prove that sinA-cosA+1/sinA+cosA-1=1/secA-tanA
Attachments:
Answers
Answered by
2
LHS
= (sinA-cosA+1)/(sinA+cos-1)
= (1 + sinA) (1 - sinA)/cos A(1-sinA)
= 1 - sin2A/cos A(1 - sinA)
= cos2A/cos A(1 - sinA)
= cosA/(1 - sinA)
= 1/ (1/cosA - sinA/cosA)
= 1/(secA - tanA)
= RHS
Hence Proved
Hope it helps dear
= (sinA-cosA+1)/(sinA+cos-1)
= (1 + sinA) (1 - sinA)/cos A(1-sinA)
= 1 - sin2A/cos A(1 - sinA)
= cos2A/cos A(1 - sinA)
= cosA/(1 - sinA)
= 1/ (1/cosA - sinA/cosA)
= 1/(secA - tanA)
= RHS
Hence Proved
Hope it helps dear
sachinanandharpeem9k:
please follow me
I followed you already
thank u
its ok
inbox me
bye see you in next question have a nice sunday
hmm
same 2 u
How did you get second step?
@FlashNish, the proof is absolutely wrong.
Answered by
7
LHS = (sin A - cos A + 1)/(sin A + cos A - 1)
Dividing numerator and denominator by cos A
= (sin A/cos A - cos A/cos A + 1/cos A)/(sin A/ cos A + cos A/cos A - 1/cos A)
= (tan A - 1 + sec A)/(tan A + 1 - sec A)
= (tan A + sec A - ( sec² A - tan² A))/(tan A + 1 - sec A)
= (tan A + sec A - sec² A + tan² A)/(tan A + 1 - sec A)
= (tan A + sec A + (tan A - sec A)(tan A + sec A))/(tan A + 1 - sec A)
= (tan A + sec A) ( 1 + tan A - sec A)/(tan A + 1 - sec A)
= tan A + sec A
= 1/ sec A - tan A
= RHS.
Dividing numerator and denominator by cos A
= (sin A/cos A - cos A/cos A + 1/cos A)/(sin A/ cos A + cos A/cos A - 1/cos A)
= (tan A - 1 + sec A)/(tan A + 1 - sec A)
= (tan A + sec A - ( sec² A - tan² A))/(tan A + 1 - sec A)
= (tan A + sec A - sec² A + tan² A)/(tan A + 1 - sec A)
= (tan A + sec A + (tan A - sec A)(tan A + sec A))/(tan A + 1 - sec A)
= (tan A + sec A) ( 1 + tan A - sec A)/(tan A + 1 - sec A)
= tan A + sec A
= 1/ sec A - tan A
= RHS.
Similar questions