Question

if tan A+sin A = m and tan A - sin A = n then prove the following m
${m}^{2} - {n}^{2} = \sqrt{mn}$
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Answer 1

Answer:

your answer attacted in the photo

Answer 2

$\bf\huge\red{\mid{\overline{\underline{AnswEr}}}\mid}$

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

L.H.S. = $(tan \:  θ  + sin θ  + tan θ  - sin θ ) \\ (tan θ  + sin θ  - tan θ  + sin θ )$

$= ( \: 2 \: tan \:  θ )(2 \: sin θ )$

$= 4 \: tan θ  \: sin θ$

$= 4 \sqrt{tan {}^{2}  θ \: sin {}^{2} \: θ }$

$= 4 \sqrt{tan {}^{2}(1 - sin {}^{2}  θ )}$

$= 4 \sqrt{tan {}^{2}  θ  - tan {}^{2}  θ \: cos {}^{2}   θ }$

$= 4 \sqrt{tan {}^{2}  θ  - sin {}^{2}  θ }$

$= 4 \sqrt{(tan θ + sin  θ )(tan θ - sin  θ )}$

$= 4 \sqrt{mn} = <strong>R</strong><strong>.</strong><strong>H</strong><strong>.</strong><strong>S</strong><strong>.</strong>$