Question

If $cosec\theta -sin\theta =a^3,sec\theta -cos\theta =b^3$, then prove that $a^2{b}^2(a^2+b^2)=1$

Answer 1

Given: $cosec\theta -sin\theta =a^3$

$sec\theta -cos\theta =b^3$

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

Since, $cosec\theta -sin\theta =a^3$

$\Rightarrow \frac{1}{sin\theta }-sin\theta =a^3(\because cosec\theta =\frac{1}{sin\theta })$

$\Rightarrow \frac{1-si{n}^2\theta }{sin\theta }=a^3$

$\Rightarrow \frac{co{s}^2\theta }{sin\theta }=a^3.[\because 1-si{n}^2\theta =co{s}^2\theta ]$

$\Rightarrow a=\frac{co{s}^{\frac{2}{3}\theta }}{sin\frac{1}{3}\theta }$

Since, $\frac{1}{cos\theta }-cos\theta =b^3(\because sec\theta =\frac{1}{cos\theta })$

$\Rightarrow \frac{1-co{s}^2\theta }{cos\theta }=b^2$

$\Rightarrow \frac{si{n}^2\theta }{cos\theta }=b^3[\because 1-co{s}^2\theta -si{n}^2\theta ]$

$\Rightarrow b=\frac{si{n}^{\frac{2}{3}\theta }}{cos\frac{1}{3}\theta }$

Now, $a^2{b}^2(a^2+b^2)$

$=\frac{co{s}^{\frac{4}{3}\theta }}{sin\frac{2}{3}\theta }\times \frac{si{n}^{\frac{4}{3}\theta }}{cos\frac{2}{3}\theta }(\frac{co{s}^{\frac{4}{3}\theta }}{sin\frac{2}{3}\theta }\times \frac{si{n}^{\frac{4}{3}\theta }}{cos\frac{2}{3}\theta })=co{s}^{\frac{2}{3}\theta }\times sin\frac{2}{3}\theta \frac{(co{s}^{\frac{6}{3}}\theta +si{n}^{\frac{6}{3}}\theta )}{sin\frac{2}{3}\theta .cos\frac{2}{3}\theta }$

$=co{s}^2\theta +si{n}^2\theta =1$ Hence proved.