Question

Prove that: (1+cot+ tan A)(sin A- cos A)= sec A/ cosec²A-cosec A/ sec² A.

Answer 1

the solution is

( 1 + tan A + cot A )
( sin A - cos A ) = (sec A/cosec^2 A) - (cosec A/sec^2 A) .............( equation 1)

we know tan A = sin A/cos A , cot A = cos A/sin A
, sec= 1/cosA and cosec = 1/sec A

put all the values in equation 1

{ 1 + (sin A/cos A) + (cos A/sin A)} ( sinA - cosA ) = { (1/cosA)/ ( 1/sin^2 A)} - { (1/sinA)/ ( 1/cos^2 A)}

{ (cos A. sin A + sin^2 A + cos^2 A ) ( sin A - cos A)}/cos A sin A = (sin^2 A / cos A) - (cos^2 A / sin A)

{ (cos A sin A + sin^2 A + cos^2 A ) ( sin A - cos A)}/cos A sin A= (sin^2 A.sin A - cos^2 A.cosA)/ cosA. sinA

simplifying both side and put the value of sin^ 2 A+ cos^2 A= 1

( cos A . sin A + 1)( sin A - cos A) = (sin^2 A .sin A - cos^2 A.cos A)

cos A. sin^2 A + sinA - cos^2 A -cos^2 A . sin A -cos A - sin^2 A.sin A + cos^2 A .cos A = 0

(sinA -cosA )- sin A(sin^2 A + cos^2 A ) + cosA (sin^2 A+ cos^2 A) =0

now

sin A - cos A- sin(A) + cos (A) = 0

sin A - cos A -sin A + cos A=0

0 = 0

hence proved






OR

use this to understand

the solution is

( 1 + tan + cot )( sin - cos ) = (sec/cosec2) - ( cosec/sec2 ) .............( equation 1)

we know tan = sin/cos , cot = cos/sin , sec= 1/cos and cosec = 1/sec

put all the values in equation 1

{ 1 + (sin/cos) + (cos/sin)} ( sin - cos ) = { (1/cos)/ ( 1/sin2 )} - { (1/sin)/ ( 1/cos2 )}

{ (cos. sin + sin2 + cos2 ) ( sin - cos)}/cos. sin = (sin2 / cos) - (cos2 / sin)

{ (cos. sin + sin2 + cos2 ) ( sin - cos)}/cos. sin = (sin2 .sin - cos2 .cos)/ cos. sin

simplifying both side and put the value of sin2+ cos2 = 1

( cos . sin + 1)( sin - cos) = (sin2 .sin - cos2.cos)

cos. sin2 + sin - cos2 -cos2 . sin -cos - sin2.sin + cos2 .cos = 0

(sin -cos )- sin (sin2 + cos2 ) + cos (sin2 + cos2) =0

now

sin - cos - sin(1) + cos (1) = 0

sin - cos -sin + cos =0

0 = 0

hence proved


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