Question

If $(cosec\theta -sin\theta )=a^3$ and $(sec\theta -cos\theta )=b^3$ , prove that

$a^2{b}^2(a^2+b^2)=1$

Answer 1

$a^3=(cosec\theta -sin\theta )=\frac{1}{sin\theta }-sin\theta =\frac{1-si{n}^2\theta }{sin\theta }{a}^3=\frac{co{s}^2\theta }{sin\theta }(a^3{)}^{\frac{2}{3}}=(\frac{co{s}^2\theta }{sin\theta }{)}^{\frac{2}{3}}{a}^2=\frac{co{s}^{\frac{4}{3}}\theta }{si{n}^{\frac{2}{3}}\theta }---\left(1\right)b^3=(sec\theta -cos\theta )=\frac{1}{cos\theta }-cos\theta =\frac{1-co{s}^2\theta }{cos\theta }{b}^3=\frac{si{n}^2\theta }{cos\theta }(b^3{)}^{\frac{2}{3}}=(\frac{si{n}^2\theta }{cos\theta }{)}^{\frac{2}{3}}{b}^2=\frac{si{n}^{\frac{4}{3}}\theta }{co{s}^{\frac{2}{3}}\theta }---\left(2\right)$

$\left(1\right)\times \left(2\right)a^2{b}^2=\frac{co{s}^{\frac{4}{3}}\theta }{si{n}^{\frac{2}{3}}\theta }\times \frac{si{n}^{\frac{4}{3}}\theta }{co{s}^{\frac{2}{3}}\theta }{a}^2{b}^2=si{n}^{\frac{2}{3}}\theta co{s}^{\frac{2}{3}}\theta ----\left(3\right)a^2+b^2=\frac{co{s}^{\frac{4}{3}}\theta }{si{n}^{\frac{2}{3}}\theta }+\frac{si{n}^{\frac{4}{3}}\theta }{co{s}^{\frac{2}{3}}\theta }=\frac{si{n}^2\theta +co{s}^2\theta }{si{n}^{\frac{2}{3}}\theta co{s}^{\frac{2}{3}}\theta }=\frac{1}{si{n}^{\frac{2}{3}}\theta co{s}^{\frac{2}{3}}\theta }\therefore a^2{b}^2(a^2+b^2)=si{n}^{\frac{2}{3}}\theta co{s}^{\frac{2}{3}}\theta \times \frac{1}{si{n}^{\frac{2}{3}}\theta co{s}^{\frac{2}{3}}\theta }=1$

Hence proved.