Question

if m= tan Theta + sin theta n=tan theta - sin theta prove that (m²-n²)²=16mn
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Answer 1

Answer:

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Answer 2

$\tt\small\underline\green{Given:-}$

$\tt{\implies m=tan\theta+sin\theta}$

$\tt{\implies n=tan\theta-sin\theta}$

$\tt\small\underline\green{To\:Prove:-}$

$\tt{\implies (m^2-n^2)^2=16mn}$

$\tt\small\underline\green{Solution:-}$

Here we will simply both sides L.H.S and R.H.S by putting the value of m and n.

$\tt\small\underline\green{Calculation\:for\:L.H.S:-}$

$\tt{\implies (m^2-n^2)^2}$

$\tt{\implies [(tan\theta+sin\theta)^2\:-(tan\theta-sin\theta)^2]^2}$

$\tt{\implies Here\: using\: identity}$

$\tt{\implies (a+b)^2=a^2+b^2+2ab}$

$\tt{\implies (a-b)^2=a^2+b^2-2ab}$

$\tt{\implies [(tan^2\theta+sin^2\theta+2tan\theta.sin\theta)-(tan^2\theta+sin^2\theta-2tan\theta.sin\theta)]^2}$

$\tt{\implies [\cancel{tan^2\theta}+\cancel{sin^2\theta}+2tan\theta.sin\theta-\cancel{tan^2\theta}-\cancel{sin^2\theta}+2tan\theta.sin\theta]^2}$

$\tt{\implies [2tan\theta.sin\theta+2tan\theta.sin\theta]^2}$

$\tt{\implies [4tan\theta.sin\theta]^2}$

$\tt{\implies 16tan^2\theta.sin^2\theta}$

$\tt\small\underline\green{Calculation\:for\:R.H.S:-}$

$\tt{\implies 16mn}$

$\tt{\implies 16[(tan\theta+sin\theta)(tan\theta-sin\theta)]\:}$

$\tt{\implies Here\: using\: identity}$

$\tt{\implies (a+b)(a-b)=a^2-b^2}$

$\tt{\implies 16[tan^2\theta-sin^2\theta]}$

$\tt{\implies 16[\dfrac{sin^2\theta}{cos^2\theta}-sin^2\theta]}$

$\tt{\implies 16[\dfrac{sin^2\theta-sin^2\theta.cos^2\theta}{cos^2\theta}]}$

$\tt{\implies 16.sin^2\theta[\dfrac{(1-cos^2\theta)}{cos^2\theta}]}$

$\tt{\implies \dfrac{16.sin^2\theta.sin^2\theta}{cos^2\theta}}$

$\tt{\implies \dfrac{16.sin^2\theta}{cos^2\theta}.sin^2\theta}$

$\tt{\implies 16tan^2\theta.sin^2\theta}$

$\sf\large{Hence\:R.H.S=L.HS}$