Find out the values of angles 120,-135,150,180,270 for all the six trigonometric ratios.

We can make use of conversions from one trigonometric ratio to another to find these values.

Answer: Table with all 6 trignometric ratios has been given below for all the angles.

So, let us see how we can solve.

Explanation:

Given below is the list containing all 6 trigonometric ratios for the above mentioned angles.

1) 120º

• sin 120° = sin (1 × 90° + 30°) = cos 30° = √3/2
• cos 120° = cos (1 × 90° + 30°) = – sin 30° = – 1/2
• tan 120° = tan (1 × 90° + 30°) = – cot 30° = – √3
• csc 120° = csc (1 × 90° + 30°) = sec 30° = 2/√3
• sec 120° = sec (1 × 90° + 30°) = – csc 30° = – 2
• cot 120° = cot (1 × 90° + 30°) = – tan 30° = – 1/√3

2) -135º

• sin (- 135°)= – sin 135°= – sin (1 × 90°+ 45°) = – cos 45° = – 1√2
• cos (- 135°)= cos 135°= cos (1 × 90°+ 45°) = – sin 45°= – 1√2
• tan (- 135°) = – tan 135° = – tan ( 1 × 90° + 45°) = – (- cot 45°) = 1
• csc (- 135°)= – csc 135°= – csc (1 × 90°+ 45°)= – sec 45° = – √2
• sec (- 135°)= sec 135°= sec (1 × 90°+ 45°)= – csc 45°= – √2
• cot (- 135°) = – cot 135° = – cot ( 1 × 90° + 45°) = – (-tan 45°) = 1 

3) 150º

• sin 150° = sin (2 × 90° – 30°) = sin 30° = 1/2
• cos 150° = cos (2 × 90° – 30°) = - cos 30° = – √3/2
• tan 150° tan (2 × 90° – 30°) = - tan 30° = – 1√3
• csc 150° = csc (2 × 90° – 30°) = csc 30° = 2
• sec 150° = sec (2 × 90° – 30°) = sec 30° = – 2√3
• cot 150° = cot (2 × 90° – 30°) = – cot 300 = – √3

4) 180º

• sin 180° = sin (2 × 90° – 0°) = sin 0° = 0
• cos 180° = cos (2 × 90° – 0°) = – cos 0° = – 1
• tan 180° = tan (2 × 90° + 0°) = tan 0° = 0
• csc 180° = csc (2 × 90° – 0°) = csc 0° = Undefined
• sec 180° = sec (2 × 90° – 0°) = – sec 0° = – 1
• cot 180° = cot (2 × 90° + 0°) = cot 0° = Undefined

5) 270º

Thus, we have calculated the given angle values for all the trigonometric ratios.