Question

Prove that : 1/(tanA+cotA) = cosAsinA​

Answer 1

$\sf \Rightarrow \quad \dfrac{1}{tan \: A + cot \: A} = cos \: A \: sin \: A \\ \\ \sf Solving \: LHS \: separately, \\ \\ \sf \Rightarrow \quad \frac{1}{tan \: A + cot \: A} \\ \\ \sf \Rightarrow \quad \frac{1}{ \frac{sin \: A}{cos \: A} + \frac{cos \: A}{sin \: A} } \qquad \bigg[ \because \: tan \: A = \frac{sin \: A}{cos \: A} \: \: ,\quad cot \: A = \frac{1}{tan \: A} \bigg] \\ \\ \sf \Rightarrow \quad \frac{1}{ \frac{ {sin}^{2} \: A + {cos}^{2} \: A }{cos \: A \: sin \: A} } \\ \\ \sf \Rightarrow \quad \frac{1}{ \frac{1}{cos \: A \: sin \: A} } \qquad \bigg[ \because \: {sin}^{2} \: A + {cos}^{2} \: A = 1 \bigg] \\ \\ \sf \Rightarrow \quad cos \: A \: sin \: A \\ \\ \sf \Rightarrow \quad LHS = RHS \\ \\ \sf Hence, \: Proved \\ \\ \underline{ \sf Some \: Formulae} \\ \\ \star \quad \sf 1 + {tan}^{2} \: A = {sec}^{2} \: A \\ \\ \star \quad \sf 1 + {cot}^{2} \: A = {cosec}^{2} \: A \\ \\ \star \quad \sf sin \: A = cos(90 - A ) \\ \\ \star \quad \sf sin \: A = \dfrac{1}{cosec \: A}$

Answer 2

Step-by-step explanation:

${\red{\underline{\underlune{\bold{To\:Prove:-}}}}}$

• $\bf \dfrac{1}{ \tan A + \cot A } = \cos A .\sin A$

${\blue{\underline{\underlune{\bold{Proof:-}}}}}$

As we know that

• $\bf \tan A = \dfrac{ \sin A}{ \cos A}$

• $\bf \cot a = \dfrac{ \cos a}{ \sin a}$

Let us prove by simplifying LHS and RHS separately

LHS =

$\bf \leadsto \dfrac{1}{ \tan A + \cot A }$

$\bf \leadsto \dfrac{1}{ \dfrac{ \sin A }{ \cos A } + \dfrac{ \cos A }{ \sin A} }$

$\bf \leadsto \dfrac{1}{ \dfrac{ { \sin }^{2}A+ { \cos }^{2}A }{ \cos A .\sin A } }$

Since; sin² A + cos² A = 1

$\bf \leadsto \dfrac{1}{ \dfrac{ 1 }{ \cos A .\sin A} }$

$\bf \leadsto \cos A . \sin A$

= RHS

Therefore:-

$\bf \dfrac{1}{ \tan A + \cot A } = \cos A .\sin A$

Hence Proved