Question

tanθ-cotθ/sinθ cosθ=sec²θ-cosec²θ=tan²θ-cot²θ,Prove it by using trigonometric identities.

Answer 1

Given :

tanθ - cotθ /sinθ cosθ = sec²θ -cosec²θ = tan²θ - cot²θ

LHS = tanθ - cot θ /sinθ cosθ

= [sinθ / cosθ - cosθ /sinθ] / sinθ cosθ

[tanθ = sinθ/cosθ]

= {sin² θ - cos²θ/sinθ .cosθ} / sinθ cosθ

= {sin² θ - cos²θ/sinθ .cosθ} × 1/ sinθ. cosθ

= sin² θ - cos²θ/sin²θ .cos²θ

= sin² θ /sin²θ .cos²θ - cos²θ/sin²θ .cos²θ

= 1/ cos²θ - 1/sin²θ

LHS = sec²θ - cosec²θ

[ 1/ cosθ = secθ , 1/sinθ = cosecθ]

= (1 + tan²θ) - (1+ cot²θ)

[ cosec²θ = 1+ cot²θ , sec²θ =1 + tan²θ]

= 1 + tan²θ - 1 - cot²θ

LHS = tan²θ - cot²θ = RHS

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

Hi ,

Here I am using 'A ' instead of theta.

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we know that ,

1 ) tanA = sinA/cosA

2 ) cotA = cosA/sinA

3 ) 1 + tan²A = sec²A

4 ) 1 + cot²A = cosec²A

5 ) 1/sinA = cosecA

6 ) 1/cosA = secA

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i ) ( tanA - cotA )/sinAcosA

multiply numerator and denominator

with sinAcosA , we get

= (sinAcosAtanA-sinAcosAcotA)/sin²Acos²A

= [ sin² A - cos²A ]/sin²Acos²A

= (sin²A/sin²Acos²A) - (cos²A/sin²Acos²A)

= 1/cos²A - 1/sin²A

= sec²A - cosec²A ---( 1 )

= ( 1 + tan²A ) - ( 1 + cot²A )

= 1 + tan²A - 1 - cot²A

= tan²A - cot²A ---( 2 )

I hope this helps you.

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