Question

secA+tanA=x then find the value of secA

Answer 1

SecA+tanA=x
1/cosA + sinA/cosA = x
1+sinA/cosA =x
Squaring both sides
(1+sinA)2/cosA2=x2
1=x2
X=1
Now,
SecA+tanA=x
SecA+tanA=1
SecA+tanA=(sec2A-tan2A)
SecA+tanA=(secA+tanA)(secA-tanA)
1=secA-tanA
SecA=tanA

Answer 2

Answer:

Given :

sec A + tan A = x

I am replacing x by ' k '

sec A + tan A = k

We know :

sec A = H / B   & tan A = P / B

H / B + P / B =  k / 1

H + P / B =  k / 1

So , B = 1

H + P = k

P = k - H

From pythagoras theorem :

H² = P² + B²

H² = ( H - k )² + 1

H² = H² + k² - 2 H k + 1

2 H k = k² + 1

H = k² + 1 / 2 k

P = k - H

P = k² - 1 / 2 k

Now write k = x we have :

Base = 1

Perpendicular P = x² - 1 / 2 x

Hypotenuse H = x² + 1 / 2 x

Value of sec A = H / B

sec A =  x² + 1 / 2 x / 1

sec A =  x² + 1 / 2 x

Therefore , we got value .