Question

If $ta{n}^2\theta =1-a^2,$ prove that $sec\theta +ta{n}^3\theta cosec\theta =(2-a^2{)}^{3/2}$

Answer 1

We have

$sec\theta +ta{n}^3\theta cosec\theta$

$=sec\theta \left[\frac{sec\theta +ta{n}^3\theta cosec\theta }{sec\theta }\right]$

$=sec\theta [1+ta{n}^3\theta \cdot \frac{cos\theta }{sin\theta }]$

$=sec\theta [1+ta{n}^3\theta \times cot\theta ]$

$=\sqrt{1+ta{n}^2\theta }[1+ta{n}^2\theta ]$

$=[1+ta{n}^2\theta {]}^{3/2}$

$=[1-(1-a^2){]}^{3/2}=(2-a^2{)}^{3/2}$

$[\because ta{n}^2\theta =1-a^2]$