Question

If sinA - tanA = p , p>0,find tanA , secA and sinA If A is acute angle.

Answer 1

Step-by-step explanation:

am using above right-angled triangle to find Trigonometric Identities :

From Pythagoras Theorem, we know that ,a2+b2=c2……………..(ii

Now, sin A= a/c……………………(iii)

cos A= b/c/…………………………(iv)

Adding the squares of equ (iii) and equ (iv), we get

sin2A+cos2A

=a2/c2+b2/c2

=(a2+b2)/c2————−[Use equ (ii)]

=c2/c2

=1…………………………………..(v)

Now,sec2A−tan2A

=1/cos2A−sin2A/cos2A

=(1−sin2A)/cos2A—————−[Use equ (v)]

=cos2A/cos2A=1………………….(vi)

sec2A−tan2A=1…………………..(vi)

,(secA+tanA)(secA−tanA)=1

or,p∗(secA−tanA)=1—————−[Use equ (i)]

or, (sec A - tan A)= 1/p…………………(vii)

Now adding equ (i) and equ (vii), we get:

secA+tanA+secA−tanA=p+1/p

or,secA=(p2+1)/2p……………………….(viii)

Substructing equ (i) and equ (vii), we get:

sec A+tan A - sec A + tan A = p-1/p

or,tanA=(p2−1)/2p……………………….(ix)

Deviding the equ (viii) and equ (ix), we get

secA/tanA=(p2+1)/2p/(p2−1)/2p

or,(1/cosA)/(sinA/cosA)=(p2+1)/(p2−1)

or,1/sinA=(p2+1)/(p2−1)

SinA=(p2−1)/(p2+1)

So the answer is,

SinA=(p2−1)/(p2+1)———————-