Question

Prove that cot a + coseca-1/cota-coseca+1=1+cosa/1+sina

Answer 1

heya friend
==============

,➡Here ur question is wrong question should be cotA+cosecA-1/cotA-co+1=1+cosA/1-sinA=1+sinA/cosA

cotA+cosecA-1/cotA-cosecA+1

=>CotA+ cosecA-(cosec^2A-cot^2A)/cotA-cosecA+(cot^2A-cosec^2A)--- As we know cosec^2A-cot^2A=1....so....

=>(cotA+cosecA)-(cosecA+cotA)(cosecA-cotA)
--------------------------------------
cotA-cosecA+(cosecA-cotA)-(-cosecA+cotA)

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

=>cotA+cosecA/cotA-cosecA

✋【here in numerator taking (cosecA-cotA)as common and in denomenator cotA-cosecA】and remaining is cancelled 】

so ,cotA+cosecA/cotA-cosecA

=>cosA+1/sina/cosA-1/SinA

=>cosA+1/cosA-1....

=>cosA+1/cosA-1 Rhs...

or second method
after cotA+cosecA-1/cotA-cosecA+1

=>cotA+coescA-1/cotA-cosecA+1

cotA+cosec-(cot^2A-cosec^2A)/cotA-cosecA+1

【since cosec^2A-cot^2A=1】

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

cotA+ cosecA

cosA/sinA+1/sinA

=>cosA+1/sinA Ans....


hope it help you.☺

#Rajukumar☺☺

Answer 2

$\frac{cotA + cosecA - 1}{cotA - cosecA + 1} \\ \\ = \frac{(cotA + cosecA) - ( {cosec}^{2}A - {cot}^{2} A)}{cotA - cosecA + 1} \\ \\ = \frac{(cosecA + cotA)(1 - cosecA + cotA) }{cotA - cosecA + 1}$

$= cosecA + cotA \\ \\ = \frac{1 + cosA}{sinA}$