Class 10 Maths Chapter 8 Previous Year Questions - Introduction to Trigonometry

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

= 1

Ans:

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)


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

Ans:



Hence, the answer is 1.

Ans: (c)
Sol:
 

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

= 4 + 6 – 1 = 9

Ans: (c)

Ans: Given, sinθ +cosθ = √3
Squaring both sides, we get (sinθ + cosθ)² = 3
⇒ sin²θ + cos²θ + 2sinθ cosθ = 3
⇒ 2sinθ cosθ = 3 - 1     ( ∵ sin²θ + cos²θ = 1)
⇒ 2sinθ cosθ = 2
⇒  sinθ cosθ = 1

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
Now, cosec α + cosec β = √2 + 2

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

= 1
= RHS
Hence proved..

Ans: (b)
(sec² θ – 1) (cosec² θ – 1) = tan² θ.cot² θ  

=

= 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°

Ans: 

= tan A
= RHS

Ans: (c)
Sol:
Given, cosθ = √3/2  = B/H

Let B = √3k and H = 2k
∴  [By Pythagoras Theorem]
⇒√k² = k

Ans: (c)
Sol: We have,

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

Ans: We have, L.H.S.

[By using 1 + tan²θ = sec²θ and 1 + cot² θ = cosec²θ ]

Hence,

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 

So, sec A = 1/cosA = 17/8

Ans:  We have, sin² 30° + cos² 60°

Ans: (c)
Sol: Given the point A (cos θ + b sin θ , 0), (0 , a sin θ − b cos θ)
By distance formula,
The distance of

[∵ cos²θ + sin²θ = 1]

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

Ans: sin θ + cos θ =√3
= (sinθ + cosθ)² = 3
= sin² θ + cos² θ + 2sin θ cos θ = 3
⇒ 2sin θ cos θ = 2
⇒ sin θ cos θ = 1
⇒ sin θ cos θ = sin²θ + cos²θ

⇒ tan θ + cot θ = 1

Ans: Given, x = a sin θ and y = b cos θ 
b²x² + a²y² = b²(a² sin² θ) + a²(b² cos² θ) 
= a²b² [sin² θ + cos² θ] 
= a²b² [sin²θ + cos²θ = 1]

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
⇒ cosec2 θ × sin² θ = -k
⇒ 1 = -k
⇒ k = -1
∴ The value of k is (-1).

Ans: 

Ans: (a)
$\sec \theta +\tan \theta +1=0$

$\sec \theta +\tan \theta +1=0$

$\Rightarrow \sec \theta +\tan \theta =-1$

Multiplying and dividing LHS by sec θ – tan θ$\sec \theta -\tan \theta$, we get

$\Rightarrow (\sec \theta +\tan \theta )\times \left(\frac{\sec }{}\right)=-1$

$\Rightarrow \left(\frac{\sec }{2}\right)=-1$

$\Rightarrow \left(\frac{1}{+}\right)=-1(\because \sec 2\theta =1+\tan 2\theta )$

$\Rightarrow \left(\frac{1}{\sec }\right)=-1$

$\Rightarrow (\sec \theta -\tan \theta )=-$Hence, the correct option is (a).