Question

What is minimim value of cosec^2x + 25sec^2x

Answer 1

Let f(x) = cosec²x + 25sec²x
we know,
sec²x - tan²x = 1 , sec²x = 1 + tan²x
cosec²x - cot²x = 1 , cosec²x = 1 + cot²x

then, f(x) = 1 + cot²x + 25(1 + tan²x)
= 1 + cot²x + 25 + 25tan²x
f(x) = 26 + cot²x + 25tan²x

we know, if a and b are two positive terms,
then, AM ≥ GM

here cot²x and 25tan²x are two positive terms,
so, (cot²x + 25tan²x)/2 ≥ √{25tan²x.cot²x}
cot²x + 25tan²x ≥ 2 × √{25 × 1} = 2×5 = 10
cot²x + 25tan²x ≥ 10
add 26 both sides,
cot²x + 25tan²x + 26 ≥ 10 + 26 = 36
f(x) = cot²x + 25tan²x + 26 ≥ 36
hence, minimum value of f(x) is 36 .
hence, minimum value of cosec²x + 25sec²x = 36

Answer 2

Answer:

Step-by-step explanation:Let f(x) = cosec²x + 25sec²x

we know,

sec²x - tan²x = 1 , sec²x = 1 + tan²x

cosec²x - cot²x = 1 , cosec²x = 1 + cot²x

then, f(x) = 1 + cot²x + 25(1 + tan²x)

= 1 + cot²x + 25 + 25tan²x

f(x) = 26 + cot²x + 25tan²x

we know, if a and b are two positive terms,

then, AM ≥ GM

here cot²x and 25tan²x are two positive terms,

so, (cot²x + 25tan²x)/2 ≥ √{25tan²x.cot²x}

cot²x + 25tan²x ≥ 2 × √{25 × 1} = 2×5 = 10

cot²x + 25tan²x ≥ 10

add 26 both sides,

cot²x + 25tan²x + 26 ≥ 10 + 26 = 36

f(x) = cot²x + 25tan²x + 26 ≥ 36

hence, minimum value of f(x) is 36 .

hence, minimum value of cosec²x + 25sec²x = 36