Question

if tan theta is equal to n sin alpha cos Alpha upon 1 minus n sin square alpha prove that tan Alpha minus beta is equal to 1 minus n tan Alpha

Answer 1

Step-by-step explanation:

$Given\:tan\beta=\frac{msin\alpha cos\alpha}{1-nsin^{2}\alpha}\:--(1)$

$LHS = tan(\alpha-\beta)\\=\frac{tan\alpha-tan\beta}{1+tan\alpha tan\beta}$

$=\frac{\frac{sin\alpha}{cos\alpha}-\frac{nsin\alpha cos\alpha}{1-sin^{2}\alpha}}{1+\frac{sin\alpha}{cos\alpha}\times \frac{nsin\alpha cos\alpha}{1-nsin^{2}\alpha}}$

$=\frac{sin\alpha\left(\frac{1}{cos\alpha}-\frac{ncos\alpha}{1-nsin^{2}\alpha}\right)}{1+\frac{nsin^{2}\alpha}{1-nsin^{2}\alpha}}$

$=\frac{sin\alpha\left(\frac{1-nsin^{2}\alpha-ncos^{2}\alpha}{cos\alpha \big(1-nsin^{2}\alpha\big)}\right)}{\frac{(1-nsin^{2}\alpha+nsin^{2}\alpha)}{(1-nsin^{2}\alpha)}}$

$=\frac{\frac{sin\alpha}{cos\alpha}[1-n(sin^{2}\alpha+cos^{2}\alpha)]}{1}$

$=tan\alpha(1-n)\\=(1-n)tan\alpha\\=RHS$

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Answer 2

hope this helps you .....................