Question

prove the following :-

tanA/1 - cotA + cotA/1 - tanA = secA . cosecA+1
​

Answer 1

Step-by-step explanation:

$L.H.S=tanA/(1-cotA) +cotA/(1-tanA)\\ \\=tanA/(1-(1/tanA)) +cotA/(1-tanA)\\ \\=tan^2A/(tanA-1) +cotA/(1-tanA)\\ \\=-tan^2A/(1-tanA) +cotA/(1-tanA)\\ \\=(-tan^2A+cotA)/(1-tanA)\\ \\=(-tan^2A+1/tanA)/(1-tanA)\\ \\=(-tan^3A+1)/tanA(1-tanA)\\ \\=(1-tan^3A)/tanA(1-tanA)\\ \\=(1-tanA)(1+tanA+tan^2A)/tanA(1-tanA)\\ \\=(1+tanA+tan^2A)/tanA\\ \\=(1+tan^2A+tanA)/tanA\\ \\=(sec^2A+tanA)/tanA\\ \\=sec^2A/tanA +tanA/tanA\\ \\=1/cos^2AtanA+1\\ \\=cosA/cos^2AsinA+1\\ \\=1/sinA cosA\\ \\=secAcosecA+1 = R.H.S$

I hope this helps!

Answer 2

Answer:

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