Question

If θ be acute angle and cosθ = 15/17, then the value of cot (90° - θ) is

Answer 1

Answer:

$\frac{8}{15}$

Step-by-step explanation:

Given :

0° < θ < 90°

cosθ = 15/17

Procedure :

sin²θ + cos²θ = 1

⇒ sin²θ = 1 - cos²θ

⇒ sin²θ = 1 - $(\frac{225}{289})$

⇒ sin²θ = $\frac{64}{289}$

∴ sinθ = ± $\frac{8}{17}$

The value of sinθ cannot be negative as given that θ must be an acute angle.

[As sin(0°) = 0

sin(90°) = 1

sin function is increasing from 0 to 1, when θ is increasing from 0° to 90°.]

∴ sinθ = $\frac{8}{17}$

Let 90° - θ be x

sin(90° - θ) = cosθ = 15/17

∴ sin(x) = $\frac{15}{17}$

⇒ csc(x) = $\frac{17}{15}$

As csc²θ - cot²θ = 1,

⇒ cot²θ = csc²θ - 1

⇒ cot²(x) = $\frac{289}{225}$ - 1

⇒ cot²(x) = $\frac{64}{225}$

⇒ cot(x) = ±$\frac{8}{15}$

[Again cot(x) cannot be negative as θ is acute}

∴ cot(x) = $\frac{8}{15}$

∴ cot(90° - θ) = $\frac{8}{15}$

Thanks !

Answer 2

Given:-

• θ is acute angle.

• cosθ = 15/17

To Find:-

• the value of cot (90° - θ)

Solution:-

$cosθ \: = \frac{15 \: \: \: \: \: = \: \: base }{17 \: \: \: \: \: = \: hypo} \: </p><p>$

$perpendicular = 8$

$≈> Cot(90° - ∅)$

$⇒tanθ \: = \: \frac{8}{15} [∴tanθ= \frac{p}{b} ]$