Question

if (cot theta + tan theta) =m and (sec theta - cos theta)=n prove that

(m^2 n)^2/3 - (mn^2)^2/3 = 1

Please solve this in a copy

Answer 1

The answer is explained in the attachment.


Hope it helps

Answer 2

Answer:

If $cot\theta+tan\theta=m$ and $sec\theta+cos\theta=n$

Prove that: $(m^{2}n)^{\frac{2}{3}}-(mn^{2})^{\frac{2}{3}}=1$

If $cot\theta+tan\theta=m$

since, $cot\theta=\frac{1}{tan\theta}$

$\frac{1}{tan\theta}+tan\theta=m$

$\frac{1+tan^{2}\theta}{tan\theta}=m$

Since $1+tan^{2}\theta=sec^{2}\theta$

$\frac{sec^{2}\theta}{tan\theta}=m$

since, $tan\theta=\frac{sin\theta}{cos\theta}$

$\frac{sec^{2}\theta}{\frac{sin\theta}{cos\theta}}=m$

$\frac{1}{sin\theta cos\theta}=m$

If $sec\theta+cos\theta=n$

since, $sec\theta=\frac{1}{cos\theta}$

$\frac{1}{cos\theta}+cos\theta=n$

$\frac{1-cos^{2}\theta}{cos\theta}=n$

$\frac{sin^{2}\theta}{cos\theta}=n$

Left hand side

$(m^{2}n)^{\frac{2}{3}}-(mn^{2})^{\frac{2}{3}}$

$((\frac{1}{sin\theta cos\theta})^{2}\frac{sin^{2}\theta}{cos\theta})^{\frac{2}{3}}-(\frac{1}{sin\theta cos\theta}\frac{sin^{2}\theta}{cos\theta})^{\frac{2}{3}}$

$(\frac{1}{cos^{2}\theta}-(\frac{sin^{2}\theta}{cos^2\theta}$

$sec^{2}\theta-tan^{2}\theta$

=1

=Right hand side

Hence proved