Question

show that cos theta/1_sin theta +1_sin theta/cos theta =2 sec theta.​

Answer 1

Correct question :

$\implies\dfrac{ \cos \theta}{1 - \sin\theta} + \dfrac{1 - \sin\theta}{ \cos\theta}=2\sec\theta$

Solution :

$\implies\dfrac{ \cos \theta}{1 - \sin\theta} + \dfrac{1 - \sin\theta}{ \cos\theta}$

By taking LCM

$\implies\frac{ { \cos}^{2}\theta + {(1 - \sin\theta)}^{2}}{(1 - \sin\theta)( \cos\theta)} \\ \\ \implies\frac{ { \cos}^{2}\theta + {1 - 2 \sin\theta + \sin}^{2} \theta }{(1 - \sin\theta)( \cos\theta)} \\ \\ \implies \frac{1 + 1 - 2 \sin \theta}{(1 - \sin\theta)( \cos\theta)} \\ \\ \implies \frac{2 - 2 \sin \theta}{(1 - \sin\theta)( \cos\theta)} \\ \\ \implies \frac{2( \cancel{1 - \sin \theta})}{( \cancel{1 - \sin\theta})( \cos\theta)} \\ \\\implies \frac{2}{ \cos \theta} \\ \\\implies \large \boxed{\boxed{2 \sec \theta}}$

LHS = RHS

Hence Proved

Answer 2

$\large\red{\underline{\boxed{\textbf{ Thanks\:For\:Ansking}}}}$

LHS = cos theta / 1-sin theta + 1-sin theta / cos theta

= cos theta * cos theta + 1-sin theta * 1-sin theta / cos theta * 1-sin theta

= cos sq theta + 1+ sin sq theta -2sin theta / cos theta * 1-sin theta

= 2 -2sin theta / cos theta * 1-sin theta

=2(1-sin theta )/ cos theta * 1-sin theta

=2/cos theta

= 2 * 1/cos theta

=2 sec theta

= RHS

Hence proved