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Answer 1

Question:

If sin²Ø + 3 cos²Ø = 4, then show that tanØ = $\sf{\dfrac{i}{\sqrt3}}$

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Solution:

sin²Ø + 3 cos²Ø = 4...Given

=> sin²Ø + cos²Ø + 2 cos²Ø = 4

But, sin²Ø + cos²Ø = 1 ... trigonometry identity

=> 1 + 2 cos²Ø = 4

=> cos²Ø = 3/2...(1)

But, sin²Ø = 1 - cos²Ø

=> sin²Ø = 1 - 3/2

=> sin²Ø = -1/2...(2)

Divide equation (2) by equation (1), we get

=> sin²Ø/cos²Ø = (-1/2) / (3/2)

=> sin²Ø/cos²Ø = -1/3

On taking square root of both the sides

=> sinØ/cosØ = $\sf{\sqrt{-1}/\sqrt3}$

But, sinØ/cosØ = tanØ and $\sqrt{-1}$ = i

$\sf{\therefore}$ tanØ = i/$\sqrt{3}$

Hence, proved.

Answer 2

CorrecTion :-

• 7sin²θ + 3cos²θ = 4

Given :-

• 7sin²θ + cos² θ = 4

To Prove :-

• tanθ = 1/√3

SoluTion :-

» 7sin²θ + 3sin²θ = 4

→ 4sin²θ + 3sin²θ + 3cos²θ = 4

→ 4sin²θ + 3 (sin²θ + cos²θ) = 4

→ 4sin²θ + 3 × 1 = 4 [ sin²θ + cos²θ = 1 ]

→ 4sin²θ = 4 - 3

→ 4sin²θ = 1

→ sin²θ = 1/4

★ As we know that,

» cos²θ = (1 - sin²θ)

→ cos²θ = 1 - 1/4

→ cos²θ = 3/4

» tan²θ = sin²θ/cos²θ

→ tan²θ = 1/4 × 4/3

→ tan²θ = 1/3

→ tanθ = √1/3

→ tanθ = 1/√3

Hence, ProVed.

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