Answers

2015-04-25T01:15:45+05:30
(cosec-sin)(sec-cos)(tan+cos)=1

=>  (cosec-sin)(sec-cos)=1/(tan-cot)
                                             [:. cosec=1/sin]
                                             [:. sec  =1/cos]
=>[1/sin -sin][1/cos -cos] =1/[tan+cot] 
when we take lcm we get
=[1--sin2/sin][1-cos2/cos] =[1/tan +cot]
when we divide this with {sin A)*(cos A)} we get

1/[{sin²A/(sin A*cos A)} + {cos²A/(sin A* cos A)}]

1/[(sin A/cos A) + cosA /sin A)] = 1/[tan(A) + cot(A)]
hence we get right side

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2015-04-25T09:48:04+05:30
LHS:
=(cosecФsecФ-cosecФcosФ-sinФsecФ+sinФcosФ)(tanФ+cotФ)
=(1/sinФcosФ - cosФ/sinФ - sinФ/cosФ + sinФcosФ)(tanФ+cotФ)
=(1-cos²Ф-sin²Ф+(sinФcosФ)²)/sinФcosФ (tanФ+cotФ)
=(1-(cos²Ф+sin²Ф)+(sinФcosФ)²)/sinФcosФ (tanФ+cotФ)..as sin²Ф+cos²Ф=1
=(1-1+(sinФcosФ)²/sinФcosФ (tanФ+cotФ)
= (sinФcosФ)²/sinФcosФ (tanФ+cotФ)
= 1/1/sinФcosФ (tanФ+cotФ)...................as 1/1/sinФcosФ = sinФcosФ
= 1/sin²Ф+cos²Ф/sinФcosФ (tanФ+cotФ)
= 1/(sin²Ф/sinФcosФ + cos²Ф /sinФcosФ) (tanФ+cotФ)
=1/(sinФ/cosФ + cosФ/sinФ)  (tanФ+cotФ)
= 1/ (tanФ+cotФ)  (tanФ+cotФ)
= 1
Hence proved.
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