Question

(1+cotA-cosecA)(1+tanA+secA)=2
Pls prove it...

Answer 1

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Answer:

We have to prove that:

(1+cotA-cosecA)(1+tanA+secA)=2

We know that

cotA= cosA/sinA

cosecA= 1/sinA

tanA= sinA/cosA

secA= 1/cosA

Now,

=(1+cosA/sinA-1/sinA)(1+sinA/cosA+1/cosA)

=(sinA+cosA-1/sinA)(cosA+sinA+1/cosA)

=1/sinAcosA(sinA+cosA-1)(cosA+sinA+1)

=1/sinAcosA[(sinA+cosA)²-1²]

We have an identity:

(a+b)²= a²+b²+2ab

So,

=1/sinAcosA[(sin²A+cos²A+2sinAcosA)-1]

Another identity:

sin²A+cos²A=1

=1/sinAcosA(1+2sinAcosA-1)

=1/sinAcosA(1-1+2sinAcosA)

=1/sinAcosA(2sinAcosA)

=1×2

=2

RHS=LHS

Hence proved.

Answer 2

Answer:

L.H.S=(1+ COTA - COSECA)(1+ TANA+SECA)

=(1 + COSA÷SINA - 1÷SINA)÷(1+SINA÷COSA +1÷COSA)

=(COSA + SINA - 1 )÷ COSA ×( SINA + COSA +1)÷SINA

=[(COS²A + SIN²A)- (1)²]÷COSA×SINA

=(COS²A + 2COSA×SINA + SIN²A - SIN²A - COS²A)÷COSA×SINA

=2SINA×COSA ÷ SINA×COSA

=2

R.H.S = L.H.S