Question

show that (1/(tanA+cotA))=cosA.sinA​

Answer 1

from L.H.S

$\frac{1}{( \tanθ + cot θ)}$

$= > \frac{1}{ \frac {sinθ} { \cosθ } + \frac{ \cosθ }{ \sinθ } }$

$(tanθ = \frac{sinθ}{cosθ} \: and \: cotθ = \frac{cosθ}{sinθ} )$

$= > \frac{1}{ \frac{ {sin}^{2}θ + {cos}^{2} θ }{cosθ.sinθ} }$

➠ sin²θ + cos²θ = 1

$= > \frac{1}{ \frac{1}{sinθ.cosθ} }$

$= > sinθ.cosθ$

$= RHS$

Hence proved!

Answer 2

*Question:-

• show that (1/(tanA+cotA))=cosA.sinA

*Answer:-

$= > \frac{1}{ \tan(A) + \cot(A) } = \sin(A) \cos(A) \\ \\ = > LHS = \frac{1}{ \tan(A) + \cot(A) } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = > \frac{1}{ \frac{ \sin(A) }{ \cos(A) } + \frac{ \cos(A) }{ \sin(A) } } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = > \frac{1}{ \frac{ { \sin(A) }^{2} + { \cos(A) }^{2} }{ \sin(A) \cos(A) } } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = > \frac{1}{ \frac{1}{ \sin(A) + \cos(A) } } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Because:-

• (Sin²A + Cos²A = 1)

Hence:-

• SinA × CosA = RHS.

Hope it helps you.

Be Brainly.