Test: Values of Trigonometric Functions - JEE MCQ

tan(660^o) cot(1200^o)
⇒ tan(720 - 60^o) cot(1080+120^o)
⇒ - tan60^o cot120^o
⇒ - tan60^o (-cot60^o)
⇒ 1

As we know that : sin(180o−x) = sin(x)
Plug in x = 30 to get
sin(180o−30°) = sin(30°)
 ⇒ 1/2
sin(90+θ) = cosθ
sin (90+60)=cos60=1/2

Since, cos(−x) = cosx
⇒ cos(−435°) = cos(435°)
= cos(360° + 75°)
= cos(75o)
= cos(90° − 15°)
=  sin15°

To solve this problem, we'll use the tangent addition formula:

• tan(A + B) = (tanA + tanB) / (1 - tanA * tanB)

We can rewrite 45° as 22° + 23°. So,

• tan(45°) = tan(22° + 23°)

Using the tangent addition formula:

• 1 = (tan22° + tan23°) / (1 - tan22° * tan23°)

Cross-multiplying, we get:

• 1 - tan22° * tan23° = tan22° + tan23°

Rearranging the terms, we get:

• tan22° + tan23° + tan22°.tan23° = 1

Therefore, the value of tan22° + tan23° + tan22°.tan23° is 1.

sec 390°
= sec(360° + 30°)
= sec 30° = 2/(√3)

cos(7π/12)
= cos(3π/12 + 4π/12)
= cos(π/4 + π/3)
Use trig identity: cos (a + b) = cos a.cos b - sin a.sin b
cosa = cos(π/4) = √2/2;
cosb = cos(π/3) = 1/2
sina = sin(π/4) = √2/2;
sina = sin(π/3) = √3/2
cos(7π/12) = cos(π/4 + π/3)
=> (√2/2)(1/2) − (√2/2)(√3/2) 
= (√2 − √6)/4
= √2(1 - √3)/2√2√2
= (1 - √3)/2√2

tan210°
= tan (180° + 30°)
= tan30°
= 1/√3

(cos135° - cos120o)/ (cos135° + cos1200)
= cos(180° - 45°) - cos(180° - 60°)/cos(180° - 45°) + cos(180°-60°)
=  (-cos45° + cos60°)/(- cos45° - cos60°)
=  (-1/√2 + 1/2)/(- 1√2 - 1/2)
=  -2 + √2 / -2 - √2 
= (-2 + √2 / -2 - √2) * (-2 + √2)/(-2 + √2)
=  (-2 + √2)2/(-2)2 - (√2)2
= (4 + 2  - 4√2) / 2
= (6 - 4√2)/2
= 3 - 2√2

Correct Answer : a

Explanation :  {1 + cotα - sec(π/2 + α)} {1 + cotα + sec(π/2 + α)}

As we know that (a-b)(a+b) = a² - b²

(1 + cotα)² - [sec(π/2 + α)]²

1 + 2cotα + cot²α - (-cosecα)²

2cotα + 1 + cot²α - cosec²α

As we know that 1 + cot²α = cosec²α

= 2cotα + cosec²α - cosec²α

= 2cotα