Question

if A is not in 2 quadrant B is not in 3 Quadrant CosA=-1/2 cos B=-1/2 then find value of 4sinB-3tanA/tanB+sinA​

Answer 1

Answer:

5/3

Step-by-step explanation:

Given,

A is not in 2nd quadrant

B is not in 3rd quadrant

cosA = - 1/2

cosB = - 1/2

To Find :-

Value of :-

$\dfrac{4sinB-3tanA}{tanB+sinA}$

How To Do :-

As they said that 'A' is not in 2nd quadrant and we can see that cosA is negative , We know that cosA will be negative in 2nd quadrant and 3rd quadrant as they said 'A' does not belongs to 2nd quadrant , So, A' belongs to 3rd quadrant. and they also said that 'B' is not in 3rd quadrant and we can see that value of 'cosB' is negative , We know that 'cos' is negative in only 2nd and 3rd quadrant , As they said that it not belongs to 3rd quadrant , So cosB belongs to 2nd quadrant. So by using the pythagoreas theorem we need to find the values of 'sinB , tanA , tanB , sinA' and we need to substitute it.

Formula Required :-

Pythagoras theorem :-

(hypotenuse side)² = (opposite side)² + (adjacent side)²

sinα = opposite side/hypotenuse side

cosα = adjacent side/hypotenuse side

tanα = opposite side/adjacent side

In 2nd quadrant :-

'cos' negative

'sin' is positive

'tan' is negative

In 3rd quadrant :-

'cos' is ngeative

'sin' is negative

'tan' is positive

Solution :-

cosA = -1/2

adjacent side/hypotenuse side = -1/2

Adjacent side = 1 , hypotenuse = 2

Let, opposite side be 'x'

[ For applying Pythagoras theorem we shouldn't take the signs]

Applying Pythagoras theorem :-

(2)^2 = (1)^2 + (x)^2

4 = 1 + x^2

4 - 1 = x^2

3 = x^2

x = √3

∴ opposite side = x = √3

As 'A' is in 3rd quadrant , sinA will be negative and 'tanA' will be positive :-

sinA = opposite side/hypotenuse side

= -√3/2

tanA = opposite side/adjacent side

= √3/1

= √3

cosB = -1/2

adjacent side/hypotenuse side = -1/2

Adjacent side = 1 , hypotenuse = 2

Let, opposite side be 'x'

[ For applying pythagoreas theorem we shouldn't take the signs]

Applying pythogoreas theorem :-

(2)^2 = (1)^2 + (x)^2

4 = 1 + x^2

4 - 1 = x^2

3 = x^2

x = √3

∴ opposite side = x = √3

As 'B' is in 2nd quadrant , tanB will be negative and sinB will be positive :-

sinB = opposite side/hypotenuse side

= √3/2

tanB = opposite side/adjacent side

= -√3/1

= -√3

$\dfrac{4sinB-3tanA}{tanB+sinA}$

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

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

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

$=\dfrac{\sqrt{3}-6\sqrt{3}}{2}\times \dfrac{2}{-3\sqrt{3}}$

$=-5\sqrt{3}\times \dfrac{1}{-3\sqrt{3}}$

$=\dfrac{5}{3}$

$\therefore \dfrac{4sinB-3tanA}{tanB+sinA}=\dfrac{5}{3}$