Given that sec{\theta}=\sqrt2, the value of \frac{1+tan{\theta}}{\sin{\theta}} is

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Solution

Q.
Given that $s e c θ = \sqrt{2}, t h e v a l u e o f \frac{1 + t a n θ}{\sin θ} i s$

1 $2 \sqrt{2}$

2 $\sqrt{2}$

3 $3 \sqrt{2}$

4 2

Ans.
(1)

Given,

$s e c θ = \sqrt{2} \Rightarrow θ = 45 ° N o w, \frac{1 + t a n θ}{\sin θ} = \frac{1 + t a n 45 °}{\sin 45 °} = \frac{1 + 1}{\frac{1}{\sqrt{2}}} = \frac{2}{\frac{1}{\sqrt{2}}} = 2 \sqrt{2}$

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