Question

Find the acute angle thetha,satisfying the equation:-sec^2thetha +Cos^2thetha=>3............❤❤❤​

Answer 1

Given

cos^2@ + sec^2@ = 3

let

cos@ = t

sec@ = 1/t

Now putting these value in above

t^2 + 1/t^2 = 3

(t + 1/t)^2 + 2 = 3

t + 1/t = -1 and +1

Now

t+ 1/t = 1

t^2 - t + 1= 0

t^2 - 2t + 1/4 - 3/4 = 0

(t - 1/2)^2 = 3/4

t = (+ sqrt3 +1)/2 and (1 - sqrt3)/2

and

t + 1/t = -1

t + 1/t + 1= 0

t^2 + t +1= 0

t^2 + 2t +1/4 - 3/4

(t + 1/2)^2 = 3/4

t = (sqrt 3 - 1)/2 and -( sqrt3 +1)/2

Now we know that

t = cos@

@ = cos^-t

we have to find acute angel which is less then 90°

so

cos ( 1/2 +sqrt3/2) = cosπ/3 cosπ/6 - sinπ/6sinπ/3

cos(1 - sqrt3)/2 = cos ( π/3 - π/6 ) = cosπ/6

@ = cos^-1cos(1-sqrt3)/2 = cos^-1cosπ/6 = π/6 = 30°

Hence 30° is less then 90°

Answer 2

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Given

cos^2@ + sec^2@ = 3

let

cos@ = t

sec@ = 1/t

Now putting these value in above

t^2 + 1/t^2 = 3

(t + 1/t)^2 + 2 = 3

t + 1/t = -1 and +1

Now

t+ 1/t = 1

t^2 - t + 1= 0

t^2 - 2t + 1/4 - 3/4 = 0

(t - 1/2)^2 = 3/4

t = (+ sqrt3 +1)/2 and (1 - sqrt3)/2

and

t + 1/t = -1

t + 1/t + 1= 0

t^2 + t +1= 0

t^2 + 2t +1/4 - 3/4

(t + 1/2)^2 = 3/4

t = (sqrt 3 - 1)/2 and -( sqrt3 +1)/2

Now we know that

t = cos@

@ = cos^-t

we have to find acute angel which is less then 90°

so

cos ( 1/2 +sqrt3/2) = cosπ/3 cosπ/6 - sinπ/6sinπ/3

cos(1 - sqrt3)/2 = cos ( π/3 - π/6 ) = cosπ/6

@ = cos^-1cos(1-sqrt3)/2 = cos^-1cosπ/6 = π/6 = 30°

Hence 30° is less then 90°

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