Question

Cot a + cosec a - 1 by cot a minus cosec a + 1 is equal to 1 + cos a by sin

Answer 1

Answer:

LHS : cot A + cosec A - 1 / cot A - cosec A + 1

[ (cosA/sinA) +(1/sinA) – 1] / [(cosA/sinA) -(1/sinA) + 1]

{ [cosA + 1-sinA ]/sinA} / { [cosA – 1+sinA ]/sinA }

{ [cosA + 1-sinA ]/sinA} * sinA/[cosA – 1+sinA ]

[cosA + 1-sinA ]/ [cosA – 1+sinA ]

(cosA- sinA) +1/ (cosA+sinA) – 1

{ (cosA- sinA) +1/ (cosA+sinA) – 1}* {(cosA+sinA) +1 /(cosA+sinA) +1}

( Rationalising it by"(cosA+sinA) +1")

{(cosA- sinA)(cosA+sinA) +(cosA- sinA) +(cosA- sinA) +1} /[(cosA+sinA) – 1][(cosA+sinA) +1]

{(cos2A- sin2A)+2cosA+sinA- sinA +1}/ (cosA+sinA)2 -1

{ cos2A-(1- cos2A) + 2cosA+1} / (cos2A+sin2A+2sinA cosA) -1

{ cos2A-1+ cos2A + 2cosA+1} / 2sinA cosA +1 – 1

2cos2A + 2cosA / 2sinA cosA

2 cosA (cosA +1) / 2cosA (sinA)

1 + cos A / sin A LHS = RHS (Hence proved)

hope this will help you

Answer 2

We know that:-

1=cosec^2A-cot^2A

1 =(cosecA+cotA) (cosecA-cotA)……….(1)

Now L.H.S.

=(cotA+cosecA-1)/(cotA-cosecA+1)

On putting the value of 1 from eq.(1) in the Nr.

=[(cotA+cosecA)-(cosecA+cotA)(cosecA-cotA)]/(cotA-cosecA+1)

=(cosecA+cotA)(1-cosecA+cotA)/(cotA-cosecA+1)

=cosecA+ cotA

= 1/sinA + cosA/sinA

= (1+cosA)/sinA

Hence proved...