217k views
4 votes
(1+sinA-cosA)/1+sinA+cosA=(1+sinA+cosA)/1+sinA-cosA=2cosecA​

User Alex Luya
by
8.6k points

1 Answer

2 votes

Answer:

Hi,

LHS = sinA-cosA+1/sinA+cosA-1

divide both numerator and denominator by cosA

LHS=(tanA−1+secA)/(tanA+1−secA)LHS=(tanA−1+secA)/(tanA+1−secA)

Now

sec2A=1+tan2Asec2A=1+tan2A

sec2A−tan2A=1sec2A−tan2A=1

Using above relation at denominator of LHS

LHS=(tanA−1+secA)/(tanA−secA+sec2A−tan2A)LHS=(tanA−1+secA)/(tanA−secA+sec2A−tan2A)

LHS=(tanA−1+secA)/((secA−tanA)(−1+secA+tanA))LHS=(tanA−1+secA)/((secA−tanA)(−1+secA+tanA))

LHS=1/(secA−tanA)LHS=1/(secA−tanA)

LHS=RHSLHS=RHS

Hence Proved.

I think above proof will clear your doubt,

All the best.

User Mr Mcwolf
by
8.3k points

Related questions

asked Apr 26, 2015 197k views
Vanevery asked Apr 26, 2015
by Vanevery
8.6k points
1 answer
1 vote
197k views
asked Jun 10, 2021 147k views
Gozwei asked Jun 10, 2021
by Gozwei
7.6k points
1 answer
1 vote
147k views
1 answer
3 votes
182k views
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories