# Discussion on Trigonometry

Posted on: 10:46:45 AM on November 9, 2015
Posted By:
SAMEER
sec A + tan A = P , then sin A = ?

REGISTER NOW & POST YOUR QUESTIONS

## Reponses To This

Posted on: 06:26:33 PM on November 10, 2015
Tan=sin/cos . substitute in the formula......

Posted on: 02:02:01 PM on November 12, 2015
PcosA-1

Posted on: 09:41:28 PM on May 29, 2016
P1/SEC-I

Posted on: 02:30:51 PM on June 7, 2016
PCOSA-1

Posted on: 03:14:56 PM on June 24, 2016
P.cosA-1

Posted on: 02:40:01 PM on August 22, 2016
here, sinA = PcosA-1

Posted on: 11:18:29 PM on September 8, 2016
sec A + tan A = p ----------------(i) sec A - tan A = 1/p ---------------(ii) Adding equations (i) and (ii), we get:- 2 sec A = (p^2+1)/p =>sec A = (p^2+1)/2p =>cos A = 2p/(p^2+1) sin A = [1-{2p/(p^2+1)}^2]^(1/2) [ Since sin A = (1-cos^2 A)^(1/2) ] = (p^2-1)/(p^2+1) ---------------(Answer)

Posted on: 08:37:43 AM on September 11, 2016
p.cosA-1

Posted on: 04:01:56 PM on September 12, 2016
PcosA-1

Posted on: 05:27:23 PM on September 23, 2016
SecA+TanA=P,SinA=PcosA-1

Posted on: 05:35:10 PM on September 26, 2016
1-p^2/1+p^2

Posted on: 08:15:21 AM on October 16, 2016
tan=sin/cos

Posted on: 04:03:52 PM on November 3, 2016
PcosA-1=sinA

Posted on: 01:13:30 PM on July 23, 2017
p square -1/p square+1

Posted on: 10:14:07 PM on August 24, 2017