Question

If m is any real number then find the number of roots of cotx - tanx= m in the first quandrant

Answer 1

it is given that, cotx - tanx = m

or, 1/tanx - tanx = m

or, (1 - tan²x)/tanx = m

or, (1 - tan²x)/2tanx = m/2

we know, tan2α = 2tanα/(1-tan²α)

so , (1 - tan²x)/2tanx = 1/tan2x = m/2

or, tan2x = 2/m

from general solution of trigonometric identities.

or, 2x = nπ + tan^-1(2/m)

or, x = nπ/2 + (1/2)tan^-1(2/m)

now a/c to question, we have to find roots in first quadrant.

if m = positive , then n = 0 gives one root of equation.

if m = negative , then n = 1 gives one root of equation.

it means , for a specific situation, given equation has only one roots.

hence, number of roots of equation = 1.

Answer 2

Answer:

1

Step-by-step explanation:

cot x - tan x = m

1 - tan²x = m tan x

tan²x - m tan x - 1 = 0

Now, let tan a & tan b be the roots of this quadratic equation.

tan a × tan b = -1

tan a = -1/tan b = - cot b

Clearly 'b' lies in either in 2nd quadrant or in 4th quadrant

And 'a' lies in either 1st quadrant or in 3rd quadrant.

Now if 'a' lies in first quadrant then number of roots in first quadrant will be = 1 i.e. ANSWER

But if 'a' lies in 3rd quadrant so a = pi + c where 0< c <pi/2

so tan a = tan c where c lies in first quadrant.

Again here tan a = tan c is the solution.

So, anyhow the 1 solution is tan function of any angle in first quadrant.

Hence, there is 1 solution in first quadrant.

Hope it helps.

Thank You.