Question

tan theta. = sin alpha - cos alpha / sin alpha + cos alpha then find sin alpha +cos alpha = pulse minus root 2

Answer 1

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Answer 2

Answer:

$sin\alpha+cos\alpha=\sqrt{2} cos\alpha$

Step-by-step explanation:

$Given \:tan\theta=\frac{sin\alpha-cos\alpha}{sin\alpha+cos\alpha}$

$Divide \: numerator \:and\\denominator \:by \:cos\alpha,\:we \:get$

$=\frac{\frac{sin\alpha}{cos\alpha}-\frac{cos\alpha}{cos\alpha}}{\frac{sin\alpha}{cos\alpha}+\frac{cos\alpha}{cos\alpha}}$

$=\frac{tan\alpha-1}{tan\alpha+1}$

$=\frac{tan\alpha-tan45\degree}{tan\alpha+tan45\degree}$

$=tan(\alpha-45)$

$\implies tan\theta =tan(\alpha-45)$

$\implies \theta = \alpha-45$

$\implies \alpha = \theta+45$

$Now,\\sin\alpha+cos\alpha\\=sin(\alpha+45)+cos(\alpha+45)\\=sin\alpha cos45+sin45 cos \alpha\\+ cos\alpha cos45-sin\alpha sin45$

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

$=2\times \frac{1}{\sqrt{2}} cos\alpha$

$=\frac{\sqrt{2}\times \sqrt{2}}{\sqrt{2}} cos\alpha$

$=\sqrt{2} cos\alpha$

Therefore,

$sin\alpha+cos\alpha=\sqrt{2} cos\alpha$

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