Question

if sinA=1/3, evaluate cosA+cosec A secA ​

Answer 1

Given

$sinA=\dfrac{1}{3}$

To find

cosA+cosec A secA

Explanation

$\sf consider \: \bigtriangleup ABC,$

$sinA=\dfrac{1}{3}$

we know that

$\boxed{ \bf sina = \dfrac{opposite}{hypotense} }$

So

$\sf in \: \bigtriangleup ABC,$

hypotenuse=3 units

opposite side =1 unit

Let us find adjacent side now

we know that

Hypotenuse ²=opposite²+adjacent²

$= > {3}^{2} = {1}^{2} + adjacent$

$= > 9 - 1 = adjacent$

$= > adjacent = 2 \sqrt{2} units$

Now

$\boxed{\bf CosA=\dfrac{adjacent}{hypotenuse}}$

$CosA=\dfrac{2\sqrt{2}}{3}$

$\boxed{\bf CosecA=\dfrac{1}{SinA}}$

$CosecA=\dfrac{3}{1}$

$\boxed{\bf SecA=\dfrac{1}{CosA}}$

$SecA=\dfrac{3}{2\sqrt{2}}$

Now

we need cosA+cosec A× secA

$= > \dfrac{2 \sqrt{2} }{3} + 3 \times \dfrac{3}{2 \sqrt{2} } \\ \\ = > \dfrac{2 \sqrt{2} }{3} + \dfrac{9}{2 \sqrt{2} } \\ \\ = > \dfrac{ {(2 \sqrt{2)} }^{2} + 9 \times 3 }{6 \sqrt{2} } \\ \\ = > \dfrac{8 + 27}{6 \sqrt{2} } \\ \\ = > \dfrac{35}{6 \sqrt{2} } units$