Question

(cosecA-sinA ) (secA - cosA ) =1÷ tanA + cot A

Answer 1

aheya friends .☺☺..

here is ur answer. .
===============
from lhs

(cosecA-sicA)(secA-cosA)

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

=>(1-sin^2 A/sinA)(1-cos^2A/cosA)

=>(cos^2A/sinA)(sin^2A/cosA)

=>(cos^2A*sin^2A/sinA*cosA)

=>cosA*sinA

now from Rhs

1/tanA+cotA

=>1
----------------------------
sinA/cosA+cosA/sinA

=>1/sin^2A+cos^2A/cosA*sinA

=>cosA*sinA/sin^2A+cos^2A【sin^2A+cos^2B=1】

=>cosA*sinA ...
>Rhs =lhs .

prooved ..
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hope it help. .

☺☺@rajukumar

Answer 2

Given,

L.H.S.
~~~~~
= ( cosec A - sin A ) ( Sec A - cos A )

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

= ( 1 - sin^2 A / Sin A ) ( 1 - cos^2 A / CosA )

= ( cos^2 A / Sin A ) ( sin^2 A / CosA )

= cos A × sin A....(1)

R.H.S
~~~~~

= 1 / tanA + cotA

= 1/tanA + cotA

= 1/(sinA / cosA) + (cos A / sinA)

= 1/(sin^2A + cos^2A)/(cosA×sinA)                

= 1/1/cos A×sinA                                                                                              
= cosA×sin A......(2)

From (1) and (2) we have,

L.H.S = R.H.S

( Hence, proved )

Hope it helps !!