If cot{\theta}=\frac{1}{\sqrt3}, the value of {sec{\ }}^2\theta+{cosec{\ }}^2\theta is

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Solution

Q.
$I f c o t θ = \frac{1}{\sqrt{3}}, t h e v a l u e o f {s e c}^{2} θ + {c o s e c}^{2} θ i s$

1 1

2 40/9

3 38/9

4 $5 \frac{1}{3}$

Ans.
(4)

Given,

$c o t θ = \frac{1}{\sqrt{3}} = c o t 60 ° \Rightarrow θ = 60 ° N o w, {s e c}^{2} θ + {c o s e c}^{2} θ = {s e c}^{2} 60 ° + {c o s e c}^{2} 60 ° = (2)^{2} + {(\frac{2}{\sqrt{3}})}^{2} = 4 + \frac{4}{3} = \frac{12 + 4}{3} = \frac{16}{3} = 5 \frac{1}{3}$

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