Question

if sin A =⅓ evaluate tanA sec A​

Answer 1

Step-by-step explanation:

In ∆ABC, sin A = 1/3

$sin \: A \: = \frac{opposite \: }{hypotenuse} = \frac{BC}{AC} = \frac{1}{3}$

Therefore, BC = 1k and AC = 3k... Let k be a positive integer....

By Pythagoras theorem, AB

$AB = \sqrt{ {(AC)}^{2} - {(BC)}^{2} }$

$= > \sqrt{ {(3k)}^{2} - {k}^{2} } \\ \\ = > \sqrt{9 {k}^{2} - {k}^{2} } = \sqrt{8} k = 2 \sqrt{2} k$

Therefore, AB = 2√2k

$tan \: A = \frac{opp}{hyp} = \frac{BC}{AB} = \frac{k}{2 \sqrt{2}k } = \frac{1}{2 \sqrt{2} }$

$sec \: A = \frac{hyp}{adj} = \frac{AC}{AB} = \frac{3k}{2 \sqrt{2} k} = \frac{3}{2 \sqrt{2} }$

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