Question

$\csc(270 {}^{0} -\beta ) = - 2.find \: the \: value \: of \: \tan( \beta). \\ a. \frac{1}{ \sqrt{3} } \\ b. - \sqrt{3} \\ c. \frac{1}{2} \\ d. \sqrt{3} .$
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Answer 1

Option d

Step-by-step explanation:

Given:-

Cosec (270°- β) = -2

To Find:-

Find the value of Tan β?

Solution:-

Given that

Cosec(270°-β) = -2

We know that

Cosec (270°-A) = - Sec A

=>Cosec (270°-β) = -Sec β

=>- Sec β = -2

=>Sec β = 2

On squaring both sides then

=>Sec^2 β = 2^2

=>Sec^2 β = 4

We know that

Sec^2 A - Tan^2 A = 1

=>Tan^2 A = Sec^2 A -1

=>Sec^2 β- 1 = 4-1

=>Sec^2 β -1 = 3

=>Tan^2 β = 3

=>Tan β = √3

Answer:-

The value of Tan β for the given problem is √3

Used formulae:-

• Cosec (270°-A) = - Sec A

• Sec^2 A - Tan^2 A = 1