Question

Que no 75
Solve completely.....

Answer 1

Answer:

because, sinA=side/incline = $\frac{x^2-1}{x^2+1}$

hope it helps, please rate it

Step-by-step explanation:

secA+tan A = x.......(1)

let's bring up this relation

sec^2 A -tan^2A = 1

(secA-tanA)(secA+tanA)=1

(secA-tanA)=1/x......(2)

let's subtitute (2) to (1)

then we will get

sec A +(secA-1/x) =x

2secA = x+1/x

2secA = (x^2 +1 )/x

sec A = (x^2+1)/2x

cosA = 2x/(x^2+1)

let's make a triangle with

bottom = 2x

incline = (x^2+1)

side = $\sqrt{incline^2 - side^2 }= \sqrt{ (x^2 +1)^2 - (2x)^2} \\\sqrt{ x^4 +2x^2 +1 - 4x^2$

side = $\sqrt{x^4-2x^2-1} = x^2-1$

because, sinA=side/incline = $\frac{x^2-1}{x^2+1}$

Answer 2

Answer is in the pic.