2006-12-19 14:47:35 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
taking (1+sinA)cosA as LCD
cos^2A+(1+sinA)^2/cosA(1+sinA)
=cos^2A+1+2sinA+sin^2A/cosA(1+sinA)
=2(1+sinA)/cosA(1+sinA)
=2/cosA
=2secA
2006-12-19 14:52:36 · answer #1 · answered by raj 7 · 0⤊ 0⤋
COSA /(1+ sin A)
= Cos A(1- sin A)/(1- sin^2 A)
= cos A(1- sin A)/ cos^2 A
= (1- sin A)/ Cos A
addion we get expression
= (1-sinA)/cos A + (1+ sin A)/cos A
= 2/cos A = 2 sec A
2006-12-19 17:19:44 · answer #2 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
Do your own work, but here is a hint, multiply the term on the left by cosA (top and bottom) and the one on the left by (1+sinA) top and bottom, and simplify. Give it a try.
2006-12-19 14:52:00 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Multiply correct and bottom with the help of (sin A + cos A + a million). (sin A - cos A + a million)(sin A + cos A + a million)/(sin A + cos A - a million)(sin A + cos A - a million) = (sin² A - cos² A + 2*sin A + a million)/(2*sin A) = (2*sin² A + 2*sin A)/(2*sin A) = sin A + a million
2016-11-30 23:48:54 · answer #4 · answered by anuj 3 · 0⤊ 0⤋
(Cos A / (1+Sin A)) + ((1+Sin A)/Cos A)
= [ Cos A . Cos A + (1+Sin A).(1+Sin A)] / [Cos A (1+Sin A)]
= [ Cos^2 A + 1 + 2 Sin A + Sin^2 A] / [Cos A (1+Sin A)]
= [2 + 2Sin A] / [Cos A (1+Sin A)]
= 2 / Cos A = 2 Sec A
2006-12-19 15:17:18 · answer #5 · answered by Srinivas c 2 · 0⤊ 0⤋
(cosA/1+sinA) + (1+sinA/cosA)
=cos^2A/[(1+sinA)(cosA)]+(1+sinA)^2/[(1+sinA)(cosA)]
=[cos^2A + 1 +2sinA+sin^2A]/[(1+sinA)cosA]
=[2(1+sinA)]/[(1+sinA)cosA]
= 2/cosA
=2secA
2006-12-19 15:13:21 · answer #6 · answered by ironduke8159 7 · 0⤊ 0⤋