If secA+tanA=p show that secA-tanA=1/p hence find the values of sinA a

1.	If secA+tanA=p show that secA-tanA=1/p hence find the values of sinA and cosA
Answer» Sec A + Tan A =p -equation 1We know thatSec^2 A - Tan^2 A = 1(Sec A -Tan A)(Sec A+ Tan A)=1[a^2 -b^2 =(a+b)(a-b)]From equation 1(Sec A - Tan A)(p) = 1Sec A - Tan A=1÷p -equation 2Adding equations 1 and 2Sec A - Tan A +Sec A+ Tan A=(1+p^2)/p2 Sec A =(1+p^2)/pSec A=(1+p^2)/2pNow you can easily find the value of sin and cos Answer,SecA+tanA=p --------------eq1Now,Sec^2A-tan^2A=1. (a^2-b^2)(SecA+tanA)(secA-tanA)=1 (a+b)(a-b)p(secA-tanA)=1. (using eq1)secA-tanA=1/p PROVED Shift tan to rhs and sq. on both sides and find value of tan in trms of p and put the values in a triangle and then find the t ratios.

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