Test: Trigonometric Identities - 2 - SAT MCQ

Given a sin 45⁰ = b cosec 30⁰

So, a/b = cosec 30⁰/ sin 45⁰

a/b = 2/( 1/√2)

a/b = 2√2/1

a⁴/b⁴ = (2√2/1)⁴

a⁴/b⁴ = 64/1

or,

a⁴/b⁴ = 4³

Given α + β = 90⁰, and α : β = 2 : 1

So, we can say that 2x + x = 90⁰

3x = 90⁰, which give

x = 30⁰

So, α = 2x = 60

β = x = 30

cos α / cos β = cos 60⁰ / cos 30⁰

=> (1/2) / (√3/2)

or, 1/2 * 2/√3

= 1/√3

Or the ratio between cos α : cos β = 1 : √3

Given tan θ - cot θ = 0
Let's put θ = 45⁰ in order to satisfy the above equation
tan 45⁰ - cot 45⁰ = 0
1 - 1 = 0 (equation satisfied with θ = 45⁰)
Now, put θ = 45⁰ in sin θ + cos θ, we will get
= sin 45⁰ + cos 45⁰
= 1/√2 + 1/√2
= √2

Given θ + φ = π/2
It can be written as, θ + φ = 90⁰ (as π = 180⁰) …….(i)
sin θ = 1/2
or, θ = 30⁰
On putting the value of θ = 30⁰ in equation (i), we will get,
30⁰ + φ = 90⁰
So, φ = 60⁰
Then, sin φ = sin 60⁰ = √3/2

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)
⇒ (1 - sin² θ) - sin² θ = 2/3
So, 1 - 2sin² θ = 2/3

Given, sin θ/(1 + cos θ) + sin θ/(1 - cos θ)
= [sin θ (1 - cos θ) + sin θ (1 + cos θ)] / [(1 - cos θ) (1 + cos θ)]
= [sin θ - sin θ cos θ + sin θ + sin θ cos θ] / [1 - cos² θ]
= 2 sin θ / sin² θ
= 2 cosec θ

(tan² θ - sec² θ)
= sin² θ/cos² θ - 1/cos² θ
= (sin² θ - 1) / cos² θ
= - cos² θ/cos² θ
= -1

Given tan7θ.tan2θ = 1
As we know, if tanA . tanB = 1 then, A + B = 90⁰
So, 7θ + 3θ = 90⁰
⇒ 9θ = 90⁰
Or, θ = 10⁰
Now, we have to find tan3θ
So, put θ = 10⁰ in tan3θ, we will get
tan 30⁰ = 1/√3

Given sin (θ + 18⁰) = cos 60⁰
sin (θ + 18⁰) = cos (90⁰ - 30⁰)
So, sin (θ + 18⁰) = sin30⁰
Then, θ = 30⁰ - 18⁰
θ = 12⁰
So, cos5θ = cos 5 x 12⁰
= cos 60⁰
= 1/2

The question is in the form of (a + b)²
So, on applying the identity, and after expanding the given equation, we will get -
⇒ sec² A sec² B + tan² A tan² B + 2 sec A sec B tan A tan B - sec² A tan² B - tan² A sec² B - 2 sec A tan B tan A sec B
⇒ Then, sec² A [sec² B - tan² B] - tan² A [sec² B - tan² B]
So, it will be written as [sec² A - tan² A] [sec² B - tan² B]
= 1 x 1
= 1.

Given sec² θ + tan² θ = 7/12
Now, here we can apply the formula -
a⁴ - b⁴ = (a² - b²) (a² + b²)
sec⁴ θ - tan⁴ θ = (sec² θ - tan² θ) (sec² θ + tan² θ)
= 1 x (sec² θ + tan² θ) {because 1 + tan² θ = sec² θ}
= 1 x 7/12
= 7/12

Given r cosθ = √3, and r sinθ = 1
r cosθ / r sinθ = 1/√3
tanθ = tan 30⁰
Or, θ = 30⁰
On putting, θ = 30⁰, we will get,
r sin 30⁰ = 1
r x ½ = 1
or r =2
Now, r² tanθ = ?
= (2)² tan 30⁰
= 4 x 1/√3
= 4/√3

Given, tan² θ + tan⁴ θ = 1 …. (i)
From equation (i),
tan² θ ( 1 + tan² θ ) = 1
tan² θ ( sec² θ ) = 1 [As according to the trigonometric identity, sec² θ - tan² θ = 1]
tan² θ = 1/ sec² θ
tan² θ = cos² θ ….(ii)
Now, cos² θ + cos⁴ θ = ?
⇒ cos² θ + (cos²)² θ
⇒ tan² θ + (tan²)² θ
⇒ tan² θ + tan⁴ θ
= 1 {from equation (i)}

It is given that sin A + cosec A = 2 ……(i)
On putting A = 90⁰, then above condition will satisfy
sin 90⁰ + cosec 90⁰ = 2
or, 1 + 1 = 2 (as the equation satisfies, so, A = 90⁰)
Now, sin⁷ A + cosec⁷ A = ?
⇒ sin⁷ 90⁰ + cosec⁷ 90⁰
⇒ 1⁷ + 1⁷
= 2

Let's see the values -
sin 30⁰ = 1/2
cos 60⁰ = 1/2
sin 60⁰ = √3/2
cos 30⁰ = √3/2
So, (1/2 + 1/2) - (√3/2 + √3/2)
= 1 - 2√3/2
Or, 1 - √3