A. cot B = 1 (when A + B = 900)
= (cot 200 . cot 700) x (cot 100 . cot 800) x cot 600
= 1 x 1 x 1/√3
= 1/√3
So, the correct value of cot 100.cot 200.cot 600.cot 700.cot 800 = 1/√3

Given tan θ + cot θ = 2
Put θ = 45⁰, above equation will satisfy as,
1 + 1 = 2
So, θ = 45⁰,
= tan¹⁰⁰ 45⁰ + cot¹⁰⁰ 45⁰
= 1¹⁰⁰ + 1¹⁰⁰
= 2

Given 7 sin² θ + 3 cos² θ = 4
=> 7 sin² θ + 3 (1 - sin² θ) = 4
=> 7 sin² θ + 3 - 3sin² θ = 4
Then, 4sin² θ = 1
Or, sin θ = 1/2
So, θ = 30⁰
Now, put θ = 30⁰ in tan θ, we will get,
tan θ = 1/√3

Alternate
We can directly check the equation by putting values of θ. Let's put θ = 30⁰
7 sin² 30⁰ + 3 cos² 30⁰= 4
Then, 7 * 1/4 + 3 * 3/4 = 4
So, 7/4 + 9/4 = 4
16/4 = 4
Or, 4 = 4 (so, it satisfy the condition)
Now, tan 30⁰ = 1/√3

Given 4 cos² θ - 1 = 0
4 cos² θ = 1
cos² θ = 1/4
cos θ = 1/2
Or, θ = 60⁰
So, tan (θ - 15⁰) = ?
⇒ tan (60⁰ - 15⁰)
= tan 45⁰
= 1

Given cos⁴ θ - sin⁴ θ = 2/3
Now, here we can apply the formula -
a⁴ - b⁴ = (a² - b²) (a² + b²)
So, (cos² θ - sin² θ) (cos² θ + sin² θ) = 2/3
So, 1 x (cos² θ - sin² θ) = 2/3 (because cos² θ + sin² θ = 1)
⇒ cos² θ - (1 - cos² θ) = 2/3 (because sin² θ = 1 - cos² θ)
So, 2cos² θ - 1 = 2/3