Question

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Answer 1

Given Equation is 7sin²x + 3cos²x = 4.

⇒ (4 + 3)sin²x + 3cos²x = 4

⇒ 4sin²x + 3sin²x + 3cos²x = 4

⇒ 4sin²x + 3(sin²x + cos²x) = 4

⇒ 4sin²x + 3 = 4

⇒ 4sin²x = 1

⇒ sin²x = (1/4)

⇒ sinx = 1/2.

We know that cosecθ = (1/sinθ)

Hence, cosecx = (1/1/2)

                         = 2.

Now,

We know that cos²x = (1 - sin²x)

⇒ cos²x = 1 - (1/2)^2

              = 1 - 1/4

              = 3/4

Then, cosx = √3/2

We know that secx = (1/cosx)

                                = (2/√3).

Therefore, the value of secx + cosecx = 2 + (2/√3)    ------ Option (A).

Hope it helps!

Answer 2

Answer:

Step-by-step explanation:

Given Equation is 7sin²x + 3cos²x = 4.

⇒ (4 + 3)sin²x + 3cos²x = 4

⇒ 4sin²x + 3sin²x + 3cos²x = 4

⇒ 4sin²x + 3(sin²x + cos²x) = 4

⇒ 4sin²x + 3 = 4

⇒ 4sin²x = 1

⇒ sin²x = (1/4)

⇒ sinx = 1/2.

We know that cosecθ = (1/sinθ)

Hence, cosecx = (1/1/2)

                         = 2.

Now,

We know that cos²x = (1 - sin²x)

⇒ cos²x = 1 - (1/2)^2

              = 1 - 1/4

              = 3/4

Then, cosx = √3/2

We know that secx = (1/cosx)

                                = (2/√3).

Therefore, the value of secx + cosecx = 2 + (2/√3)    ------ Option (A).

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