Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 67345 by Aditya789 last updated on 26/Aug/19

3sinx+5cosx=5 then prove that   5sinx−3cox= +3

$$\mathrm{3}{sinx}+\mathrm{5}{cosx}=\mathrm{5}\:{then}\:{prove}\:{that}\: \\ $$$$\mathrm{5}{sinx}−\mathrm{3}{cox}=\:+\mathrm{3} \\ $$

Answered by $@ty@m123 last updated on 26/Aug/19

(3/5)sin x+cos x=1  ((1−cos x)/(sin x))=(3/5)  ((1−cos x)/(sin x))×((1+cos x)/(1+cos x))=(3/5)  ((1−cos^2 x)/(sin x(1+cos x)))=(3/5)  ((sin^2 x)/(sin x(1+cos x)))=(3/5)  ((sin x)/(1+cos x))=(3/5)  5sin x=3+3cos x  5sinx−3cosx= 3

$$\frac{\mathrm{3}}{\mathrm{5}}\mathrm{sin}\:{x}+\mathrm{cos}\:{x}=\mathrm{1} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}×\frac{\mathrm{1}+\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}\left(\mathrm{1}+\mathrm{cos}\:{x}\right)}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\frac{\mathrm{s}{in}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}\left(\mathrm{1}+\mathrm{cos}\:{x}\right)}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\mathrm{5sin}\:{x}=\mathrm{3}+\mathrm{3cos}\:{x} \\ $$$$\mathrm{5}{sinx}−\mathrm{3}{cosx}=\:\mathrm{3} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com