Question

If $cosecx-sinx=a^{3}andsecx-cosx=b^3;thenwhatisthevalueof{a}^2{b}^2\left(a^2+b^2\right)?$

• A. 0
• B. 1
• C. -1
• D. 2
• E. -2

Answer 1

The correct option is B

1

$\begin{array}{c}{a}^3=cosecx-sinxand{b}^3=secx-cosx\left(given\right)\\ \Rightarrow a^3=1/sinx-sinxand{b}^3=1/cosx-cosx\\ \Rightarrow a^3=\left(1-si{n}^{2}x\right)/sinxand{b}^3=\left(1-co{s}^{2}x\right)/cosx\\ \Rightarrow a^3=\left(co{s}^{2}x/sinx\right)and{b}^3=\left(si{n}^{2}x/cosx\right)\\ \Rightarrow a=\sqrt[3]{3\left(co{s}^{2}x/sinx\right)}andb=\sqrt[3]{3\left(si{n}^{2}x/cosx\right)}\\ Nowthegivenexpression-\\ a^2{b}^2\left(a^2+b^2\right)\\ =a^4{b}^2+a^2{b}^4\end{array}$

$\begin{array}{c}Nowbyputtingthevaluesofaandbinitthentheexpression\\ =\left[{\left(co{s}^{2}x/sinx\right)}^{\left(4/3\right)}\right]\left[{\left(si{n}^{2}x/cosx\right)}^{\left(2/3\right)}\right]+\left[{\left(co{s}^{2}x/sinx\right)}^{\left(2/3\right)}{\left(si{n}^{2}x/cosx\right)}^{\left(4/3\right)}\right]\\ =co{s}^{2}x+si{n}^{2}x\\ =1\end{array}$