Find the values of each of the following trigonometric ratios.(i) sin

1.	Find the values of each of the following trigonometric ratios.(i) sin 300° (ii) cos (-210°) (iii) sec 390° (iv) tan (-855°) (v) cosec 1125°
Answer» (i) sin 300° = sin(360° – 60°) [For 360° – 60°. No change in T-ratio. 300° lies in 4th quadrant ‘sin’ is negative] = -sin 60° = - \(\frac{\sqrt{3}}{2}\) (ii) cos(-210°) = cos 210° (∵ cos(-θ) = cos θ) [∵ 180 + 30°. No change in T-ratio. 210° lies 3rd quadrant ‘cos’ is negative] = cos(180° + 30°) = -cos 30° = - \(\frac{\sqrt{3}}{2}\) (iii) sec 390° = sec(360° + 30°) = sec 30° = \(\frac{1}{cos 30°}\) = \(\frac{1}{\frac{\sqrt{3}}{2}}\) = \(\frac{2}{\sqrt{3}}\) (iv) tan(-855°) = -tan 855° (∵ tan(-θ) = – tan θ) [∵ Multiplies of 360° are dropped out. For 180° – 45°. No change in T-ratio. 180° – 45° lies in 2nd quadrant ‘tan’ is negative] = -tan(2 x 360° + 135°) = -tan 135° = -tan(180° – 45°) = -(-tan 45°) = -(-1) = 1 (v) cosec 1125° = cosec(3 x 360°+ 45°) = cosec 45° = \(\frac{1}{sin45°}\) = \(\frac{1}{\frac{1}{\sqrt{2}}}\) = √2

Find the values of each of the following trigonometric ratios.(i) sin 300° (ii) cos (-210°) (iii) sec 390° (iv) tan (-855°) (v) cosec 1125°

Answer»

(i) sin 300° = sin(360° – 60°)

[For 360° – 60°. No change in T-ratio. 300° lies in 4th quadrant ‘sin’ is negative]

= -sin 60°

= - \(\frac{\sqrt{3}}{2}\)

(ii) cos(-210°) = cos 210° (∵ cos(-θ) = cos θ)

[∵ 180 + 30°. No change in T-ratio. 210° lies 3rd quadrant ‘cos’ is negative]

= cos(180° + 30°)

= -cos 30°

= - \(\frac{\sqrt{3}}{2}\)

(iii) sec 390° = sec(360° + 30°)

= sec 30°

= \(\frac{1}{cos 30°}\)

= \(\frac{1}{\frac{\sqrt{3}}{2}}\)

= \(\frac{2}{\sqrt{3}}\)

(iv) tan(-855°) = -tan 855° (∵ tan(-θ) = – tan θ)

[∵ Multiplies of 360° are dropped out. For 180° – 45°. No change in T-ratio. 180° – 45° lies in 2nd quadrant ‘tan’ is negative]

= -tan(2 x 360° + 135°)

= -tan 135°

= -tan(180° – 45°)

= -(-tan 45°)

= -(-1)

= 1

(v) cosec 1125° = cosec(3 x 360°+ 45°)

= cosec 45°

= \(\frac{1}{sin45°}\)

= \(\frac{1}{\frac{1}{\sqrt{2}}}\)

= √2