Question

SOLVE THIS QUESTION..... ​

Answer 1

Answer:

$\huge\underline \bold \red{solution \: - }$

Given that ,

secθ = x+1/4x

1+tan²θ = sec²θ

or tan²θ = sec²θ – 1

on expanding,

tan²θ = ( x+1/4x)² -1

or tan²θ = ( x²+1/16x² + 1/2 -1 )

or tan²θ = (x² +1/16x² -1/2 )

or tan²θ = x² +1/16x²-1/2

or tan²θ = (x-1/4x)^2

or tan²θ = +(x-1/4x) or – (x-1/4x)

when ,

tanθ = (x-1/4x)

we get,

secθ+tanθ = x+1/4x + x-1/4x = 2x

when ,

tanθ = -(x-1/4x)

secθ+tanθ = (x+1/4x) – (x-1/4x) = 1/2x

Hence,

secθ+tanθ = (x+1/4x) – (x-1/4x) = 1/2x

and

secθ+tanθ = x+1/4x + x-1/4x = 2x......

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hope it helps u didi...

Answer 2

Answer:

secθ = x+1/4x

1+tan²θ = sec²θ

or tan²θ = sec²θ – 1

on expanding,

tan²θ = ( x+1/4x)² -1

or tan²θ = ( x²+1/16x² + 1/2 -1 )

or tan²θ = (x² +1/16x² -1/2 )

or tan²θ = x² +1/16x²-1/2

or tan²θ = (x-1/4x)^2

or tan²θ = +(x-1/4x) or – (x-1/4x)

when ,

tanθ = (x-1/4x)

we get,

secθ+tanθ = x+1/4x + x-1/4x = 2x

when ,

tanθ = -(x-1/4x)

secθ+tanθ = (x+1/4x) – (x-1/4x) = 1/2x

Hence,

secθ+tanθ = (x+1/4x) – (x-1/4x) = 1/2x

and

secθ+tanθ = x+1/4x + x-1/4x = 2x......

Step-by-step explanation:

I hope it's help you!!!