Question

set 0 < θ < $\frac{\pi }{2}$ and sin θ = $\frac{1}{3}$
find the value of cos θ - tan θ

Answer 1

Solution

Concept ①.

Given,

$0<\theta<\dfrac{\pi }{2}$,

the angle exists in the first quadrant.

Then,

$\sin \theta >0, \cos \theta>0, \tan\theta>0$.

Step ①: Finding the values of $\cos \theta,\tan \theta$.

By trigonometric identity,

$\sin^{2} \theta +\cos^{2}\theta=1$

$\implies \left(\dfrac{1}{3} \right)^{2}+\cos^{2}\theta = 1$

$\implies \cos^{2}=\dfrac{8}{9}$

$\implies \cos\theta=\dfrac{2\sqrt{2} }{3}$

And,

$\tan\theta=\dfrac{\sin \theta }{\cos \theta }$

$\implies \tan\theta=\dfrac{1}{2\sqrt{2} }$

$\implies \tan \theta =\dfrac{\sqrt{2} }{4}$

Step ②: Evaluation.

Hence,

$\cos \theta -\tan \theta$

$=\left(\dfrac{2\sqrt{2} }{3} \right)-\left(\dfrac{\sqrt{2} }{4} \right)$

$=\dfrac{8\sqrt{2} }{12} -\dfrac{3\sqrt{2} }{12}$

$=\boxed{\dfrac{5\sqrt{2} }{12} }$

This is the required answer.

Answer 2

Given :-

0 < θ < π/2

sin θ = 1/3

To Find :-

Value of cosθ - tanθ

Solution :-

As we know that

sin²θ + cos²θ = 1

Given,

sinθ = 1/3

(1/3)² + cos²θ = 1

(1 × 1/3 × 3) + cos²θ = 1

1/9 + cos²θ = 1

cos²θ = 1 - 1/9

cos²θ = 9 - 1/9

cos²θ = 8/9

cosθ = √(8/9)

cosθ = √8/3

cosθ = 2√2/3

Now

tanθ = 1/2√2

tanθ = 1 × √2/4

tanθ = √2/4

Now

cosθ - tanθ

2√2/3 - √2/4

8√2 - 3 × √2/12

8√2 - 3√2/12

5√2/12