Question

If tan x = -3/4 and x is in the second quadrant, then what is the value of
sin x. cos x?​

Answer 1

Answer :-

-12/25

Given :-

• tan x = -3/4
• x is in 2nd Quadrant

To find :-

• sinx . cos x

SOLUTION :-

In the 2nd Quadrant ,

• sinA is positive
• cosA is negative

So,

tanx = -3/4

From this we shall find the cosA, sinA

Let take ,

tanx = 3/4

We know that,

tanA = opposite side/adjacent side

tanA = 3/4

So,

• Opposite side = 3
• Adjacent side = 4

From Pythagoras theorem we find the hypotenuse

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

(3)² + (4)² = (hyp)²

9 + 16 = (hyp)²

25 = (hyp)²

(5)² = (hyp)²

Hypotenuse = 5

So,

sinA = opposite side/hypotenuse

cosA = adjacent side/hypotenuse

sinx = 3/5

cosx = 4/5

But , x belongs to 2nd Quadrant and in 2nd Quadrant "sin" is positive and "cos" is negative .

So,

sinx = 3/5

cosx = -4/5

(sinx ) (cosx )

(3/5 ) (-4/5 )

(3×-4)/(5×5)

-12/25

So,

sinx . cosx = -12/25

Know more :-

Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigonometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonometric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj

Answer 2

Answer:

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