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  •  If    <img alt="\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}," src="https://learn.careers360.com/latex-image/?%5Ctan%20%5Ctheta%3D%5Cfrac%7B%5Csin%20%5Ca

 If    \tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha},  \sin \alpha+\cos \alpha \text { and } \sin \alpha-\cos \alpha  must be equal to ?        

Option: 1

\sqrt{2} \cos \theta,\sqrt{2} \sin \theta


Option: 2

\sqrt{2} \sin \theta,\sqrt{2} \cos \theta


Option: 3

\sqrt{2} \sin \theta,\sqrt{2} \sin \theta


Option: 4

\sqrt{2} \cos \theta, \sqrt{2} \cos \theta


Answers (1)

best_answer

Trigonometric Ratio for Compound Angles (Part 1) -

Trigonometric Ratio for Compound Angles (Part 1)

 

The sum or difference of two or more angles is called a compound angle. If A, B and C are any angle then A + B, A - B, A + B + C, A + B - C etc all are compound angles.

1. cos (α - β) = cos α cos β + sin α sin β

2. cos (α+ β) = cos α cos β - sin α sin β

3. sin (α - β) = sin α cos β - cos α sin β 

4. sin (α + β) = sin α cos β + cos α sin β 

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\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha},\ \ \ \ \text{multiply and devide by } \frac{1}{\sqrt{2}}\\ = \frac{\sin(\alpha - \frac{\pi}{4})}{\cos (\alpha -\frac{\pi}{4})}\\ =\tan (\alpha - \frac{\pi}{4})\\ \rightarrow \theta =\alpha - \frac{\pi}{4}\\ \text{Now}\\ \sin \alpha + \cos \alpha = \sin( \theta + \frac{\pi}{4})+\cos( \theta + \frac{\pi}{4})\\ =\frac{1}{\sqrt2}\sin \theta+\frac{1}{\sqrt2}\cos \theta+\frac{1}{\sqrt2}\cos \theta-\frac{1}{\sqrt2}\sin \theta\\=\sqrt{2}cos \theta \\ \sin \alpha - \cos \alpha = \sin( \theta + \frac{\pi}{4})-\cos( \theta + \frac{\pi}{4})\\ =\frac{1}{\sqrt2}\sin \theta+\frac{1}{\sqrt2}\cos \theta-\frac{1}{\sqrt2}\cos \theta+\frac{1}{\sqrt2}\sin \theta\\=\sqrt{2} \sin \theta

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Ritika Kankaria

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