Question

If a,b,y are acute angles such that cos a=tan b,cos b=tan y,cos y=tan a ..find the value of (2sin y+1)^2​

Answer 1

Answer:

Correct option is

A

7998

Given

cosx=tany⋯(1)

cosy=tanz⋯(2)

cosz=tanx⋯(3)

Let x=y=z

⟹cosx=tanx

⟹cosx=

cosx

sinx

⟹sinx=cos

2

x

⟹sinx=1−sin

2

x

⟹sin

2

x+sinx−1=0

⟹sinx=

2

−1±

1+4

⟹sinx=

2

−1±

5

⟹sinx=2sin18

∘

⟹1234sinx+1342sinx+1423sinx

⟹1234(2sin18)+1342(2sin18)+1423(2sin18)

⟹7998sin18

⟹k=7998

Answer 2

Answer:

The value of (2sin y + 1)² is 5.

Step-by-step explanation:

Given,

that a, b, and y are acute angles.

cos a=tan b,cos b=tan y,cos y=tan a

To find,

The value of (2sin y+1)^2​

Assumption,
Let the angles a, b, and y be equal

i.e. a = b = y

Calculation,

As cos a = tan b

⇒ cos a = tan a

⇒ cos a = sin a/cos a (As tanA = sinA/cosA )

⇒ cos² a = sin a

⇒ 1 - sin² a = sin a  (sin² A + cos² A =1)

⇒ sin² a + sin a - 1 = 0

$\implies \sin a = \frac{-1 \pm\sqrt{1 + 4} }{2}$

$\implies \sin a = \frac{-1 \pm\sqrt{5} }{2}$

⇒ Now the value of (2sin y + 1)² = (2sin a + 1)²

$\implies (2\times \frac{-1 \pm\sqrt{5} }{2} +1)^2$

⇒ (-1 ± √5 +1)²

⇒ 5

Therefore, the value of (2sin y + 1)² is 5.

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