Question

prove that: (sin - cosec)(cos-sec)=1/tan+cot

Answer 1

Sol:

LHS = (Cosec A - Sin A) (Sec A - Cos A)

= (1/sin A - Sin A)(1/cos A - cos A)

= [(1 - sin2 A)/sin A] [(1 - cos2 A)/cos A]

= [(cos2 A)/sin A][sin2 A/cos A]

= sin A cos A

RHS = 1 / (Tan A + Cot A)

= 1 / (sin A/cos A + cos A/sin A)

= 1 / [(sin2 A + cos2 A)/sin A cos A]

= 1 / [1/sin A cos A]

= sin A cos A

LHS = RHS

Hence proved.v

Answer 2

Hey!!!!

Good Evening

We have to prove

(sinA - cosecA) (cosA - secA) = 1/(tanA + cotA)

LHS = (sinA - cosecA) (cosA - secA)

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

= [ (sin²A - 1)/sinA ] [ (cos²A - 1)/cosA ]

= (cos²A/sinA) (sin²A/cosA)

After cancelling

= sinAcosA

Multiply and divide by tanA + cotA

= (sinAcosA)(tanA + cotA)/(tanA + cotA)

= (sin²A + cos²A)/(tanA + cotA)

= 1/(tanA + cotA)

= RHS

HENCE PROVED

Hope this helps