Question

Prove :-

( Class 10)

Answer 1

Hey there,

Cmon, lets break the LHS into 2 terms

A=(sinA/1+cosA+1+cosA/sinA) and, 
B=(sinA/1-cosA-1-cosA/sinA0

Now let's solve A
sin²A+(1+cosA)²/sinaA(1+cosA)
=sin²A+cos²A+1+2cosA/sinA(1+cosA)
=2(1+cosA)/sinA(1+cosA)

Cancelling 1+cosA we get 2/sinA=2cosecA

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Now coming on to B,

sinA/1-cosA-1-cosA/sinA
Now, dividing by cosA, we get
tanA/secA-1-secA-1/tanA
=tan²A-(sec-1)²/tanA(secA-1)
=2(secA-1)/tanA(secA-1)
Cancelling secA-1, we get-
2/tanA=2cotA

Now A*B=2cosecA*2cotA=4cosecAcotA=R.H.S.

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