(ix) (cosec A - sin A)(sec A - cos A) =1/
tan A + cot A
answer me fast!
Answers
Step-by-step explanation:
( cosecA - sinA) (secA - cosA ) = 1 / ( tanA+ cotA)
Taking L.H.S.
(cosecA - sinA ) (secA - cosA )
({1/ sinA} - sinA) ( {1/cosA} - cosA )
[ cosecA = 1/sinA and secA =1/cosA ]
( {1- sin^2 A}/cosA)({1 - cos^2} / sinA )
(cos^2 A / sinA ) ( sin^2 A / cosA )
[ sin^2 A + cos^2 A = 1]
Here at above the equation cos square A will cancel from cos A
sin square A will cancel with sin A
Remaining after solving
sinAcosA ..............................................(1)
Taking R.H.S.
1 / ( tanA + cotA )
1 / ( {sinA / cosA} + { cosA / sinA}
[tanA = sinA/cosA and cotA = cosA/sinA]
sinAcosA/(sin^2 A + cos^2 A)
After taking L.C.M.
sinAcosA .......................................................(2)
From eq.(1) and eq(2)
L.H.S. = R.H.S. Proved