Question

If 0° < theta < 90° and sin theta + cos theta = a, then the value of tan theta + cot theta is

a) a² + 1

b) 1/(a²+2)

c) 2/(a²-1)

d) 4/(a²-1)

Please show the full calculation.​​

Answer 1

Given that:

• 0° < θ < 90° and sin θ + cos θ = a

To Find:

• The value of tan θ + cot θ.

According to the question.

↠ sin θ + cos θ = a ______(i)

Squaring both sides.

↠ (sin θ + cos θ)² = a²

Using (a + b)² = a² + b² + 2ab

↠ sin²θ + cos²θ + 2sin θ•cos θ = a²

Value of sin²θ + cos²θ = 1

↠ 1 + 2sin θ•cos θ = a²

↠ 2sin θ•cos θ = a² - 1

↠ sin θ•cos θ = (a² - 1)/2 ______(ii)

Again,

↠ sin θ + cos θ = a

Dividing by cos θ both sides.

↠ sin θ/cos θ + cos θ/cos θ = a/cos θ

↠ tan θ + 1 = a/cos θ ______(iii)

Again,

↠ sin θ + cos θ = a

Dividing by sin θ both sides.

↠ sin θ/sin θ + cos θ/sin θ = a/sin θ

↠ 1 + cot θ = a/sin θ ______(iv)

Adding eqⁿ (iii) and eqⁿ (iv).

↠ tan θ + 1 + 1 + cot θ = a/cos θ + a/sin θ

↠ tan θ + cot θ = {a/cos θ + a/sin θ} - 2

↠ tan θ + cot θ = {(asin θ + acos θ)/sin θ•cos θ} - 2

↠ tan θ + cot θ = {a(sin θ + cos θ)/sin θ•cos θ} - 2

Putting all the values.

↠ tan θ + cot θ = {a(a)/(a² - 1)/2} - 2

↠ tan θ + cot θ = {2a²/(a² - 1)} - 2

↠ tan θ + cot θ = (2a² - 2a² + 2)/(a² - 1)

↠ tan θ + cot θ = 2/(a² - 1)

Hence,

• The value of tan θ + cot θ is c) 2/(a²-1).

Answer 2

Answer:

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Step-by-step explanation:

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