Question

tanA/1-cotA+cotA/1-tanA=sinAcosA+1/sinAcosA​

Answer 1

Answer:

tanA/(1-cotA) +cotA/(1-tanA)

=tanA/(1–1/tanA) +cotA/(1-tanA)

=tan^2A/(tanA-1) +cotA/(1-tanA)

=-tan^2A/(1-tanA) +cotA/(1-tanA)

=(-tan^2A+cotA)/(1-tanA)

=(-tan^2A+1/tanA)/(1-tanA)

=(-tan^3A+1)/tanA(1-tanA)

=(1-tan^3A)/tanA(1-tanA)

=(1-tanA)(1+tanA+tan^2A)/tanA(1-tanA)

=(1+tanA+tan^2A)/tanA

=(1+tan^2A+tanA)/tanA

=(sec^2A+tanA)/tanA

=sec^2A/tanA +tanA/tanA

=1/cos^2AtanA+1

=cosA/cos^2AsinA+1

=1/cosAsinA+1

=secAcosecA+1

Hope it helps you..

Answer 2

Given :-

$\sf \dfrac{tanA}{1-cotA}+\dfrac{cotA}{1-tanA}=\dfrac{sinAcosA+1}{sinAcosA}$

Proof :-

• Convert this into SinA and CosA

$\bullet\sf tanA= \dfrac{sinA}{cosA}\ \ ;\ \ \bullet \sf cotA=\dfrac{cosA}{sinA}$

Taking LHS :-

$\dashrightarrow\sf \dfrac{tanA}{1-cotA}+\dfrac{cotA}{1-tanA}\\ \\ \\ \dashrightarrow\sf \dfrac{\bigg\{\dfrac{sinA}{cosA}\bigg\}}{\bigg\{1-\dfrac{cosA}{sinA}\bigg\}}\ + \dfrac{\bigg\{\dfrac{cosA}{sinA}\bigg\}}{\bigg\{1-\dfrac{sinA}{cosA}\bigg\}}\\ \\ \\ \dashrightarrow\sf \dfrac{\bigg\{\dfrac{sinA}{cosA}\bigg\}}{\bigg\{\dfrac{sinA-cosA}{sinA}\bigg\}} \ + \dfrac{\bigg\{\dfrac{cosA}{sinA}\bigg\}}{\bigg\{\dfrac{cosA-sinA}{cosA}\bigg\}} \\ \\ \\ \dashrightarrow\sf \bigg\{\dfrac{sinA}{cosA}\bigg\}\times \bigg\{\dfrac{sinA}{sinA-cosA}\bigg\}\ + \bigg\{\dfrac{cosA}{sinA}\bigg\}\times \bigg\{\dfrac{cosA}{cosA-sinA}\bigg\}\\ \\ \\ \dashrightarrow\sf \bigg\{\dfrac{sin^2A}{cosA\big(sinA - cosA\big)}\bigg\}\ + \bigg\{\dfrac{cos^2A}{sinA\big(cosA-sinA\big)}\bigg\}$

• Take (-) as common

$\dashrightarrow\sf \bigg\{\dfrac{sin^2A}{cosA\big(sinA - cosA\big)}\bigg\} \ -\bigg\{ \dfrac{cos^2A}{sinA\big(sinA-cos A\big)}\bigg\}\\ \\ \\ \dashrightarrow\sf \dfrac{sin^3A-cos^3A}{sinA\ cosA\big(sinA-cosA\big)}$

• Formula used here:-

$\underline{\star{\scriptsize{\sf\ \ \ a^3-b^3= (a-b)(a^2+b^2+ab)}}}\\ \\ \\ \dashrightarrow\sf \dfrac{\cancel{\big(sinA-cosA\big)}\big(sin^2A+cos^2A+sinA. cosA\big)}{sinA cosA\cancel{\big(sinA-cosA\big)}}\\ \\ \\ \underline{\star{\scriptsize {\sf\ \ sin^2A+cos^2A=1}}}\\ \\ \\ \dashrightarrow\sf \bigg\lgroup\dfrac{1+ sinA \ cosA}{sinA\ cosA}\bigg\rgroup\ \ \ \ \ Hence\ Proved !!$