Previous Year Questions: Introduction to Trigonometry

Ans: We have, tanA + cotA = 6 
On squaring both sides, we get (tanA + cotA)² = 36 
⇒ tan²A + cot²A + 2 tanA cotA = 36 
Since tan A cot A = 1
⇒ tan²A + cot²A + 2 = 36
⇒ tan ²A + cot ²A = 36 - 2 = 34
∴ tan²A + cot²A - 4 = 34 - 4 = 30

Ans: (d) 
We have, tan3θ = √3 
⇒ tan3θ = tan 60° 

Ans: (d) 
Given, sin 4θ = √3/2 = sin 60°.
⇒ 4θ = 60°
⇒ θ = 15°.

Ans: (b) 
α + β = 90° and α = 2β     ...(i) 
∴ 2β + β = 90° ⇒ 3β = 90° ⇒ β = 30° 
∴ α = 2 × 30° = 60° [From (i)] 
∴ cos²α + sin²β = cos²60° + sin²30°

Ans: (c) 

Ans: Given, 4k = tan²60° - 2 cosec²30° - 2 tan²30° 

Ans: We have, x cos60° + y cos0° + sin30° - cot45° = 5 

Ans: 

Ans: (a) 
We have,
[∵ 1 - cos²θ = sin²θ]

Ans: (b) 
We have, 
(tan A · cosec A)² - (sin A . sec A)²  
Using identities:
tan A = sin A / cos A,
cosec A = 1 / sin A,
sec A = 1 / cos A.
sec² A - tan² A = 1

Ans: (a)
We have, cotθ + tanθ 
cot θ = cos θ / sin θ,
tan θ = sin θ / cos θ.

Ans: (c)

Ans: (d) 
Given, sin B = 1/4

Ans: 
We have, a sec θ + b tan θ = m
and b sec θ + a tan θ = n
Taking, m² - n²
= (a sec θ + b tan θ)² - (b sec θ + a tan θ)²
= a²sec²θ + b²tan²θ + 2ab tan θ sec θ - b²sec²θ - a²tan²θ - 2ab tan θ sec θ
= a²(sec²θ - tan²θ) + b²(tan²θ - sec²θ)
= a² × 1 + b² × (-1)
= a² - b²
∴ m² - n² = a² - b²
⇒  a² + n² = b² + m²
Hence, proved.

Ans: 
To prove: tan²A + 1 = sec²A 
Taking L.H.S., tan² A+ 1 

∴ L.H.S = R.H.S

Ans: 

Ans: 

Ans: 

Ans: 

Factor numerator and denominator:

Using

So, 

Cancel the common factor:

Hence proved.

Ans: sin θ + cos θ = x ...(i) 
Squaring both sides in equation (i), 
sin² θ + cos² θ + 2sin θ cos θ = x² 
2sin θ cos θ = x² - 1 ...(ii) 
[∵ sin² θ + cos² θ = 1] 
Also, sin² θ + cos² θ = (sin θ + cos θ)² - 2sin θ cos θ

Ans: 

Ans: (c)
sin α = √3/2, ⇒ sin α  = sin 60º
⇒ α = 60º
∵ cos β = √3/2, 
⇒ cos β = cos 30º 
⇒ β = 30º 
tan α. tan β = tan 60º. tan 30º
= √3 x 1√3

= 1

Ans:

5 tan 60°(sin² 60° + cos² 60°) tan 30° = 5 × √31 × 1√3

= 5 × √3 × √3

= 5 × 3

= 15

Ans:
L.H.S. = (cosec θ - sin θ) (sec θ - cos θ) (tan θ + cot θ)
= (cosec θ - sin θ) (sec θ - cos θ) (tan θ + cot θ)
Since, cosec θ = 1/sin θ, sec θ =  1/cos θ, tan θ = sin θ/cos θ, cot θ = cos θ/sin θ
= (1/sin θ - sin θ) (1/cos θ - cos θ) (sin θ/cos θ + cos θ/sin θ)

= 1 - sin²θsin θ × 1 - cos²θcos θ × sin²θ + cos²θsin θ . cos θ

= cos²θ × sin²θsin θ × cos θ × 1sin θ . cos θ

= sin θ . cos θ1 × 1sin θ . cos θ    [ : sin²θ + cos²θ = 1 ]

= 1 = R.H.S.
Hence, proved.

