(i) sin A tan A / 1-cos A = 1+sec A
Share
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
[tex]\mathfrak{\huge{\green{\underline{\underline{Answer :}}}}}[/tex]
LHS = Sin A.Tan A.(1/1-Cos A)
=Sin A. (SinA/CosA).(1/1-CosA)
=(Sin^2A)/(CosA.(1-CosA)
=(1-Cos^2A)/( CosA.(1-CosA)
=(1+CosA)(1-CosA)/ CosA.(1-CosA)
=(1/cosA)+(CosA/CosA)
=SecA+1=RHS
Hence Proved
Verified answer
Answer:
[tex](sinA*sinA/cosA) /(1-cosA)[/tex]
[tex]sin^2A/(cosA (1-cosA ))[/tex]
[tex](1-cos^2A)/ (cosA (1-cosA))[/tex]
[tex](1+cosA)(1-cosA) / (cosA (1-cosA))[/tex]
[tex](1+cosA)/cosA[/tex]
[tex](1/cosA)+(cosA/cosA)[/tex]
[tex]secA+1[/tex]
[tex]1+secA proved[/tex]