Question

plzzzzz solve it.... ​

Answer 1

Answer:-

$\displaystyle{\boxed{\red{\sf\:\sin\:A\:=\:\dfrac{1}{\sqrt{10}}}}}$

$\displaystyle{\boxed{\pink{\sf\:\cos\:A\:=\:\dfrac{3}{\sqrt{10}}}}}$

$\displaystyle{\boxed{\green{\sf\:\tan\:A\:=\:\dfrac{1}{3}}}}$

$\displaystyle{\boxed{\blue{\sf\:\sec\:A\:=\:\dfrac{\sqrt{10}}{3}}}}$

$\displaystyle{\boxed{\red{\sf\:\cot\:A\:=\:3}}}$

Step-by-step-explanation:-

We have given that,

$\displaystyle\sf\:cosec\:A\:=\:\sqrt{10}$

We have to find all the remaining trigonometric ratios.

Now,

$\displaystyle\sf\:cosec\:A\:=\:\sqrt{10}$

$\displaystyle{\pink{\sf\:\sin\:A\:=\:\dfrac{1}{cosec\:A}}\sf\:\:\:-\:-\:-\:[\:Trigonometric\:identity\:]}$

$\displaystyle{\implies\boxed{\red{\sf\:\sin\:A\:=\:\dfrac{1}{\sqrt{10}}}}}$

─────────────────────

Now,

$\displaystyle{\pink{\sf\:1\:+\:\cot^2\:A\:=\:cosec^2\:A}\sf\:\:\:-\:-\:[\:Trigonometric\:identity\:]}$

$\displaystyle{\implies\sf\:1\:+\:\cot^2\:A\:=\:\left(\:\sqrt{10}\:\right)^2}$

$\displaystyle{\implies\sf\:1\:+\:\cot^2\:A\:=\:10}$

$\displaystyle{\implies\sf\:\cot^2\:A\:=\:10\:-\:1}$

$\displaystyle{\implies\sf\:\cot^2\:A\:=\:9}$

$\displaystyle{\implies\sf\:\cot\:A\:=\:\sqrt{9}}$

$\displaystyle{\implies\boxed{\pink{\sf\:\cot\:A\:=\:3}}}$

─────────────────────

Now,

$\displaystyle{\pink{\sf\:\tan\:A\:=\:\dfrac{1}{\cot\:A}}\sf\:\:\:-\:-\:-\:[\:Trigonometric\:identity\:]}$

$\displaystyle{\implies\boxed{\green{\sf\:\tan\:A\:=\:\dfrac{1}{3}}}}$

─────────────────────

Now,

$\displaystyle{\pink{\sf\:\tan\:A\:=\:\dfrac{\sin\:A}{\cos\:A}}\sf\:\:\:-\:-\:[\:Trigonometric\:identity\:]}$

$\displaystyle{\implies\sf\:\dfrac{1}{3}\:=\:\dfrac{\dfrac{1}{\sqrt{10}}}{\cos\:A}}$

$\displaystyle{\implies\sf\:\cos\:A\:=\:\dfrac{\dfrac{1}{\sqrt{10}}}{\dfrac{1}{3}}}$

$\displaystyle{\implies\sf\:\cos\:A\:=\:\dfrac{1}{\sqrt{10}}\:\times\:\dfrac{3}{1}}$

$\displaystyle{\implies\boxed{\blue{\sf\:\cos\:A\:=\:\dfrac{3}{\sqrt{10}}}}}$

─────────────────────

Now,

$\displaystyle{\pink{\sf\:\sec\:A\:=\:\dfrac{1}{\cos\:A}}\sf\:\:\:-\:-\:[\:Trigonometric\:identity\:]}$

$\displaystyle{\implies\sf\:\sec\:A\:=\:\dfrac{1}{\dfrac{3}{\sqrt{10}}}}$

$\displaystyle{\implies\sf\:\sec\:A\:=\:1\:\times\:\dfrac{\sqrt{10}}{3}}$

$\displaystyle{\implies\boxed{\red{\sf\:\sec\:A\:=\:\dfrac{\sqrt{10}}{3}}}}$

Answer 2

Answer:

We know that, cosec θ = AB/BC = hypotenuse/perpendicular = (k√10)/k

(where k is a positive)

By Pythagoras Theorem:

AC² = AB² + BC²

= 10k² + k²

= 9k²

AC = 3k

Find other T-rations using their definitions:

Sin θ = BC/AB = 1/√10

cos θ = AC/AB = (3k)/(k√10) = 3/√10

tan θ = BC/AC = sin θ /cos θ = 1/3

sec θ = AB/AC = 1/cos θ = √10/3

cot θ = AC/BC = 1/tan θ = 3

hope this helps you