Question

If sin θ = - 3 / 5 and θ lies in 3rd quadrant then tan θ = ​

Answer 1

Answer- The above question is from the chapter 'Trigonometric Functions'.

Trigonometry- The branch of Mathematics which helps in dealing with measure of three sides of a right-angled triangle is called Trigonometry.

Trigonometric Ratios:

sin θ  = Perpendicular/Hypotenuse

cos θ = Base/Hypotenuse

tan θ = Perpendicular/Base

cosec θ = Hypotenuse/Perpendicular

sec θ = Hypotenuse/Base

cot θ = Base/Perpendicular

Also, tan θ = sin θ/cos θ and cot θ = cos θ/sinθ

Trigonometric Identities:

1. sin² θ + cos² θ = 1

2. sec² θ - tan² θ = 1

3. cosec² θ - cot² θ = 1

Given question: If sin θ = -3/5 and θ lies in 3rd quadrant then tan θ = ​___.

Solution: sin θ = $<strong> </strong>\frac{-3}{5}$

x lies in 3rd quadrant.

We know that sin² θ + cos² θ = 1.

(-3/5)² + cos² θ = 1

$cos^{2} \: \theta = 1 - \dfrac{9}{25}\\\\cos^{2} \: \theta = \dfrac{25 \: - \: 9}{25}\\\\cos^{2} \: \theta = \dfrac{16}{25}\\\\cos \: \theta = \pm \: \dfrac{4}{5}$

We know that cos θ is negative in thired quadrant.

So, cos θ = $\frac{-4}{5}$

tan θ = sin θ/cos θ

$tan \: \theta = \dfrac {\dfrac{-3}{5}} {\dfrac{-4}{5}}$

$\implies tan \: \theta = \dfrac{3}{4}$

∴ If sin θ = -3/5 and θ lies in 3rd quadrant then tan θ = 3/4.

Answer 2

Answer:

Answer- The above question is from the chapter 'Trigonometric Functions'.

Trigonometry- The branch of Mathematics which helps in dealing with measure of three sides of a right-angled triangle is called Trigonometry.

Trigonometric Ratios:

sin θ = Perpendicular/Hypotenuse

cos θ = Base/Hypotenuse

tan θ = Perpendicular/Base

cosec θ = Hypotenuse/Perpendicular

sec θ = Hypotenuse/Base

cot θ = Base/Perpendicular

Also, tan θ = sin θ/cos θ and cot θ = cos θ/sinθ

Trigonometric Identities:

1. sin² θ + cos² θ = 1

2. sec² θ - tan² θ = 1

3. cosec² θ - cot² θ = 1

Given question: If sin θ = -3/5 and θ lies in 3rd quadrant then tan θ = ___.

Solution: sin θ = < strong > < /strong > \frac{-3}{5}<strong></strong>

5

−3

x lies in 3rd quadrant.

We know that sin² θ + cos² θ = 1.

(-3/5)² + cos² θ = 1

\begin{gathered}cos^{2} \: \theta = 1 - \dfrac{9}{25}\\\\cos^{2} \: \theta = \dfrac{25 \: - \: 9}{25}\\\\cos^{2} \: \theta = \dfrac{16}{25}\\\\cos \: \theta = \pm \: \dfrac{4}{5}\end{gathered}

cos

2

θ=1−

25

9

cos

2

θ=

25

25−9

cos

2

θ=

25

16

cosθ=±

5

4

We know that cos θ is negative in thired quadrant.

So, cos θ = \frac{-4}{5}

5

−4

tan θ = sin θ/cos θ

tan \: \theta = \dfrac {\dfrac{-3}{5}} {\dfrac{-4}{5}}tanθ=

5

−4

5

−3

\implies tan \: \theta = \dfrac{3}{4}⟹tanθ=

4

3

∴ If sin θ = -3/5 and θ lies in 3rd quadrant then tan θ = 3/4.