1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If θ = $\frac{a}{b}$, then $\frac{\left(a\mathrm{sin\theta }-b\mathrm{cos\theta }\right)}{\left(a\mathrm{sin}\theta +b\mathrm{cos\theta }\right)}$ = ? (a) $\frac{\left({a}^{2}+{b}^{2}\right)}{\left({a}^{2}-{b}^{2}\right)}$ (b) $\frac{\left({a}^{2}-{b}^{2}\right)}{\left({a}^{2}+{b}^{2}\right)}$ (c) $\frac{{a}^{2}}{\left({a}^{2}+{b}^{2}\right)}$ (d) $\frac{{b}^{2}}{\left({a}^{2}+{b}^{2}\right)}$

Open in App
Solution

(b) $\frac{\left({a}^{2}-{b}^{2}\right)}{\left({a}^{2}+{b}^{2}\right)}$ We have tan θ = $\frac{a}{b}$ Now, dividing the numerator and denominator of the given expression by cos θ, we get: $\frac{a\mathrm{sin}\theta -b\mathrm{cos}\theta }{a\mathrm{sin}\theta +b\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\frac{\frac{1}{\mathrm{cos}\theta }\left(a\mathrm{sin}\theta -b\mathrm{cos}\theta \right)}{\frac{1}{\mathrm{cos}\theta }\left(a\mathrm{sin}\theta +b\mathrm{cos}\theta \right)}=\frac{a\mathrm{tan}\theta -b}{a\mathrm{tan}\theta +b}=\frac{\frac{{a}^{2}}{b}-b}{\frac{{a}^{2}}{b}+b}=\frac{{a}^{2}-{b}^{2}}{{a}^{2}+{b}^{2}}$

Suggest Corrections
6
Join BYJU'S Learning Program
Related Videos
Range of Trigonometric Ratios from 0 to 90
MATHEMATICS
Watch in App
Join BYJU'S Learning Program