If tan θ+ sin θ=m and tan θ- sin θ=n, then

Question

If tan θ+ sin θ=m and tan θ- sin θ=n, then (A) m2-n2=4 m n (B) m2+n2=4 m n (C) m2-n2=m2+n2 (D) m2-n2=4 √m n

Tardigrade Admin · Accepted Answer

From the given relations, m+ n=2 tan θ, m-n=2 sin θ. Thus, m2-n2=4 tan θ sin θ ...(i) Also √m n=4 √ tan 2 θ- sin 2 θ ...(ii) =4 sin θ tan θ From Eqs. (i) and (ii), we get m2-n2=4 √m n.

Q. If $tan θ + sin θ = m$ and $tan θ - sin θ = n$ , then

$m^{2} - n^{2} = 4 mn$

$m^{2} + n^{2} = 4 mn$

$m^{2} - n^{2} = m^{2} + n^{2}$

$m^{2} - n^{2} = 4 mn$

From the given relations,
$m + n = 2 tan θ, m - n = 2 sin θ$ .
Thus, $m^{2} - n^{2} = 4 tan θ sin θ$ ...(i)
Also $mn = 4 tan^{2} θ - sin^{2} θ$ ...(ii)
$= 4 sin θ tan θ$
From Eqs. (i) and (ii), we get $m^{2} - n^{2} = 4 mn$ .

Q. If tanθ+sinθ=m and tanθ−sinθ=n, then

m2−n2=4mn

m2+n2=4mn

m2−n2=m2+n2

m2−n2=4mn​

From the given relations, m+n=2tanθ,m−n=2sinθ. Thus, m2−n2=4tanθsinθ ...(i) Also mn​=4tan2θ−sin2θ​ ...(ii) =4sinθtanθ From Eqs. (i) and (ii), we get m2−n2=4mn​.

Q. If $tan θ + sin θ = m$ and $tan θ - sin θ = n$ , then

$m^{2} - n^{2} = 4 mn$

$m^{2} + n^{2} = 4 mn$

$m^{2} - n^{2} = m^{2} + n^{2}$

$m^{2} - n^{2} = 4 mn$

From the given relations,
$m + n = 2 tan θ, m - n = 2 sin θ$ .
Thus, $m^{2} - n^{2} = 4 tan θ sin θ$ ...(i)
Also $mn = 4 tan^{2} θ - sin^{2} θ$ ...(ii)
$= 4 sin θ tan θ$
From Eqs. (i) and (ii), we get $m^{2} - n^{2} = 4 mn$ .