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Relation between Trigonometric Ratios

If tanθ+sinθ=...

Question

If $t a n θ + s i n θ = m$ and $t a n θ - s i n θ = n$

Show that $m^{2} - n^{2} = 4 \sqrt{m n}$ [4 MARKS]

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Solution

Concept: 1 Mark
Substitution: 1 Mark
Application: 2 Marks

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

$= (t a n θ + s i n θ + t a n θ - s i n θ) (t a n θ + s i n θ - t a n θ + s i n θ)$

$= (2 t a n θ) (2 s i n θ)$

$= 4 t a n θ s i n θ$ ......... (1)

$4 \sqrt{m n} = 4 \sqrt{(t a n θ + s i n θ) (t a n θ - s i n θ)}$

$= 4 \sqrt{t a n^{2} θ - s i n^{2} θ}$

$= 4 \sqrt{\frac{s i n^{2} θ}{c o s^{2} θ} - s i n^{2} θ}$

$= 4 \sqrt{\frac{s i n^{2} θ - s i n^{2} θ c o s^{2} θ}{c o s^{2} θ}}$

$= 4 \sqrt{\frac{s i n^{2} θ (1 - c o s^{2} θ)}{c o s^{2} θ}}$

$= 4 t a n θ s i n θ$ .......... (2)

From (1) and (2)

$m^{2} - n^{2} = 4 \sqrt{m n}$

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Similar questions

m = tan θ + sin θ

n = tan θ + sin θ

m^{2} - n^{2} = 4 \sqrt{m n}

tan θ + sin θ = m

tan θ - sin θ = n

(m^{2} - n^{2}) = \pm 4 \sqrt{m n}

cos θ > 0, tan θ + sin θ = m

tan θ - sin θ = n

, then show that

m^{2} - n^{2} = 4 \sqrt{m n}

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