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Question

(i) Prove that cotθ+cosecθ1cotθcosecθ+1=cosecθ+cotθ=1+cosθsinθ

(ii) Prove that 1secAtanA1cosA=1cosA1secA+tanA.

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Solution

(i)cotθ+cscθ1cotθcscθ+1

=cotθ+cscθ(csc2θcot2θ)cotθcscθ+1

=cotθ+cscθ(cscθcotθ)(cscθ+cotθ)cotθcscθ+1

=(cotθ+cscθ)(1cscθ+cotθ)cotθcscθ+1

=cotθ+cscθ .........Result1

=cosθsinθ+1sinθ

=1+cosθsinθ .........Result2

Hence from result1 and result2 we have

cotθ+cscθ1cotθcscθ+1=cotθ+cscθ=1+cosθsinθ

Hence proved.

(ii)L.H.S=1secAtanA1cosA

Multiplying by secA+tanA in the numerator and denominator of first term,we get

=secA+tanA(secAtanA)(secA+tanA)secA

=secA+tanA(sec2Atan2A)secA

=secA+tanAsecA since sec2Atan2A=1

=tanA

Adding and subtracting secA, we get

=secA+tanAsecA

=1cosA(secAtanA)

Multiplying by secA+tanA in the numerator and denominator of second term,we get

=1cosA(secAtanA)(secA+tanA)(secA+tanA)

=1cosAsec2Atan2A(secA+tanA)

=1cosA1(secA+tanA) since sec2Atan2A=1

=R.H.S

1secAtanA1cosA=1cosA1(secA+tanA)

Hence proved.

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