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 62347 by rajesh4661kumar@gamil.com last updated on 20/Jun/19

Answered by Kunal12588 last updated on 20/Jun/19

$$\frac{csc\theta -sin\theta }{sec\theta -cos\theta }=\frac{a^3}{b^3}$$$$\Rightarrow cot\theta \frac{1-{sin}^2\theta }{1-{cos}^2\theta }=\frac{a^3}{b^3}$$$$\Rightarrow cot\theta =\frac{a}{b}$$$$\Rightarrow {cot}^2\theta =\frac{a^2}{b^2}$$$$\Rightarrow 1+{cot}^2\theta =\frac{a^2+b^2}{b^2}$$$$\Rightarrow {csc}^2\theta =\frac{a^2+b^2}{b^2}$$$$\frac{{csc}^2\theta -1}{csc\theta }=a^3$$$$\Rightarrow \frac{\frac{a^2+b^2-b^2}{b^2}}{\frac{\sqrt{a^2+b^2}}{b}}=a^3$$$$\Rightarrow a^2=a^{3}b\sqrt{a^2+b^2}$$$$\Rightarrow 1=ab\sqrt{a^2+b^2}$$$$\Rightarrow a^2{b}^2(a^2+b^2)=1$$

Commented by peter frank last updated on 20/Jun/19

$$nice$$

Answered by Kunal12588 last updated on 20/Jun/19

$$ANOTHERWAY$$$$\mathrm{cosec}\theta -\sin \theta =a^3$$$$\underset{{\tan }^{-1}t=\theta }{\Rightarrow }\frac{\sqrt{1+t^2}}{t}-\frac{t}{\sqrt{1+t^2}}=a^3$$$$\Rightarrow 1+t^2-t^2=a^{3}t\sqrt{1+t^2}$$$$\Rightarrow \frac{1}{a^{3}t}=\sqrt{1+t^2}$$$$\sec \theta -\cos \theta =b^3$$$$\underset{{\tan }^{-1}t=\theta }{\Rightarrow }\sqrt{1+t^2}-\frac{1}{\sqrt{1+t^2}}=b^3$$$$\Rightarrow 1+t^2-1=b^3\sqrt{1+t^2}$$$$\Rightarrow \frac{t^2}{b^3}=\sqrt{1+t^2}$$$$\frac{1}{a^{3}t}=\frac{t^2}{b^3}$$$$\Rightarrow t=\frac{b}{a}$$$$puttingthevalueoft$$$$\frac{\frac{b^2}{a^2}}{b^3}=\sqrt{1+\frac{b^2}{a^2}}$$$$\Rightarrow \frac{1}{a^{2}b}=\frac{\sqrt{a^2+b^2}}{a}$$$$\Rightarrow ab\sqrt{a^2+b^2}=1$$$$\Rightarrow a^2{b}^2(a^2+b^2)=1$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com