Question

If $ta{n}^{2}x=1-a^2$, prove that $secx+ta{n}^{3}x.cosecx=(2-a^2{)}^{\frac{3}{2}}$ . Also find the values of a for which the above result holds true .

Answer 1

We have,

$secx+ta{n}^{3}xcosecx$

$=secx\left(\frac{secx+ta{n}^{2}xcosecx}{secx}\right)$

$=secx(1+ta{n}^{3}x.\frac{cosx}{sinx})$

$=secx(1+ta{n}^{3}x\times cotx)$

$=\sqrt{1+ta{n}^{2}x}(1+ta{n}^{2}x)$

$=(1+ta{n}^{2}x{)}^{\frac{3}{2}}$

$=(1+(1-a^2){)}^{\frac{3}{2}}=(2-a^2{)}^{\frac{3}{2}}$

The above result holds true for all the values of a.