Ans:

Given:

2 tan A = 3 ⇒ tan A = 3/2

Using sin² A + cos² A = 1, let:

sin A = 3/√13, cos A = 2/√13

Substituting in the given expression:

4 sin A + 5 cos A6 sin A + 2 cos A

= 4 × 3/√13 + 5 × 2/√136 × 3/√13 + 2 × 2/√13

= 12/√13 + 10/√1318/√13 + 4/√13

= 22/√1322/√13

= 1

Ans: (c)
Sol:
58 × (2)² - (√3)² + (1√2)² = 5/8 × 4 - 3 + 12 = 0 

Ans: Since θ = 45°, sec 45° = √2, cosec 45° = √2, sin 45° = 1/√2 cos 45° = 1/√2
2sec² θ + 3 cosec² θ - 2 sin θ cos θ
= 2 (√2)² + 3 (√2)² - 2 (1√2) × (1√2)

= 2 × 2 + 3 × 2 - 2 × 12

= 4 + 6 - 1

= 9

Ans: (c)

Option (a): cos²θ - sin²θ - 1
Using the identity: cos²θ - sin²θ = cos 2θ, we get cos²θ - sin²θ - 1 = cos 2θ - 1, which is not always true. So, this option is incorrect.

Option (b): cosec²θ - sec²θ - 1
Using the identities cosec²θ = 1 + cot²θ and sec²θ = 1 + tan²θ, 
we get cosec²θ - sec²θ - 1 = (1 + cot²θ) - (1 + tan²θ) - 1 = cot²θ - tan²θ - 1, which is not always zero. So, this option is incorrect.

Option (c): sec²θ - tan²θ - 1
Using the identity sec²θ = 1 + tan²θ, 
we get sec²θ - tan²θ - 1 = (1 + tan²θ) - tan²θ - 1 = 0, which is always true. 
So, this option is correct.

Option (d): cot²θ - tan²θ = 1
Using the identity cot²θ = 1/tan²θ, we get cot²θ - tan²θ = (1/tan²θ) - tan²θ, which is not always equal to 1. So, this option is incorrect.

Ans: 

Given: sinθ + cosθ = √2

Squaring both sides:

(sinθ + cosθ)² = (√2)²

sin²θ + cos²θ + 2sinθ cosθ = 2

Using identity: sin²θ + cos²θ = 1

1 + 2sinθ cosθ = 2

2sinθ cosθ = 1

sinθ cosθ = 1/2

Ans: Given, sin α = 1/√2 and cot β = √3
We know that, cosec α = 1/sinα = √2
Also, 1 + cot²β = cosec²β
⇒ cosec²β = 4
⇒ cosec β = √4 = 2 
Now, cosec α + cosec β = √2 + 2

Ans: LHS = sec A(1 + sin A )( sec A - tan A)

= 1cos A (1 + sin A) 1cos A - sin Acos A

= 1cos A (1 + sin A) (1 - sin Acos A)

= 1 - sin² Acos² A = cos² Acos² A

= 1

= RHS

Hence proved..

