Question

if tan theta + sin theta is equal to M and 10 theta minus sin theta is equal to N show that n square minus n square is equal to 4 root MN​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Correct Question :-

If tanθ + sinθ = m and tanθ - sinθ = n show that

• (m² - n²) = 4√mn

Solution :-

$\tt\:LHS=({m}^{2}-{n}^{2})$

$\tt\longrightarrow\:{(tan\theta+sin\theta)}^{2}-{(tan\theta-sin\theta)}^{2}$

$\tt\longrightarrow\:4tan\theta\:sin\theta$

$\tt\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:[{(a+b)}^{2}-{(a-b)}^{2}=4ab]$

Now,

$\tt\:RHS=4\sqrt{mn}$

$\tt\longrightarrow\:4\sqrt{(tan\theta+sin\theta)(tan\theta-sin\theta)}$

$\tt\longrightarrow\:4\sqrt{(ta{n}^{2}\theta-si{n}^{2}\theta})$

$\tt\longrightarrow\:4\sqrt{\bigg(\dfrac{si{n}^{2}\theta}{co{s}^{2}\theta}-si{n}^{2}\theta\bigg)}$

$\tt\longrightarrow\:4\sqrt{\dfrac{si{n}^{2}\theta-si{n}^{2}\theta\:co{s}^{2}\theta}{cos\theta}}$

$\tt\longrightarrow\:4\:\dfrac{sin\theta}{cos\theta}\:\sqrt{1-co{s}^{2}\theta}$

$\tt\longrightarrow\:4\:tan\theta\:\sqrt{si{n}^{2}\theta}$

$\tt\longrightarrow\:4tan\theta\:sin\theta$

$\tt\:Hence,LHS=RHS$

$\tt\therefore({m}^{2}-{n}^{2})=4\sqrt{mn}$