Question

Cos θ = √3/2 then find the value of 1−secθ/1+cosecθ
Please answer this question fast.

Answer 1

• The value of cosθ = √3/2.
• Now using the identity, sin²θ + cos²θ = 1

sin²θ = 1 - (3/4)

sin²θ = 1/4

sinθ = (1/2)

Now, we have to find the value of

1-sec theta/1+cosec theta

Substituting the values in the given expression , we get

$\frac{1-sec\theta}{1+cosec\theta}= \frac{1-(1/cos\theta)}{1+(1/sin\theta)}$

$= \frac{1-2/(\sqrt{3})}{1+(2)}$

$= \frac{(\sqrt{3})-2}{3(\sqrt{3})}$

$= \frac{3-2\sqrt{3}}{9}$

Answer 2

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value of cosθ = √3/2.

Now using the identity, sin²θ + cos²θ = 1

sin²θ = 1 - (3/4)

sin²θ = 1/4

sinθ = (1/2)

Now, we have to find the value of

1-sec theta/1+cosec theta

Substituting the values in the given expression , we get

\frac{1-sec\theta}{1+cosec\theta}= \frac{1-(1/cos\theta)}{1+(1/sin\theta)}

1+cosecθ

1−secθ

=

1+(1/sinθ)

1−(1/cosθ)

= \frac{1-2/(\sqrt{3})}{1+(2)}=

1+(2)

1−2/(

3

)

= \frac{(\sqrt{3})-2}{3(\sqrt{3})}=

3(

3

)

(

3

)−2

= \frac{3-2\sqrt{3}}{9}=

9 is answer

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