Question

If $cosec\theta -sin\theta =a^3$ and $sec\theta -cos\theta =b^3$, then $a^2{b}^2(a^2+b^2)=$ ______.

• A. 1
• B. -1
• C. 2
• D. None of these

Answer 1

The correct option is A 1

Given: $cosec\theta -sin\theta =a^{3}andsec\theta -cos\theta =b^3\Rightarrow a=(\frac{co{s}^2\theta }{sin\theta }{)}^{\frac{1}{3}}and{b}^3=\frac{1}{cos\theta }-cos\theta =\frac{si{n}^2\theta }{cos\theta }\Rightarrow b=(\frac{si{n}^2\theta }{cos\theta }{)}^{\frac{1}{3}}Now,wehave,a^2{b}^2(a^2+b^2)\therefore {\left(\frac{{\cos }^2\theta }{\sin \theta }\right)}^{\frac{2}{3}}\times {\left(\frac{{\sin }^2\theta }{\cos \theta }\right)}^{\frac{2}{3}}\left[\right(\frac{co{s}^2\theta }{sin\theta }{)}^{\frac{2}{3}}+\left(\frac{si{n}^2\theta }{cos\theta }{)}^{\frac{2}{3}}\right]\Rightarrow \left(\frac{co{s}^2\theta si{n}^2\theta }{sin\theta cos\theta }{)}^{\frac{2}{3}}\right[\frac{co{s}^{\frac{4}{3}}\theta }{si{n}^{\frac{2}{3}}\theta }+\frac{si{n}^{\frac{4}{3}}\theta }{co{s}^{\frac{2}{3}}\theta }]\Rightarrow (sin\theta cos\theta {)}^{\frac{2}{3}}\left[\frac{co{s}^2\theta +si{n}^2\theta }{si{n}^{\frac{2}{3}}\theta co{s}^{\frac{2}{3}}\theta }\right]=1$