Question

1. If sin (x-2y)=cos(4y-x) then the value of cot 2y is?
2. If cot A = 8/15, then what's the value of √1-cos A/1+cos A ?

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Answer 1

Step-by-step explanation:

1) sin (x-2y) = cos (4y-x)

or, sin (x-2y) = sin (90 - 4y + x)

or, x-2y = 90-4y+x

or, 4y-2y+x-x = 90

or, 2y = 90

or, y = 45

So, cot 2y = cot 2×45 = cot 90 = ∞

THE SECOND PART IS ALSO EASY BUT IT IS DIFFICULT TO TYPE IT AND SHOW THE SUM. SO, I DID ONLY THE FIRST PART.

HOPE IT HELPS

Answer 2

Answer:

1) cot (2y) = 0

2) $\tt{ \implies \dfrac{\sqrt{1-cos\,A}}{\sqrt{1+cos\,A}} = \dfrac{3}{5} }$

Step-by-step explanation:

1) Given: sin (x-2y) = cos (4y-x)

To find: cot (2y)

Solution:

sin (x-2y) = cos(4y-x)

⇒ sin (x-2y) = sin (90° - (4y - x))

⇒ sin (x-2y) = sin (90° - 4y + x)

⇒ x - 2y = 90° - 4y + x

⇒ x - 2y + 4y - x = 90°

⇒ 2y = 90°

⇒ y = 45°

Now, cot (2y) = cot (2×45°)

⇒ cot (2y) = cot 90°

⇒ cot (2y) = 0

2) Given: cot A = 8/15

To find: $\tt{ \dfrac{\sqrt{1-cos\,A}}{\sqrt{1+cos\,A}} }$

Solution:

cot A = 8/15

⇒ Base = 8 and Perpendicular = 15

⇒ Hypotenuse = √(8² + 15²) = √(64 + 225) = √289 = 17

⇒ cos A = 8/17

$\text{Now, } \tt{ \dfrac{\sqrt{1-cos\,A}}{\sqrt{1+cos\,A}} = \dfrac{\sqrt{1-\dfrac{8}{17}}}{\sqrt{1+\dfrac{8}{17}}} }$

$\tt{ \implies \dfrac{\sqrt{1-cos\,A}}{\sqrt{1+cos\,A}} = \dfrac{\sqrt{\dfrac{17-8}{17}}}{\sqrt{\dfrac{17+8}{17}}} }$

$\tt{ \implies \dfrac{\sqrt{1-cos\,A}}{\sqrt{1+cos\,A}} = \dfrac{\sqrt{9}}{\sqrt{25}} }$

$\tt{ \implies \dfrac{\sqrt{1-cos\,A}}{\sqrt{1+cos\,A}} = \dfrac{3}{5} }$

Formulas used :-

1) sin A = cos (90° - A)

2) If sin A = sin B ⇒ A = B

3) cot A = Base / Perpendicular

4) cos A = Base / Hypotenuse

5) (Hypotenuse)² = (Perpendicular)² + (Base)²

or $\tt{ Hypotenuse = \sqrt{(Perpendicular)^{2} + (Base)^{2}} }$