Ans: (b)

(sec²θ - 1) (cosec²θ - 1) = tan²θ . cot²θ

tan²θtan²θ       [ ∵ sec²θ - 1 = tan²θ & cosec²θ - 1 = cot²θ ]

= 1

Ans: Given, 
sin θ - cos θ = 0 
sin θ = cos θ 
tan θ = 1 
tan θ = tan 45° 
⇒ θ = 45° 
Now, sin⁴ θ + cos⁴ θ = sin⁴ 45° + cos⁴ 45°

= (1√2)⁴ + (1√2)⁴

= 14 + 14 = 12

Ans:

LHS = sin A - 2 sin³ A2 cos³ A - cos A

= sin A (1 - 2 sin² A)cos A (2 cos² A - 1)

= sin A (1 - 2 (1 - cos² A)cos A (2 cos² A - 1)

= tan A 1 - 2 + 2 cos² A2 cos² A - 1

= tan A 2 cos² A - 12 cos² A - 1

= tan A

= RHS

Ans: (c)
Sol:

Ans: (c)
Sol: Part 1 simplify
Part 2 simplify
Add both parts:

Ans: (a)
Sol: Given, 2 sin2θ = 1 ⇒ sin2θ = 1/2
⇒ 2θ = 30°
⇒ θ = 15°

Ans: (b)
Sol: Given, sin²θ + sinθ = 1 ---(i)
sinθ = 1 - sin²θ
⇒ sinθ = cos²θ ---(ii)
∴ cos²θ + cos⁴θ
= sinθ + sin²θ [From (ii)]
= 1        [From (i)]

Ans: We have, 3 sin A = 1
∴ sin A = 1/3
Now by using cos² A = 1 - sin² A, we get

cos² A = 1 - 19 = 89
⇒ cos A = 2√23

∴ sec A = 1cos A = 32√2  = 3√24

Ans: We have, L.H.S.
1 + cot²θ1 + tan²θ = cosec²θsec²θ
[By using 1 + tan²θ = sec²θ and 1 + cot² θ = cosec²θ ]
⇒ 1sin²θ = cos²θsin²θ = cot²θ
Hence,
1 + cot²θ1 + tan²θ = cot²θ 

Ans: (a)
Sol: We have, sin θ = cos θ
or sin θ / cos θ = 1
⇒ tan θ = 1 and cot θ = 1     [∵ cot θ = 1/tanθ]
∴ tan² θ + cot² θ = 1²  + 1²  = 2
Hence, A option is correct.

Ans: In right angle ΔABC we have
15 cot A = 8
⇒ cot A = 8/15

Since, cot A = AB/BC
∴ AB/BC = 8/15
Let AB = 8k and BC = 15k
By using Pythagoras theorem, we get
AC² = AB² + BC²
⇒ (8k)² + (15)² = 64k² + 225k² = 289k² = (17k)² 

⇒ AC = √((17k)²) = 17k

∴ sin A = BCAC = 15k17k = 1517

and cos A = ABAC = 8k17k = 817

So, sec A = 1cos A = 178

Ans:  We have, sin² 30° + cos² 60°

Ans: (c)
Sol: Given the point (acos θ + b sin θ , 0), (0 , a sin θ - b cos θ)
By distance formula,

The distance of

AB = √(x₂ - x₁)² + (y₂ - y₁)²

So,

AB = √(a cos θ + b sin θ - 0)² + (0 - a sin θ + b cos θ)²

= √ a²(sin²θ + cos²θ) + b²(sin²θ + cos²θ)

But according to the trigonometric identity,

sin²θ + cos²θ = 1

Therefore,

AB = √ a² + b²

Ans: We have 5(tan²θ - sec²θ)
= 5(-1) = - 5 [By using 1 + tan²θ = sec² θ ⇒ tan²θ - sec²θ = - 1]

Ans: Given: sinθ + cosθ = √2

Squaring both sides:

(sinθ + cosθ)² = 2

sin²θ + cos²θ + 2sinθ cosθ = 2

1 + 2sinθ cosθ = 2

sinθ cosθ = 1/2

Now,

tanθ + cotθ = (sinθ / cosθ) + (cosθ / sinθ)

= (sin²θ + cos²θ) / (sinθ cosθ)

= 1 / (1/2)

= 2

Ans: Given, x = a sin θ and y = b cos θ 
Putting the values of x and y in  (b²x² + a²y²)
We get, 
= b²(a sin θ)² + a²_{(b cos θ)}²
= a²b² [sin² θ + cos² θ]   [Also, sin²θ + cos²θ = 1]
= a²b² [1]  
= a²b²

Ans: We know that, 
sin² θ + cos² θ = 1 
So, (sin² θ + cos² θ) ² = 1² 
⇒ sin⁴ θ + cos⁴ θ + 2sin² θ cos² θ = 1 
i.e., sin⁴ θ + cos⁴ θ = 1 - 2 sin² θ cos² θ ...(i) 
Also, (sin² θ + cos² θ) ³ = 1³
⇒ sin⁶ θ + cos⁶ θ + 3 sin² θ cos² θ (sin² θ + cos² θ) = 1 
⇒ sin⁶ θ+ cos⁶ θ+ 3sin2 θ cos² θ (1) = 1 
i.e., sin⁶ θ + cos⁶ θ = 1 - 3 sin² θ cos² θ ...(ii) 
Now, 
LHS = 2(sin⁶ θ + cos⁶ θ) - 3(sin⁴ θ + cos⁴ θ) + 1 
= 2(1 - 3 sin² θ cos² θ) - 3(1 - 2 sin² θ cos² θ) + 1 
= 2 - 3 + 1 
= 0 
Hence, proved.

Ans: L.H.S. = (sin⁴ θ - cos⁴ θ + 1) cosec² θ 
= [(sin² θ + cos² θ) (sin² θ - cos² θ) + 1] cosec² θ
[(1) (sin² θ - cos² θ) + 1] cosec² θ         as [ sin² θ + cos² θ = 1] 
= [sin² θ + (1 - cos² θ)] cosec² θ 
= (sin² θ + sin² θ) cosec²θ
= (2 sin² θ) cosec² θ
=

= 2 × 1
= 2 = R.H.S.
Hence, proved.

Ans: Given,
⇒ sin x + cos y = 1
⇒ sin 30° + cos y = 1
⇒ 1/2 + cos y = 1
⇒ cos y = 1 - 1/2
⇒ cos y = 1/2
⇒ cos y = cos 60°.
Hence, y = 60°.

Ans:  Given:
cosec² θ (cos θ - 1)(1 + cos θ) = k
Concept used:
Cosec α = 1/Sin α
Sin² α + Cos² α = 1
(a + b)(a - b) = a² - b²
Calculation:
cosec² θ (cos θ - 1)(1 + cos θ) = k
⇒ cosec² θ (1 - cos θ)(1 + cos θ) = - k
⇒ cosec² θ (1 - cos² θ) = -k      [Also, sin² θ + cos² θ = 1]
⇒ cosec² θ × sin² θ = -k
⇒ 1 = -k
⇒ k = -1
∴ The value of k is (-1).

Ans: 

(1 + cos Asin A - 1sin A ) (1 + sin Acos A + 1cos A )

= sin A + cos A - 1sin A × cos A + sin A + 1cos A

= (sin A + cos A)² - 1sin A . cos A

= sin²A + cos²A + 2 sin A . cos A - 1sin A . cos A

= sin²A + cos²A - 1 + 2 sin A . cos Asin A . cos A

= 2

Ans: (a)

sec θ + tan θ + 1 = 0

⇒ sec θ + tan θ = -1

Multiplying and dividing LHS by sec θ - tan θ, we get

⇒ (sec θ + tan θ) × sec θ - tan θsec θ - tan θ = -1

⇒ sec² θ - tan² θsec θ - tan θ = -1

⇒ 1 + tan² θ - tan² θsec θ - tan θ = -1 (∵ sec² θ = 1 + tan² θ)

⇒ 1sec θ - tan θ = -1

⇒ sec θ - tan θ = -1

Hence, the correct option is (a).